# Coordinate Geometry : Previous Year's Questions

Very short answer type questions : -
Q.1      (a) Find the value of a so that the point (3, a) lies on the line represented by 2x – 3 y = 5.
[Delhi 2009]

(b) Find the distance between the points $\left( { - {8 \over 5},2} \right)$ and $\left( {{2 \over 5},2} \right)$
[Delhi 2009]

(c) If the mid-point of the line segment joining the points P(6, b – 2) and Q(–2, 4) is (2, –3), find the value of b.
[Foreign 2009]

(d) If P(2, p) is the mid-point of the line segment joining the points A(6, –5) & B(–2, 11), find the value of p.
[Delhi 2010]

(e) If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D.
[Delhi 2010]

(f) What is the distance between the points A(c, 0) and B(0, –c) ?                                     [AI 2010]
(g) Find the distance between the points, A(2a, 6a) and B$\left( {2a + \sqrt {3a} ,5a} \right)$.
[Foreign 2010]

(h) Find the value of k if P(4, –2) is the mid-point of the line segment joining the points A(5k, 3) and B(–k, –7).
[Foreign 2010]

Sol.        (a) Given 2x – 3y = 5 ….. (1)
On putting point (3, a) in equation (1),We get
$2\left( 3 \right) - 3\left( a \right) = 5$
$\Rightarrow$ 6 – 3a = 5
$\Rightarrow$ –3a = 5 – 6
$\Rightarrow$ –3a= – 1
$\Rightarrow$ a= ${{ - 1} \over { - 3}}$
Hence, value of $a = {1 \over 3}$
(b) Let A $\left( {{{ - 8} \over 5},2} \right)$ and $B\left( {{2 \over 5},2} \right)$ be the given points.

Distance $AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
where,${\rm{x}}_{\rm{1}} = {{ - 8} \over 5}$,$x_2 = {2 \over 5}$,$y_1 = 2$,$y_2 = 2$
$\Rightarrow AB = \sqrt {{{\left[ {{2 \over 5} - \left( {{{ - 8} \over 5}} \right)} \right]}^2} + {{\left( {2 - 2} \right)}^2}}$
$\Rightarrow AB = \sqrt {{{\left[ {{2 \over 5} + {8 \over 5}} \right]}^2} + 0}$
$\Rightarrow AB = \sqrt {{{\left( {{{10} \over 5}} \right)}^2}}$
$\Rightarrow AB = \sqrt {{{\left( 2 \right)}^2}}$
$\Rightarrow AB{\rm{ }} =\pm 2$
AB = 2 [Neglecting AB = -2]

Hence, the distance between $\left( {{{ - 8} \over 5},2} \right)$ and $\left( {{2 \over 5},2} \right)$ is 2 units.
(c)  Let point R (x, y) be the mid-point of line segment joining P(6, b –2) and Q(–2, 4).

Therefore, R(x, y) = $\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)$
where ${x_1} = 6,{y_1} = b - 2,{x_2} = - 2\,and\,{y_2} = 4$
$\Rightarrow R(x , y) = \left( {{{6 + \left( { - 2} \right)} \over 2},{{b - 2 + 4} \over 2}} \right)$
But mid-point is (2,-3)

$\Rightarrow (2, - 3) = \left( {{{6 - 2} \over 2},{{b + 2} \over 2}} \right)$
$\Rightarrow (2, - 3) = \left( {{4 \over 2},{{b + 2} \over 2}} \right)$
$\Rightarrow (2,-3) = \left( {2,{{b + 2} \over 2}} \right)$
On Equating y co-ordinate, We get
$- 3 = {{b + 2} \over 2}$
$\Rightarrow$ b + 2 = - 6
$\Rightarrow$ b = - 8
Hence, the value of b = - 8.
(d)  It is given that P (2, p) is the mid –point of the line segment joining the points A(6, - 5) and B (-2 , 11).

Mid-point of A (6, -5) and B(-2, 11) is $= \left[ {{{6 + \left( { - 2} \right)} \over 2},{{ - 5 + 11} \over 2}} \right] = \left[ {{4 \over 2},{6 \over 2}} \right] = [2,3]$
But mid-point of A(6,-5) and B(-2,11) is P(2,p)
On Equating y co-ordinate ,We get

Let D(x,y) be the fourth vertex of parallelogram ABCD.
It is given that A(1, 2), B (4, 3) and C(6, 6) are the vertices of parallelogram ABCD.
We know that diagonals of a parallelogram bisect each-other so mid-points of AC and BD are same.
Mid-point of AC $= O\left( {{{1 + 6} \over 2},{{2 + 6} \over 2}} \right)$
$= O\left( {{7 \over 2},{8 \over 2}} \right)$
$= O\left( {{7 \over 2},4} \right)............(1)$
Mid-point of BD $= O\left( {{{4 + x} \over 2},{{y + 3} \over 2}} \right)..............(2)$
From (1) and (2)
$\left( {{7 \over 2},4} \right) = \left( {{{4 + x} \over 2},{{y + 3} \over 2}} \right)$
On Equating x and y co-ordinate ,We get
${{4 + x} \over 2} = {7 \over 2}$
$\Rightarrow 4 + x = 7$
$\Rightarrow x = 7 - 4$
$\Rightarrow x = 3$
And
${{y + 3} \over 2} = 4$
$\Rightarrow y + 3 = 8$
$\Rightarrow y = 8 - 3$
$\Rightarrow y = 5$
So, the fourth vertex of parallelogram ABCD is (3, 5).
(f) Given points are A(c, 0) and B(0, – c)

Distance between A and B is : -
$AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
Where,${x_1} = c,{y_1} = 0,{x_2} = 0, {y_2} = - c$
$\Rightarrow AB = \sqrt {{{\left( {0 - c} \right)}^2} + {{\left( { - c - 0} \right)}^2}}$

$\Rightarrow AB = \sqrt {{{\left( { - c} \right)}^2} + {{\left( { - c} \right)}^2}}$
$\Rightarrow AB = \sqrt {{c^2} + {c^2}}$
$\Rightarrow AB = \sqrt {2{c^2}}$
$\Rightarrow AB = \pm \sqrt 2 c$
$AB = \sqrt 2 C$[Neglecting AB = $- \sqrt 2 C$]

Hence, the distance between A (c, 0) and B(0, – c) is $\sqrt 2 c$ units.
(g) It is given that point A (2a, 6a) and B $\left( {2a + \sqrt 3 a,5a} \right)$.

Distance between A and B is :
$AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

Where, ${x_1} = 2a$, ${y_1} = 6a$ , ${x_2} = 2a +\sqrt {3a}$ and ${y_2} = 5a$
$\Rightarrow AB = \sqrt {{{\left( {2a +\sqrt {3a} - 2a} \right)}^2} + {{\left( {5a - 6a} \right)}^2}}$
$\Rightarrow AB = \sqrt {{{\left( {\sqrt 3 a} \right)}^2} + {{\left( { - a} \right)}^2}}$
$\Rightarrow AB = \sqrt {3{a^2} + {a^2}}$
$\Rightarrow AB = \sqrt {4{a^2}} = \pm 2a$
$\Rightarrow AB{\rm{ }} = {\rm{ }}2a$ [Neglecting AB = -2a]
Hence, the distance between A(2a, 6a) and B $\left( {2a + \sqrt {3a} ,5a} \right)$ is 2a units.
(h) It is given that P(4, -2) is the mid-point of line segment joining the points A(5k, 3) and B(-k, - 7).

Mid-point of A(5k, 3) and B(-k, -7) is :
$= \left( {{{5k - k} \over 2},{{3 - 7} \over 2}} \right) = \left( {{{4k} \over 2},{{ - 4} \over 2}} \right) = \left( {2k, - 2} \right)$
But P(4, - 2) is mid-point of A and B
$\Rightarrow \left( {4, - 2} \right) = \left( {2k, - 2} \right)$
$\Rightarrow 4 = 2k$
$\Rightarrow k = {4 \over 2}$
$\Rightarrow k = 2$
Hence, the value of k = 2

Short answer type questions –I : -
Q.1       Find the value of p for which the points (–1, 3), (2, p) and (5, –1) are collinear.
[Delhi 2006]

Sol.       Let A(-1, 3), B(2, p) and C(5, -1) be the given points.
Condition for collinear points:
${x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0$
Where ${x_1} = - 1,{y_1} = 3,{x_2} = 2,{y_2} = p,{x_3} = 5$ and${y_3} = - 1$
$\Rightarrow - 1\left\{ {p - \left( { - 1} \right)} \right\} + 2\left( { - 1 - 3} \right) + 5\left( {3 - p} \right) = 0$
$\Rightarrow - 1\left( {p + 1} \right) + 2\left( { - 4} \right) + 5\left( {3 - p} \right) = 0$
$\Rightarrow - p - 1 - 8 + 15 - 5p = 0$
$\Rightarrow - 6p - 9 + 15 = 0$
$\Rightarrow - 6p + 6 = 0$
$\Rightarrow - 6p = - 6$
$\Rightarrow 6p = 6$
$\Rightarrow p = {6 \over 6}$
$\Rightarrow p = 1$
Hence, value of p = 1

Q.2      If (–2, –1); (a, 0); (4, b) and (1, 2) are the vertices of a parallelogram, find the values of a and b.
[AI 2006C]

Sol.      Let points A(-2, -1), B (a, 0), C(4, b) and D(1, 2) be the given vertices of a parallelogram
We know that the diagonals of parallelogram ABCD bisect each other
$\Rightarrow {{{x_1} + {x_3}} \over 2} = {{{x_2} + {x_4}} \over 2}$
Where, ${x_1} = - 2,{x_2} = a,{x_3} = 4$ and ${x_4} = 1{\rm{ }}$
$\Rightarrow {{ - 2 + 4} \over 2} = {{a + 1} \over 2}$
$\Rightarrow - 2 + 4 = a + 1$
$\Rightarrow 2 - 1 = a$
$\Rightarrow a = 1$
Now, ${{{y_1} + {y_3}} \over 2} = {{{y_2} + {y_4}} \over 2}$
Where,${y_1} = - 1,{\rm{ }}{y_2} = 0,{y_3} = b$ and ${y_4} = 2{\rm{ }}$
$\Rightarrow {{ - 1 + b} \over 2} = {{0 + 2} \over 2}$
$\Rightarrow - 1 + b = 2$
$\Rightarrow b = 2 + 1$
$\Rightarrow b = 3$
Hence, values of a and b are 1 and 3 respectively.

Q.3      In what ratio does the line $x - y - 2 = 0$ divide the line segment joining (3, –1) and (8, 9) ?
[Delhi 2007]

Sol.       Let the points A(3, -1) and B(8, 9) be the given points of line segment AB and
line x – y – 2 = 0 divide the line segment AB at point C (x, y) in the ratio m : n

By section formula :
$C(x,y) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$ ……….. (1)
Where ${x_1} = 3,{x_2} = 8,{y_1} = - 1,{y_2} = 9$
$\Rightarrow C(x,y) = \left[ {{{m \times 8 + n \times 3} \over {m + n}},{{m \times 9 + n \times \left( { - 1} \right)} \over {m + n}}} \right]$
$\Rightarrow C(x,y) = \left[ {{{8m + 3n} \over {m + n}},{{9m - n} \over {m + n}}} \right]$

As point C(x,y) lie on the line $x - y - 2 = 0$
$\Rightarrow {{8m + 3n} \over {m + n}} - \left( {{{9m - n} \over {m + n}}} \right) - 2 = 0$
$\Rightarrow 8m + 3n - 9m + n - 2m - 2n = 0$
$\Rightarrow - 3m + 2n = 0\Rightarrow 3m = 2n$
$\Rightarrow$ ${m \over n} = {2 \over 3}$
Hence, line x – y – 2 = 0 divide the line segment joining A (3, –1 and B(8, 9) in the ratio2 : 3 .

Q.4       If the point C(–1, 2) divides the line segment AB in the ratio 3 : 4, where the coordinates of A are (2, 5), find the coordinates of B.
[AI 2007]

Let the co-ordinates of point B be $({x_2},\,{y_2})$
Given that C divide AB in the ratio 3 : 4
By section formula :
$C( - 1,2) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$
Where,$m = 3,n = 4,{x_1} = 2$ and $y1 = 5$
$\Rightarrow C\left( { - 1,2} \right) = \left[ {{{3 \times {x_2} + 4 \times 2} \over {3 + 4}},{{3 \times {y_2} + 4 \times 5} \over {3 + 4}}} \right]$
$\Rightarrow C\left( { - 1,2} \right) = \left[ {{{3{x_2} + 8} \over 7},{{3{y_2} + 20} \over 7}} \right]$

On Equating x and y co-ordinates ,We get
$- 1 = {{3{x_2} + 8} \over 7}$
$\Rightarrow - 7 = 3{x_2} + 8$
$\Rightarrow 3{x_2} = - 7 - 8$
$\Rightarrow 3{x_2} = - 15$
$\Rightarrow {x_2} = {{ - 15} \over 3}$
$\Rightarrow {x_2} = - 5$
And $2 = {{3{y_2} + 20} \over 7}$
$\Rightarrow 14 = 3{y_2} + 20$
$\Rightarrow 3{y_2} = 14 - 20$
$\Rightarrow 3{y_2} = - 6$
$\Rightarrow {y_2} ={{ - 6} \over 3}$
$\Rightarrow {y_2} = - 2$
Hence, the co-ordinates of B are (–5,–2).

Q.5       For what value of p are the points (2, 1), (p, –1) and (–1, 3) collinear ?
[Delhi 2008]

Sol.        Let points A(2, 1), B(p, –1) and C(–1,3) be the given points.
Condition of collinear points is :
$\Rightarrow {x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0$ ……. (1)
On putting ${x_1} = 2,{y_1} = 1,{x_2} = b,{y_2} = - 1,{x_3}{\rm{ = }} - 1$ and ${y_3} = 3$ in equation (1),We get

$\Rightarrow 2\left( { - 1 - 3} \right) + p\left( {3 - 1} \right) + \left( { - 1} \right)\left\{ {1 - \left( { - 1} \right)} \right\} = 0$
$\Rightarrow 2\left( { - 4} \right) + p\left( 2 \right) - 1\left( {1 + 1} \right) = 0$
$\Rightarrow - 8 + 2p - 2 = 0$
$\Rightarrow 2p - 10 = 0$
$\Rightarrow 2p = 10$
$\Rightarrow p = {{10} \over 2}$
$\Rightarrow p = 5$
Hence, the value of p = 5.

Q.6       If the distances of P(x, y) from the points A(3, 6) and B(–3, 4) are equal prove that 3x + y = 5.
[AI 2008, Delhi 2008]

Sol.        It is given that p(x, y) has equal distance from the points A(3, 6) and B(–3, 4) i.e.PA = PB
$\Rightarrow \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 6} \right)}^2}} = \sqrt {{{\left\{ {x - \left( { - 3} \right)} \right\}}^2} + {{\left( {y - 4} \right)}^2}}$

Squaring both the sides
$\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x + 3} \right)^2} + {\left( {y - 4} \right)^2}$
$\Rightarrow {x^2} + 9 - 6x + {y^2} + 36 - 12y = {x^2} + 9 + 6x + {y^2} + 16 - 8y$
$\Rightarrow 45 - 6x - 12y = 25 + 6x - 8y$
$\Rightarrow - 6x - 6x - 12y + 8y = 25 - 45$
$\Rightarrow - 12x - 4y = - 20$
$\Rightarrow - \left( {12x + 4y} \right) = - 20$
$\Rightarrow 12x + 4y = 20$
$\Rightarrow 12x + 4y - 20 = 0$
$\Rightarrow 4\left( {3x + y - 5} \right) = 0$
$\Rightarrow 3x + y = 5$
Hence proved.

Q.7        Find the area of the $\Delta ABC$ with vertices A(–5, 7), B(–4, –5) and C(4, 5).
[AI 2008]

Sol.        It is given that point A(–5, 7), B(–4,–5) and C(4, 5) are the vertices of a triangle
Area of $\Delta ABC$ = ${1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$
Where, ${x_1} = - 5,{y_1} = 7,{x_2} =- 4,{y_2} = - 5,{x_3} = 4$ and${y_3} = 5$

$\Rightarrow$ Area of $\Delta ABC$ $= {1 \over 2}\left| {\left( { - 5} \right)\left( { - 5 - 5} \right) + \left( { - 4} \right)\left( {5 - 7} \right) + 4\left\{ {7 - \left( { - 5} \right)} \right\}} \right|$
$\Rightarrow$ Area of $\Delta ABC$ $= {1 \over 2}\left| {\left( { - 5} \right)\left( { - 10} \right) + \left( { - 4} \right)\left( { - 2} \right) + 4\left( {7 + 5} \right)} \right|$

$\Rightarrow$ Area of $\Delta ABC$ $= {1 \over 2}\left| {50 + 8 + 48} \right|$
$\Rightarrow$ Area of $\Delta ABC$ $= {1 \over 2}\left| {106} \right| = 53$
Hence, the area of triangle = 53 sq. units.

Q.8      The point R divides the line segment AB where A(–4, 0), B(0, 6) are such that $AR = {3 \over 4}AB$. Find the coordinates of R.
[AI 2008]

It is given that R divides line segment AB. Let coordinates of R be (x, y).
$AR = {3 \over 4}AB$ [Given]..............(1)
AB = AR + RB
$\Rightarrow {4 \over 3}AR = AR + RB$
$\Rightarrow {4 \over 3}AR - AR = RB$
$\Rightarrow AR = 3RB \Rightarrow {{AR} \over {RB}} = 3$
So, point R divides AB into 3 : 1 ratio.
By section Formula:

R(x,y) = $\left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$
Where$m = 3,n = 1,{x_1} = - 4,{x_2} = 0,{y_1} = 0$ and ${y_2} = 6$
$\Rightarrow R(x,y) = \left[ {{{3 \times 0 + 1 \times \left( { - 4} \right)} \over {3 + 1}},{{3 \times 6 + 1 \times 0} \over {3 + 1}}} \right]$
$\Rightarrow R(x,y) = \left[ {{{0 - 4} \over 4},{{18 + 0} \over 4}} \right]$
$\Rightarrow R(x,y) = \left[ {{{ - 4} \over 4},{{18} \over 4}} \right]$
$\Rightarrow R(x,y) = \left[ { - 1,{9 \over 2}} \right]$
Hence, the coordinates of $R = \left( { - 1,{9 \over 2}} \right)$

Q.9        If A(4, –8), B(3, 6) and C(5, –4) are the vertices of $\Delta ABC$, D is the mid point of BC and P is a point on AD joined such that ${{AP} \over {PD}} = 2$, find the coordinates of P.
[AI 2008]

Sol.

Given
A(4, –8), B(3, 6) and C (5, –4) are the vertices of $\Delta ABC$
and D is mid-point of BC
So, the co-ordinates of $D = \left( {{{3 + 5} \over 2},{{6 - 4} \over 2}} \right)$
$\Rightarrow D = \left( {{8 \over 2},{2 \over 2}} \right)$
$\Rightarrow D = \left( {4,1} \right)$
Now,${{AP} \over {PD}} = {2 \over 1}$ [Given]
Means, P divides AD into 2 : 1 ratio
By section formula:
$P(x,y) = [{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}]$
Where, $m = 2,n = 1,{x_1} = 4,{x_2} = 4,{y_1} = - 8$ and ${y_2} = 1$

$\Rightarrow P(x,y) = \left[ {{{2 \times 4 + 1 \times 4} \over {2 + 1}},{{2 \times 1 + 1 \times ( - 8)} \over {2 + 1}}} \right]$
$\Rightarrow P(x,y) = \left[ {{{8 + 4} \over {2 + 1}},{{2 - 8} \over {2 + 1}}} \right]{\rm{ }}$
$\Rightarrow P(x,y) = \left[ {{{12} \over 3},{{ - 6} \over 3}} \right]$
$\Rightarrow P(x,y) = \left[ {4, - 2} \right]$

Hence, the co-ordinates of P= (4, –2).

Q.10      Find the ratio in which the line 3x + y –9 = 0 divides the line segment joining the points (1, 3) and (2, 7).
Sol.        Let A (1, 3) and B(2, 7) be the given points
And line 3x + y – 9 = 0 divides line segment AB at C(x,y) in the ratio m : n.
By section formula: $\Rightarrow C(x,y) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$
Where, ${x_1} = 1,{x_2} = 2,{y_1} = 3$ and ${y_2} = 7$
$\Rightarrow C(x,y) = \left[ {{{m \times 2 + n \times 1} \over {m + n}},{{m \times 7 + n \times 3} \over {m + n}}} \right]$
$\Rightarrow C(x,y) = \left[ {{{2m + n} \over {m + n}},{{7m + 3n} \over {m + n}}} \right]{\rm{ }}$

As point C(x,y) lie on the line $3x + y - 9 = 0$
$\Rightarrow 3\left( {{{2m + n} \over {m + n}}} \right) + {{7m + 3n} \over {m + n}} - 9 = 0$
$\Rightarrow 6m + 3n + 7m + 3n - 9m - 9n = 0$
$\Rightarrow 4m - 3n = 0$

$\Rightarrow 4m = 3n$
$\Rightarrow {m \over n} = {3 \over 4}$
Hence, 3x + y – 9 = 0 divides line segment AB in the ratio 3 : 4 .

Q.11      Find the area of the quadrilateral whose vertices are A(0, 0), B(6, 0), C(4, 3) and D(0, 3).
[Delhi 2008 C]

A(0, 0), B(6, 0), C(4, 3) and D(0, 3) are the given vertices of quadrilateral ABCD.
Area of quadrilateral ABCD = Area of $\left( {\Delta ABC + \Delta ACD} \right)$ ………. (1)
Area of $\Delta ABC$ $= {1 \over 2}\left| {\left( {0 + 6 \times 3 + 4 \times 0 - 0 - 0 - 0} \right)} \right|$
$= {1 \over 2}\left| {18} \right|$ sq. units
$= 9$ sq. units
Area of $\Delta ACD$ $= {1 \over 2}\left| {0 \times 3 + 4 \times 3 + 0 - 0 - 0 - 0} \right|$

$= {1 \over 2}\left| {12} \right|$
$= 6$ sq. units
From (1) –

Area of quadrilateral ABCD  = Area of $\Delta ABC$ + Area of $\Delta ACD$
= 9 + 6 sq. units = 15 sq. units
Hence, area of quadrilateral ABCD is 15 square units.

Q.12     Find the area of a rhombus whose vertices taken in order are the points (3, 0), (4, 5), (–1, 4) and (–2, –1).
[AI 2008 C, Delhi 2008 C]

Let A (3, 0,) B(4, 5), C(–1, 4) and D(–2, –1) be the given points of Rhombus ABCD
Area of Rhombus ABCD = Area of $\Delta ABC$ + Area of $\Delta ACD$ ……… (1)

Area of $\Delta ABC$

$= {1 \over 2}\left| {\left( {3 \times 5} \right) + \left( {4 \times 4} \right) + \left( { - 1 \times 0} \right) - \left( {4 \times 0} \right) - \left( {- 1 \times 5} \right) - \left( {3 \times 4} \right)} \right|$
$= {1 \over 2}\left| {36 - 12} \right|$

$= {1 \over 2}\left| {24} \right|$
= 12 sq. units
Area of $\Delta ACD$ $= {1 \over 2}\left| {\left( {3 \times 4} \right) + \left\{ { - 1 \times \left( { - 1} \right)} \right\} + \left( { - 2 \times 0} \right) - \left( { - 1 \times 0} \right) - \left( { - 2 \times 4} \right) - \left\{ {3 \times \left( { - 1} \right)} \right\}} \right|$
$= {1 \over 2}\left| {12 + 1 - 0 - 0 + 8 + 3} \right|$
$= {1 \over 2}\left| {24} \right|$
= 12 sq. units
Area of Rhombus ABCD  = (12 + 12) sq. units

= 24 sq. units
Hence, area of Rhombus ABCD is 24 square units.

Q.13     Find the coordinates of the point of trisection of the line segment joining (1, –2) and (–3, 4).
[AI 2008 C]

Sol.        Let A(1, –2), B(–3, 4) be the given points of line segment AB.
And C and D be the points which trisect AB i.e. AC = CD = DB = K (let)
${{AC} \over {CB}} = {{AC} \over {CD + DB}}$
$\Rightarrow {{AC} \over {CB}} = {K \over {K + K}}$
$\Rightarrow {{AC} \over {CB}} = {K \over {2K}}$
$\Rightarrow {{AC} \over {CB}} = {1 \over 2}$
So, C divides AB in the ratio 1 : 2 .
By section formula:
$C(x,y) = \left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
Where, $m = 1,n = 2,{x_1} = 1,{x_2} = - 3,{y_1} = -2$ and ${y_2} = 4$
$\Rightarrow C(x,y) = \left[ {{{1 \times \left( { - 3} \right) + 2 \times 1} \over {1 + 2}},{{1 \times 4 + 2\left( { - 2} \right)} \over {1 + 2}}} \right]$
$\Rightarrow C(x,y) = \left[ {{{ - 3 + 2} \over {1 + 2}},{{4 - 4} \over {1 + 2}}} \right]$
$\Rightarrow C(x,y) = \left[ {{{ - 1} \over 3},{0 \over 3}} \right]$
Now,${{AD} \over {DB}} = {{AC + CD} \over {DB}}$
$\Rightarrow {{AD} \over {DB}} = {{K + K} \over K}$
$\Rightarrow {{AD} \over {DB}} = {{2K} \over K}$
$\Rightarrow {{AD} \over {DB}} = {2 \over 1}$
So, point D divides AB in the ratio 2 : 1.
By section formula: $D(x,y) = \left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
Where, $m = 2,n = 1,{x_1} = 1,{x_2} = - 3,{y_1} = -2$ and ${y_2} = 4$

$\Rightarrow D(x,y) = \left[ {{{2 \times \left( { - 3} \right) + 1\left( 1 \right)} \over {2 + 1}},{{2 \times \left( 4 \right) + 1\left( { - 2} \right)} \over {2 + 1}}} \right]$
$\Rightarrow D(x,y) = \left[ {{{ - 6 + 1} \over {2 + 1}},{{8 - 2} \over {2 + 1}}} \right]$
$\Rightarrow D(x,y) = \left[ {{{ - 5} \over 3},{6 \over 3}} \right]$
$\Rightarrow D(x,y) = \left[ {{{ - 5} \over 3},2} \right]$
Hence, the co-ordinates of C and D are$\left( {{{ - 1} \over 3},0} \right)$ and $\left( {{{ - 5} \over 3},2} \right)$ respectively.

Q.14      Find the point on y-axis which is equidistant from the point (5, –2) and (–3, 2).
OR
The line segment joining the points A(2, 1) and B(5, –8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k.
[Delhi 2009]

Sol.         Let point A(5, -2) and B(-3, 2) be the given points and C be the point on y-axis which is equidistant from A & B.
So,
Coordinates of C = (0, y)

Now, AC = BC

$\Rightarrow \sqrt {{{\left( {5 - 0} \right)}^2} + {{\left( { - 2 - y} \right)}^2}} = \sqrt {{{\left( { - 3 - 0} \right)}^2} + {{\left( {2 - y} \right)}^2}}$
On Squaring both the sides, We get
$\Rightarrow {\left( {5 - 0} \right)^2} + {\left( { - 2 - y} \right)^2} = {\left( { - 3 - 0} \right)^2} + {\left( {2 - y} \right)^2}$
$\Rightarrow {\left( 5 \right)^2} + {\left( { - 2 - y} \right)^2} = {\left( { - 3} \right)^2} + {\left( {2 - y} \right)^2}$
$\Rightarrow 25 + 4 + {y^2} + 4y = 9 + 4 + {y^2} - 4y$
$\Rightarrow 29 + 4y = 13 - 4y$
$\Rightarrow 4y + 4y = 13 - 29$
$\Rightarrow 8y =-16$
$\Rightarrow y = {{ - 16} \over 8}$
$\Rightarrow y = - 2$
Hence, the point C is (0, –2).
OR
It is given that line segment joining the points A(2, 1) and (5, -8) is trisected at P and Q. So, AP = PQ = QB = K (let)
${{AP} \over {PB}} = {{AP} \over {PQ + QB}}$

$\Rightarrow {{AP} \over {PB}} = {K \over {K + K}}$
$\Rightarrow {{AP} \over {PB}} = {K \over {2K}}$
$\Rightarrow {{AP} \over {PB}} = {1 \over 2}$
So, P divides AB in the ratio 1 : 2 .
By Section formula:
$P(x,y) = \left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
Where, $m = 1,n = 2,{x_1} = 2,{x_2} = 5,{y_1} = 1$ and ${y_2} = - 8$

$\Rightarrow P(x,y) = \left[ {{{1 \times 5 + 2 \times 2} \over {1 + 2}},{{1 \times \left( { - 8} \right) + 2\left( 1 \right)} \over {1 + 2}}} \right]$
$\Rightarrow P(x,y) = \left[ {{{5 + 4} \over 3},{{ - 8 + 2} \over 3}} \right]$
$\Rightarrow P(x,y) = \left[ {{9 \over 3},{{ - 6} \over 3}} \right]$
$\Rightarrow P(x,y) = \left[ {3, - 2} \right]$
It is given that point P lies on line$2x - y + k = 0$.
$\Rightarrow 2\left( 3 \right) - \left( { - 2} \right) + k = 0$
$\Rightarrow 6 + 2 + k = 0$
$\Rightarrow 8 + k = 0$
$\Rightarrow k = - 8$
Hence, the value of k = – 8.

Q.15     If P(x, y) is any point on the line joining the points A(a, 0) and B(0, b) then show that ${x \over a} + {y \over b} = 1$.
[Delhi 2009]

Sol.        It is given that point P(x, y) is any point on the line joining the points A(a, 0) and B(0, b).
So we can say that P, A and B are collinear points, means area of $\Delta PAB$ is zero.

Area of $\Delta PAB = {1 \over 2}\left| {\left( {x \times 0} \right) + \left( {a \times b} \right) + \left( {0 \times y} \right) - \left( {a \times y} \right) - \left( {0 \times 0} \right) - \left( {x \times b} \right)} \right|$
$\Rightarrow 0 = {1 \over 2}\left| {0 + ab - 0 - ay - xb} \right|$

$\Rightarrow 0 = {1 \over 2}\left| {ab - ay - xb} \right|$
$\Rightarrow 0 = \left| {ab - ay - xb} \right|$
$\Rightarrow ab = ay + xb$
On dividing both sides by ab,We get

${{ab} \over {ab}} = {{ay} \over {ab}} + {{xb} \over {ab}}$
$\Rightarrow 1 = {y \over b} + {x \over a}$
$\Rightarrow {x \over a} + {y \over b} = 1$
“Hence proved”

Q.16      Find the point on x-axis which is equidistant from the points (2, –5) and (–2, 9).
[Delhi 2009]

Sol.        Let point A (2, –5) and B(–2, 9) be the given points and point C on x-axis
which is equidistant from A and B.

Coordinates of point C = (x, 0) AC = BC
$\Rightarrow \sqrt {{{\left( {2 - x} \right)}^2} + {{\left( { - 5 - 0} \right)}^2}} =\sqrt {{{\left\{ {x - \left( { - 2} \right)} \right\}}^2} + {{\left( {0 - 9} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( {2 - x} \right)}^2} + {{\left( { - 5} \right)}^2}} =\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( { - 9} \right)}^2}}$
Squaring both the sides

$\Rightarrow {\left( {2 - x} \right)^2} + {\left( { - 5} \right)^2} = {\left( {x + 2} \right)^2} + {\left( { - 9} \right)^2}$
$\Rightarrow 4 + {x^2} - 4x + 25 = {x^2} + 4 + 4x + 81$
$\Rightarrow 29 - 4x = 4x + 85$
$\Rightarrow - 4x - 4x = 85 - 29$
$\Rightarrow - 8x = 56$
$\Rightarrow x = {{ - 56} \over 8}$
$\Rightarrow x = - 7$
Hence, the point C is (–7, 0)

Q.17     If the points A(4, 3) and B(x, 5) are on the circle with the centre O(2, 3), find the value of x.
[AI 2009]

Sol.       Point A(4, 3) and B(x, 5) are the given points on circle with centre O(2, 3). We know that points on the circle are equidistant from the centre.
$\Rightarrow AO = OB$
$\Rightarrow \sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {3 - 3} \right)}^2}} = \sqrt {{{\left( {2 - x} \right)}^2} + {{\left( {3 - 5} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} =\sqrt {{{\left( {2 - x} \right)}^2} + {{\left( { - 2} \right)}^2}}$
Squaring both the sides
$\Rightarrow 4 + 0 = {\left( {2 - x} \right)^2} + 4$
$\Rightarrow 4 = 4 + {x^2} - 4x + 4$
$\Rightarrow {x^2} - 4x + 4 = 0$
$\Rightarrow {x^2} - 2x - 2x + 4 = 0$
$\Rightarrow \left( {x - 2} \right)\left( {x - 2} \right) = 0$

$\Rightarrow x - 2 = 0$ or $x - 2 = 0$
$\Rightarrow x = 2$
Hence, the value of x = 2.

Q.18     Find the ratio in which the point (2, y) divides the line segment joining the points A(–2, 2) and B(3, 7). Also find the value of y.
[AI 2009]

Sol.        Let point P(2, y) divides the line segment joining the points
A(–2, 2) and B(3, 7) in the ratio m : n .

By section formula: $P\left( {2,y} \right) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$
Where, ${x_1} = - 2,{x_2} = 3,{y_1} = 2$ and ${y_2} = 7$
$\Rightarrow P\left( {2,y} \right) = \left[ {{{m \times 3 + n \times ( - 2)} \over {m + n}},{{m \times7+ n \times 2} \over {m + n}}} \right]$

$\Rightarrow P\left( {2,y} \right) = \left[ {{{3m - 2n} \over {m + n}},{{7m + 2n} \over {m + n}}} \right]$
On Equating x and y co-ordinate ,We get ${{3m - 2n} \over {m + n}} = 2$
$\Rightarrow 3m - 2n = 2m + 2n$

$\Rightarrow m = 4n$................(1)
$\Rightarrow {m \over n} = {4 \over 1}$
And $y = {{7m + 2n} \over {m + n}}$
$\Rightarrow y = {{7(4n) + 2n} \over {4n + n}}$ [Using (1)]
$\Rightarrow y = {{28n + 2n} \over {4n + n}} = {{30n} \over {5n}} = 6$
Hence, point (2, y) divides line segment AB into 4 : 1 ratio and value of y = 6.

Q.19       Find the ratio in which the point (x, 2) divides the line segment joining the points (–3, –4) and (3, 5). Also find the value of x.
[AI 2009]

Sol.         Let the point P(x, 2) divides the line segment joining the points A(–3, –4) and B(3, 5) in the ratio m : n .
By section formula: $P(x,2) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$
Where,${x_1} = - 3,{x_2} = 3,{y_1} = - 4$ and ${y_2} = 5$
$\Rightarrow P(x,2) = \left[ {{{m \times 3 + n( - 3)} \over {m + n}},{{m \times 5 + n \times ( - 4)} \over {m + n}}} \right]{\rm{ }}$
$\Rightarrow P(x,2) = \left[ {{{3m - 3n} \over {m + n}},{{5m - 4n} \over {m + n}}} \right]{\rm{ }}$
On Equating x and y co-ordinate,We get ${{5m - 4n} \over {m + n}} = 2$
$\Rightarrow 5m - 4n = 2m + 2n$
$\Rightarrow 3m = 6n$....................(1)
$\Rightarrow {m \over n} = 2$
And $x= {{3m - 3n} \over {m + n}}$
$\Rightarrow x = {{6n - 3n} \over {2n + n}} = {{3n} \over {3n}} = 1$ [Using(1)]
Hence, the value of x = 1 and P(x,2) divides the line segment joining A(-3,-4)and B(3,5) in the ratio 2:1.

Q.20       Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3).
[AI 2009]

Sol. Let point A(0, –1), B(2, 1) and C(0, 3) be the given vertices of
$\Delta ABC$ and D, E, F be the mid points of sides BC, CA and AB respectively.

Area of $\Delta ABC = {\kern 1pt} {1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$
$= {\kern 1pt} {1 \over 2}\left| {0\left( {1 - 3} \right) + 2\left\{ {3 - \left( { - 1} \right)} \right\} + 0\left( { - 1 - 1} \right)} \right|$
$= {\kern 1pt} {1 \over 2}\left| {0 \times \left( { - 2} \right) + 2\left( 4 \right) + 0\left( { - 2} \right)} \right|$
$= {\kern 1pt} {1 \over 2}\left| {0 + 8 + 0} \right|$
$= {\kern 1pt} {1 \over 2}\left| 8 \right|$
= 4 sq. units
Area of $\Delta DEF = {1 \over 4} \times Area\,\,of\,\Delta ABC$
$= \left( {{1 \over 4} \times 4} \right)Sq.units$
= 1 sq. units
Hence, area of triangle formed by mid points is 1 square units.

Q.21       The centre of a circle is $\left( {2\alpha - 1,7} \right)$ and it passes through the point (–3, –1). If the diameter of the circle is 20 units, then find the value of $\alpha$.
[Foreign 2009]

Sol.          Let $O\left( {2\alpha - 1,7} \right)$ be the centre of the circle and it basses through the point P(–3, –1). Diameter of circle = 20 units
Radius of circle $= {{diameter} \over 2}$ $= {{20} \over 2}$ = 10 units
We know that any point on the circle has distance equal to radius from centre, so – OP = 10
$\Rightarrow \sqrt {{{\left\{ {2\alpha - 1 - \left( 3 \right)} \right\}}^2} + {{\left\{ {7 - \left( { - 1} \right)} \right\}}^2}} = 10$
$\Rightarrow \sqrt {{{\left\{ {2\alpha - 1 - \left( 3 \right)} \right\}}^2} + {{\left\{ {7 - \left( { - 1} \right)} \right\}}^2}} = 10$
$\Rightarrow 4{\alpha ^2} + 4 + 8\alpha + 64 = 100$
$\Rightarrow 4{\alpha ^2} + 68 - 100 + 8\alpha = 0$
$\Rightarrow 4{\alpha ^2} + 8\alpha - 32 = 0$
$\Rightarrow 4\left( {{\alpha ^2} + 2\alpha - 8} \right) = 0$
$\Rightarrow {\alpha ^2} + 2\alpha - 8 = 0$
$\Rightarrow {\alpha ^2} + 4\alpha - 2\alpha - 8 = 0$
$\Rightarrow \alpha \left( {\alpha + 4} \right) - \alpha \left( {\alpha + 4} \right) = 0$
$\Rightarrow \left( {\alpha + 4} \right)\left( {\alpha - 2} \right) = 0$
$\alpha {\rm{ }} + {\rm{ }}4{\rm{ }} = {\rm{ }}0{\rm{ }}and{\rm{ }}\alpha {\rm{ }} = {\rm{ }}-2$
$\alpha = -4,{\rm{ }}\alpha {\rm{ }} = {\rm{ }}2$
Hence, the volume of $\alpha = -4{\rm{ }}and{\rm{ }}\alpha {\rm{ }} = {\rm{ }}2$

Q.22        If C is a point lying on the line segment AB joining A(1, 1) and B(2, –3) such that 3AC = CB, then find the coordinates of C.
[Foreign 2009]

OR
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
[Foreign 2009]

Sol.          It is given that point C lying on line segment joining the points A(1, 1) and B(2, –3) such that 3AC = CB
$\Rightarrow {{AC} \over {CB}} = {1 \over 3}$ Therefore, C divides AB in the ratio 1 : 3.
By section formula: $C(x,y) = \left[ {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right]$
Where,$m = 1,n = 3,{x_1} = 1,{x_2} = 2,{y_1} = 1$ and ${y_2} = - 3$
$\Rightarrow C(x,y) = \left[ {{{1 \times 2 + 3 \times 1} \over {3 + 1}},{{1 \times ( - 3) + 3 \times 1} \over {3 + 1}}} \right]$
$\Rightarrow C(x,y) = \left[ {{{1 + 3} \over 4},{{ - 3 + 3} \over 4}} \right]{\rm{ }}$
$\Rightarrow C(x,y) = \left[ {{4 \over 4},{0 \over 4}} \right]{\rm{ }}$
$\Rightarrow C(x,y) = \left[ {1,0} \right]{\rm{ }}$

Hence, Co-ordinates of C = (1,0).

Q.23        If the points (–2, 1), (a, b) and (4, –1) are collinear and a –b = 1 then find the value of a and b.
[Foreign 2009]

Sol.          Let points A(–2, 1), B(a, b) and C(4,–1) be the given points.
It is given that points A(–2, 1), B(a, b) and C(4,–1) are collinear means Area of $\Delta ABC = 0$ Area of $\Delta ABC = {1 \over 2}\left| {\left( { - 2 \times b} \right) + a \times \left( { - 1} \right) + 4 \times 1 - 1 \times a - b \times 4 - \left( { - 1} \right)\left( { - 2} \right)} \right|$
$\Rightarrow 0 = {1 \over 2}\left| { - 2b - a + 4 - a - 4b - 2} \right|$
$\Rightarrow 0 = \left| { - 6b - 2a + 2} \right|$
$\Rightarrow 6b + 2a = 2$
$\Rightarrow 2\left( {3b + a} \right) = 2$
$\Rightarrow a + 3b = 1..........(1)$
Also given a – b = 1...................(2)
Put a = b + 1 in equation (1)

$\Rightarrow$ b + 1 + 3b = 1
$\Rightarrow$ 4b = 1 – 1
$\Rightarrow b = {0 \over 4}$
$\Rightarrow$ b = 0
Put b = 0 in equation (2) –
$\Rightarrow$ a – 0 = 1
$\Rightarrow$ a = 1
Hence, the value of a = 1 and b = 0

Short answer type question – II –

Q.1         Show that the points A(1, 2), B(5, 4), C(3, 8) and D(–1, 6) are the vertices of a square.
[Delhi 2006]

Sol.          Points A(1, 2), B(5, 4), C(3, 8) and D(–1, 6) are the given vertices. of a quadrilateral. By using distance formula –
$AB = \sqrt {{{\left( {1 - 5} \right)}^2} + {{\left( {2 - 4} \right)}^2}}$
$\Rightarrow AB = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 2} \right)}^2}}$
$\Rightarrow AB = \sqrt {16 + 4}$
$\Rightarrow AB = \sqrt {20}$ units...............(1)
$BC = \sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {4 - 8} \right)}^2}}$

$\Rightarrow BC = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2}}$
$\Rightarrow BC = \sqrt {4 + 16}$
$\Rightarrow BC = \sqrt {20}$ units........................(2)
$CD = \sqrt {{{\left\{ {3 - \left( { - 1} \right)} \right\}}^2} + {{\left( {8 - 6} \right)}^2}}$

$\Rightarrow CD = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( 2 \right)}^2}}$
$\Rightarrow CD = \sqrt {{{\left( 4 \right)}^2} + 4}$
$\Rightarrow CD = \sqrt {16 + 4}$
$\Rightarrow CD = \sqrt {20}$ units...........(3)
$DA = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {6 - 2} \right)}^2}}$

$\Rightarrow DA = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2}}$
$\Rightarrow DA = \sqrt {4 + 16}$
$\Rightarrow DA = \sqrt {20}$ units...............(4)
From (1),(2),(3)and(4)

AB = BC = CD = DA (All sides are equal)..................(5)
Now $AC = \sqrt {{{\left( {1 - 3} \right)}^2} + {{\left( {2 - 8} \right)}^2}}$

$\Rightarrow AC = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 6} \right)}^2}}$
$\Rightarrow AC = \sqrt {4 + 36}$
$\Rightarrow AC = \sqrt {40}$ units................(6)
$BD = \sqrt {{{\left( { - 1 - 5} \right)}^2} + {{\left( {6 - 4} \right)}^2}}$

$\Rightarrow BD = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( 2 \right)}^2}}$
$\Rightarrow BD = \sqrt {36 + 4}$
$\Rightarrow BD = \sqrt {40}$ units....................(7)
From (6) and (7)

AC=BD (diagonals are equal)....................(8)
Therefore all four sides are equal and diagonals are also equal so it is proved that A,B,C and D are the vertices of a square.

Q.2         If the points (10, 5), (8, 4) and (6, 6) are the mid-points of the sides of a triangle, find its vertices.
[Foreign 2006]

Sol.         Let D(10, 5), E(8, 4) and F(6, 6) be the given points of $\Delta ABC$ and $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$
be the vertices of $\Delta ABC$.

D is mid-point of BC
$\Rightarrow D(10,5) = \left[ {{{{x_2} + {x_3}} \over 2},{{{y_2} + {y_3}} \over 2}} \right]$
Compairing x co-ordinate -
$\Rightarrow 10 = {{{x_2} + {x_3}} \over 2}$
$\Rightarrow {x_2} + {x_3} = 20$....................(1)
And, compairing y co-ordinate -
$\Rightarrow$ $5 = {{{y_2} + {y_3}} \over 2}$
$\Rightarrow {y_2} + {y_3} = 10$....................(2)
E is mid-point of AC

$\Rightarrow E(8,4) = \left[ {{{{x_1} + {x_3}} \over 2},{{{y_1} + {y_3}} \over 2}} \right]$
Compairing x co-ordinate -
$\Rightarrow {{{x_1} + {x_3}} \over 2} = 8$
$\Rightarrow {x_1} + {x_3} = 16$
And ,compairing y co-ordinate -
$\Rightarrow$ ${{{y_1} + {y_3}} \over 2} = 4$
$\Rightarrow {y_1} + {y_3} = 8$.....................(4)
F is mid-point of AB –

$\Rightarrow F(6,6) = \left[ {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right]$
Compairing x co-ordinate -
$\Rightarrow {{{x_1} + {x_2}} \over 2} = 6$
$\Rightarrow {x_1} + {x_2} = 12$.......................(5)
And ,compairing y co-ordinate -
$\Rightarrow$ ${{{y_1} + {y_2}} \over 2} = 6$
$\Rightarrow {y_1} + {y_2} = 12$.................(6)
Adding equation (1), (3) and (5)

$\Rightarrow \left( {{x_2} + {x_3}} \right) + \left( {{x_3} + {x_1}} \right) + \left( {{x_1} + {x_2}} \right) = 20 + 16 + 12$
$\Rightarrow 2\left( {{x_1} + {x_2} + {x_3}} \right) = 48$
$\Rightarrow \left( {{x_1} + {x_2} + {x_3}} \right) = {{48} \over 2}$
$\Rightarrow \left( {{x_1} + {x_2} + {x_3}} \right) = 24..........(7)$
Adding equation (2), (4) and (6)

$\Rightarrow \left( {{y_2} + {y_3}} \right) + \left( {{y_1} + {y_3}} \right) + \left( {{y_1} + {y_2}} \right) = 10 + 8 + 12$
$\Rightarrow 2\left( {{y_1} + {y_2} + {y_3}} \right) = 30$
$\Rightarrow \left( {{y_1} + {y_2} + {y_3}} \right) = {{30} \over 2}$
$\Rightarrow \left( {{y_1} + {y_2} + {y_3}} \right) = 15..........(8)$
Put${x_1} + {x_2} = 12$ in equation (7)

$\Rightarrow 12 + {x_3} = 24$
$\Rightarrow {x_3} = 24 - 12$
$\Rightarrow {x_3} = 12$
Put ${x_1} + {x_3} = 16$ in equation (7) –

$\Rightarrow {x_2} + 16 = 24$
$\Rightarrow {x_2} = 24 - 16$
$\Rightarrow {x_2} = 8$
Put${x_2} + {x_3} = 20$ in equation (7) –

$\Rightarrow {x_1} + 20 = 24$
$\Rightarrow {x_1} = 24 - 20$
$\Rightarrow {x_1} = 4$
Put ${y_1} + {y_2} = 12$ in equation (8) –

$\Rightarrow 12 + {y_3} = 15$
$\Rightarrow {y_3} = 15 - 12$
$\Rightarrow {y_3} = 3$
Put${y_2} + {y_3} = 10$ in equation (8) –

$\Rightarrow {y_1} + 10 = 15$
$\Rightarrow {y_1} = 15 - 10$
$\Rightarrow {y_1} = 5$
Put ${y_1} + {y_3} = 8$in equation (8) –

$\Rightarrow {y_2} + 8 = 15$
$\Rightarrow {y_2} = 15 - 8$
$\Rightarrow {y_2} = 7$
Hence, the co-ordinates of A(4, 5) , B(8, 7) and C(12, 3).

Q.3         Show that the points (7, 10), (–2, 5) and (3, –4) are the vertices of an isosceles right triangle.
[Delhi 2007]

Sol.         Let A(7, 10), B(–2, 5) and C(3, –4) be the given vertices. By using distance formula –
$AB = \sqrt {{{\left\{ {7 - \left( { - 2} \right)} \right\}}^2} + {{\left( {10 - 5} \right)}^2}}$
$\Rightarrow$ $AB = \sqrt {{{\left( {7 + 2} \right)}^2} + {{\left( 5 \right)}^2}}$

$\Rightarrow$$AB = \sqrt {{{\left( 9 \right)}^2} + {{\left( 5 \right)}^2}}$
$\Rightarrow$$AB = \sqrt {81 + 25}$
$\Rightarrow$$AB = \sqrt {106}$ units
$\Rightarrow$$BC = \sqrt {{{\left( { - 2 - 3} \right)}^2} + {{\left\{ {5 - \left( { - 4} \right)} \right\}}^2}}$

$\Rightarrow$$BC = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( {5 + 4} \right)}^2}}$
$\Rightarrow$$BC = \sqrt {25 + {{\left( 9 \right)}^2}}$
$\Rightarrow$$BC = \sqrt {25 + 81}$
$\Rightarrow$$BC = \sqrt {106}$ units
$\Rightarrow$$CA = \sqrt {{{\left( {3 - 7} \right)}^2} + {{\left( { - 4 - 10} \right)}^2}}$

$\Rightarrow$$CA = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 14} \right)}^2}}$
$\Rightarrow$$CA = \sqrt {16 + 196}$
$\Rightarrow$$CA = \sqrt {212}$
$AB = BC \ne CA$ means A, B and C are the vertices of isosceles triangle.

Now using Pythagoras theorem in $\Delta ABC$ -
${\left( {CA} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}$
$\Rightarrow {\left( {\sqrt {212} } \right)^2} = {\left( {\sqrt {106} } \right)^2} + {\left( {\sqrt {106} } \right)^2}$
$\Rightarrow 212 = 106 + 106$
$\Rightarrow$ L.H.S. = R.H.S.
Hence, we can say that A, B and C are the vertices of an isosceles right triangle.

Q.4          Show that A(–3, 2), B(–5, –5), C(2, –3) and D(4, 4) are the vertices of a rhombus.
[Foreign 2008]

Sol.           A(–3, 2), B(–5, –5), C(2, –3) and D(4, 4) are the given vertices of quadrilateral ABCD. By using distance formula – $AB = \sqrt {{{\left\{ { - 3 - \left( { - 5} \right)} \right\}}^2} + {{\left\{ {2 - \left( { - 5} \right)} \right\}}^2}}$
$\Rightarrow$$AB = \sqrt {{{\left( { - 3 + 5} \right)}^2} + {{\left( {2 + 5} \right)}^2}}$
$\Rightarrow$$AB = \sqrt {{{\left( 2 \right)}^2} + {{\left( 7 \right)}^2}}$
$\Rightarrow$$AB = \sqrt {4 + 49}$
$\Rightarrow$$AB = \sqrt {53}$ unit
$\Rightarrow$$BC = \sqrt {{{\left\{ {2 - \left( { - 5} \right)} \right\}}^2} + {{\left\{ { - 3 - \left( { - 5} \right)} \right\}}^2}}$

$\Rightarrow$$BC = \sqrt {{{\left( {2 + 5} \right)}^2} + {{\left( { - 3 + 5} \right)}^2}}$
$\Rightarrow$$BC = \sqrt {{{\left( 7 \right)}^2} + {{\left( 2 \right)}^2}}$
$\Rightarrow$$BC = \sqrt {49 + 4}$
$\Rightarrow$$BC = \sqrt {53}$ unit
$\Rightarrow$$CD = \sqrt {{{\left( {2 - 4} \right)}^2} + {{\left( { - 3 - 4} \right)}^2}}$

$\Rightarrow$$CD = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 7} \right)}^2}}$
$\Rightarrow$$CD = \sqrt {4 + 49}$
$\Rightarrow$$CD = \sqrt {53}$ unit
$DA = \sqrt {{{\left\{ {4 - \left( { - 3} \right)} \right\}}^2} + {{\left( {4 - 2} \right)}^2}}$

$DA = \sqrt {{{\left( {4 + 3} \right)}^2} + {{\left( 2 \right)}^2}}$
$DA = \sqrt {49 + 4}$
$DA = \sqrt {53}$ unit
AB = BC = CD = DA (all the sides are equal)
Diagonal $AC = \sqrt {{{\left( { - 3 - 2} \right)}^2} + {{\left\{ {2 - \left( { - 3} \right)} \right\}}^2}}$

$\Rightarrow$ $AC = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( {2 + 3} \right)}^2}}$
$\Rightarrow$ $AC = \sqrt {{{\left( 5 \right)}^2} + {{\left( 5 \right)}^2}}$
$\Rightarrow$ $AC = \sqrt {25 + 25}$
$\Rightarrow$ $AC = \sqrt {50}$ unit
Diagonal $BD = \sqrt {{{\left( { - 5 - 4} \right)}^2} + {{\left( { - 5 - 4} \right)}^2}}$

$\Rightarrow$$BD = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( { - 9} \right)}^2}}$
$\Rightarrow$ $BD = \sqrt {81 + 81}$
$\Rightarrow$ $BD = \sqrt {162}$ unit
So, diagonal $AC \ne BD$ Hence, A(–3, 2), B(–5,–5), C(2,–3) and D(4, 4) are the vertices of Rhombus.

Q.5            Point P divides the line segment joining the points A(2, 1) and B(5, –8) such that ${{AP} \over {AB}} = {1 \over 3}$. If P lies on the line 2x –y + k = 0, find the value of k.
[Delhi 2010]

Sol.            Let P(x,y) be the point which divides the line segment joining A(2,1) and B(5,-8).
${{AP} \over {AB}} = {1 \over 3}$ (given) AB = AP + PB
$\Rightarrow$ 3 = 1 + PB
$\Rightarrow$ PB = 3 – 1
$\Rightarrow$PB = 2
So, P divides AB into 1 : 2 ratio.
The co-ordinates of P(x, y) = $\left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
Put $m = 1,n = 2,{x_1} = 2,{y_1} = 1,{x_2} = 5{\mkern 1mu} and{\mkern 1mu} {y_2} = - 8$ in above formula –
$P(x,y) = \left( {{{1 \times 5 + 2 \times 2} \over {1 + 2}},{{1 \times \left( { - 8} \right) + 2\left( 1 \right)} \over {1 + 2}}} \right)$
$= \left( {{{5 + 4} \over 3},{{ - 6} \over 3}} \right)$
$= \left( {{9 \over 3}, - 2} \right)$
$= \left( {3, - 2} \right)$
$2x - y + k = 0$ ……….. (1)
Put the co-ordinates of P in equation (1) –

$\Rightarrow 2 \times 3 - \left( { - 2} \right) + k = 0$
$\Rightarrow 6 + 2 + k = 0$
$\Rightarrow 8 + k = 0$
$\Rightarrow k = - 8$
Hence, the value of k = –8.

Q.6          If R(x, y) is a point on the line segment joining the points P(a, b) and Q(b, a), then prove that x + y = a + b.
[Delhi 2010]

Sol.          If R(x, y) is a point on the line segment joining the points P(a, b) and Q(b, a). Then, points P, Q and R on a line. $\Rightarrow {x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0...........(1)$
Put $x_1 = a,y_1 = b,x_2 = x,\,y_2 = y,x_3 = b\,and\,y_3 = a\,$ in equation (1) –
$\Rightarrow a\left( {y - a} \right) + x\left( {a - b} \right) + b\left( {b - y} \right) = 0$
$\Rightarrow ay - {a^2} + xa - xb + {b^2} - by = 0$
$\Rightarrow ay + ax - xb - by = {a^2} - {b^2}$ $\Rightarrow \left( {a - b} \right)\left( {x + y} \right) = \left( {a + b} \right)\left( {a - b} \right)$
$\Rightarrow \left( {x + y} \right) = {{\left( {a + b} \right)\left( {a - b} \right)} \over {\left( {a - b} \right)}}$
$\Rightarrow \left( {x + y} \right) = \left( {a + b} \right)$
“Hence proved”.

Q.7        Point P divides the line segment joining the points A(–1, 3) and B(9, 8) such that ${{AP} \over {PB}} = {k \over 1}$. If P lies on the line x – y +2 = 0, find the value of k.
[AI 2010]

Sol.        Let point P(x, y) divides the line segment joining the points A(–1, 3) and B(9, 8)
${{AP} \over {PB}} = {K \over 1}$ (given)
So, P divides AB into K : 1 ratio –
$\Rightarrow$ $x = {{m{x_2} + n{x_1}} \over {m + n}}$ ……….. (1)
Put $x_1 = - 1,x_2 = 9,\,m = K\,and\,n = 1\,$ in equation (1) –
$\Rightarrow x = {{K \times 9 + 1 \times \left( { - 1} \right)} \over {K + 1}}$
$\Rightarrow$ $x = {{9K - 1} \over {K + 1}}$
$\Rightarrow$ $y = {{m{y_2} + n{y_1}} \over {m + n}}$ ………….(2)

Put $y_1 =3 ,y_2 = 8,\,m = K\,and\,n = 1\,$in equation (2) –
$\Rightarrow y = {{K \times 8 + 1 \times 3} \over {K + 1}}$
$\Rightarrow y = {{8K + 3} \over {K + 1}}$
According to the question –

P lies on the line x – y + 2 = 0 ……… (3)
So, put the value of x and y in equation (3) –
$\Rightarrow {{9K - 1} \over {K + 1}} - \left( {{{8K + 3} \over {K + 1}}} \right) + 2 = 0$
$\Rightarrow {{9K - 1 - 8k - 3 + 2k + 2} \over {\left( {K + 1} \right)}} = 0$
$\Rightarrow 11K - 8K - 4 + 2 = 0$
$\Rightarrow 3K - 2 = 0$
$\Rightarrow 3K = 2$
$\Rightarrow K = {2 \over 3}$
Hence, the value of $K = {2 \over 3}$.

Q.8        If the points (p, q), (m, n) and (p – m, q – n) are collinear, show that pn = qm.
[AI 2010]

Sol.       To prove : - pn = qn
Let P(p, q), Q(m, n) and R(p–m, q–n) be the given points, which are collinear. So, area of $\Delta PQR$ = 0 Area of $\Delta PQR$ = 0
$\Rightarrow {1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}({y_1} - {y_2}} \right| = 0$
$\Rightarrow \left| {p\left\{ {n - \left( {q - n} \right)} \right\} + m\left\{ {\left( {q - n} \right) - q} \right\} + \left( {p - m} \right)\left( {q - n} \right)} \right| = 0$
$\Rightarrow \left| {p\left( {n - q + n} \right) + m\left( {q - n - q} \right) + pq - pn - mq + mn} \right| = 0$
$\Rightarrow \left| {pn - pq + pn + mq - mn - qm + pq - pn - mq + mn} \right| = 0$
$\Rightarrow \left| {2pn - pn - pq + pq + mq - 2mq - mn + mn} \right| = 0$
$\Rightarrow \left| {pn - mq} \right| = 0$
$\Rightarrow pn = mq$
“Hence proved”

Q.9         If point $\left( {{1 \over 2},y} \right)$ lies on the line segment joining the points A(3, –5) and B(–7, 9), then find the ratio in which P divides AB. Also find the value of y.
[Foreign 2010]

Sol.          Let point P(½, y) lies on the line segment joining the points A(3, –5) and B(–7, 9). Let P divides AB into m : n ratio.
$x = {{m{x_2} + n{x_1}} \over {m + n}}..........(1)$
Put $x = 1/2,x_1 = 3\,and\,x_2 =- 7$ in equation (1) –
$\Rightarrow {1 \over 2} = {{m \times \left( { - 7} \right) + n\left( 3 \right)} \over {m + n}}$
$\Rightarrow m + n = - 14m + 6n$
$\Rightarrow m + 14m = 6n - n$
$\Rightarrow 15m = 5n$
$\Rightarrow {m \over n} = {5 \over {15}}$
$\Rightarrow {m \over n} = {1 \over 3}$
$\Rightarrow$$y = {{m{y_2} + n{y_1}} \over {m + n}}.........(2)$

Put ${y_1} = - 5,m = 1,n = 3{\mkern 1mu} and{\mkern 1mu} {y_2} = 9$ in equation (2) –
$\Rightarrow$ $y = {{1 \times 9 + 3\left( { - 5} \right)} \over {1 + 3}}$
$\Rightarrow$ $y = {{9 - 15} \over 4}$
$\Rightarrow$ $y = {{ - 6} \over 4}$
$\Rightarrow$ $y = {{ - 3} \over 2}$
Hence, p divides line segment AB into 1 : 3 ratio, and the value of y = ${{ - 3} \over 2}$.

Q.10       Find the coordinates of the points which divide the line segment joining A(2, –3) and B(–4, –6) into three equal parts.
OR
Show that the points A(3, 5), B(6, 0), C(1, –3) and D(–2, 2) are the vertices of a square ABCD.
[Foreign 2011]

Sol.          Let point P, Q be the points which divide the line segment joining A(2, –3)
and B(–4, –6) into three equal parts. AP = PQ = QB = K (let)
Now, we take point P which divides AB into m : n ratio. ${{AP} \over {PB}} = {{AP} \over {PQ + QB}}$
$= {K \over {K + K}}$
$= {K \over {2K}}$
$= {1 \over 2}$
Means, p divide AB into 1 : 2 ratio. So, the coordinates of p are –

$\Rightarrow$$p = \left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
$\Rightarrow$$p = \left( {{{1 \times \left( { - 4} \right) + 2\left( 2 \right)} \over {1 + 2}},{{1 \times \left( { - 6} \right) + 2\left( { - 3} \right)} \over {1 + 2}}} \right)$
$\Rightarrow$$p = \left( {{{ - 4 + 4} \over 3},{{ - 6 - 6} \over 3}} \right)$
$\Rightarrow$$p = \left( {0,{{ - 12} \over 3}} \right)$
$\Rightarrow$$p = \left( {0, - 4} \right)$
Now, we take point Q which is also divide AB into m : n ratio. –
${{AQ} \over {QB}} = {{AP + PQ} \over {QB}}$
$= {{K + K} \over K}$
$= {{2K} \over K}$
$= {2 \over 1}$
Means, Q divides AB into 2 : 1 ratio, so, the co-ordinates of Q are –
$\Rightarrow$$Q = \left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
$\Rightarrow Q = \left( {{{2 \times \left( { - 4} \right) + 1\left( 2 \right)} \over {2 + 1}},{{2 \times \left( { - 6} \right) + 1\left( { - 3} \right)} \over {2 + 1}}} \right)$
$\Rightarrow$$Q = \left( {{{ - 8 + 2} \over 3},{{ - 12 - 3} \over 3}} \right)$
$\Rightarrow$$Q = \left( {{{ - 6} \over 3},{{ - 15} \over 3}} \right)$
$\Rightarrow$$Q = \left( { - 2, - 5} \right)$
Hence, the co-ordinates of P and Q are (0, –4) and (–2, –5) respectively.
OR
Point A(3, 5), B(6, 0), C(1,–3) and D(–2, 2) are the given vertices of a quadrilateral ABCD. By using distance formula –
$AB = \sqrt {{{\left( {3 - 6} \right)}^2} + {{\left( {5 - 0} \right)}^2}}$
$\Rightarrow$$AB = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 5 \right)}^2}}$
$\Rightarrow$$AB = \sqrt {9 + 25}$
$\Rightarrow$$AB = \sqrt {34}$ unit
$\Rightarrow$$BC = \sqrt {{{\left( {6 - 1} \right)}^2} + {{\left\{ {0 - \left( { - 3} \right)} \right\}}^2}}$

$\Rightarrow$$BC = \sqrt {{{\left( 5 \right)}^2} + {{\left( 3 \right)}^2}}$
$\Rightarrow$$BC = \sqrt {25 + 9}$
$\Rightarrow$$BC = \sqrt {34}$ unit
$\Rightarrow$$CD = \sqrt {{{\left\{ {1 - \left( { - 2} \right)} \right\}}^2} + {{\left( { - 3 - 2} \right)}^2}}$

$\Rightarrow$$CD = \sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( { - 5} \right)}^2}}$
$\Rightarrow$$CD = \sqrt {{{\left( 3 \right)}^2} + {{\left( { -5} \right)}^2}}$
$\Rightarrow$$CD = \sqrt {9 + 25}$
$\Rightarrow$$CD = \sqrt {34}$ unit
$\Rightarrow DA = \sqrt {{{\left( { - 2 - 3} \right)}^2} + {{\left( {2 - 5} \right)}^2}}$

$\Rightarrow$$DA = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( { - 3} \right)}^2}}$
$\Rightarrow$$DA = \sqrt {25 + 9}$
$\Rightarrow$$DA = \sqrt {34}$ unit
So, AB = BC = CD = DA (all sides are equal)
Now diagonal AC and BD –

$\Rightarrow$$AC = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left\{ {5 - \left( { - 3} \right)} \right\}}^2}}$
$\Rightarrow$$AC = \sqrt {{{\left( 2 \right)}^2} + {{\left( {5 + 3} \right)}^2}}$
$\Rightarrow$$AC = \sqrt {4 + {{\left( 8 \right)}^2}}$
$\Rightarrow$$AC = \sqrt {4 + 64}$
$\Rightarrow$$AC = \sqrt {68}$ unit
$\Rightarrow$$BD = \sqrt {{{\left\{ {6 - \left( { - 2} \right)} \right\}}^2} + {{\left( {0 - 2} \right)}^2}}$
$\Rightarrow$$BD = \sqrt {{{\left( {6 + 2} \right)}^2} + {{\left( { - 2} \right)}^2}}$
$\Rightarrow$$BD = \sqrt {{{\left( 8 \right)}^2} + {{\left( { - 2} \right)}^2}}$
$\Rightarrow$$BD = \sqrt {64 + 4}$
$\Rightarrow$$BD = \sqrt {68}$ unit
Diagonal AC = BD

So, we can say that A(3, 5), B(6, 0), C(1, –3) and D(–2, 2) are the vertices of a square ABCD.

Q.11          For what value of k, (k > 0), is the area of the triangle with vertices (–2, 5), (k, –4) and (2k +1, 10) equal to 53 sq. units ?
[Foreign 2012]

Sol.            Let A(–2, 5), B(k, –4) and C(2k + 1, 10) be the given vertices of a $\Delta ABC$
Area of $\Delta ABC$ = 53 square units Area of $\Delta ABC$ = 53 sq. units
$\Rightarrow {1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right| = 53$
$\Rightarrow \left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right| = 53 \times 2.........(1)$
Put $x_1 =- 2,y_1 = 5,x_2 = k,y_2 =- 4,x_3 = 2k + 1\,and\,y_3 = 10$ in equation (2) –
$\Rightarrow \left| {\left( { - 2} \right)\left( { - 4 - 10} \right) + k\left( {10 - 5} \right) + \left( {2k + 1} \right)\left\{ {5 - \left( { - 4} \right)} \right\}} \right| = 106$
$\Rightarrow \left| {\left( { - 2} \right)\left( { - 14} \right) + k\left( 5 \right) + \left( {2k + 1} \right)\left( 9 \right)} \right| = 106$
$\Rightarrow \left| {28 + 5k + 18k + 9} \right| = 106$
$\Rightarrow \left| {37 + 23k} \right| = 106$
$\Rightarrow 37 + 23k = 106$ and $37 + 23k = -106$
$\Rightarrow 23k = 106 - 37$ and $\Rightarrow 23k = - 106 - 37$
$\Rightarrow k = {{69} \over {23}}$ and $\Rightarrow 23k = - 143$
$\Rightarrow k = 3$ and $\Rightarrow k = {{ - 143} \over {23}}$
k > 0 (given)
so, the value of k = 3

Q.12           The line segment AB joining the points A(3, –4) and B(1, 2) is trisected at the points P(p,–2) and Q(5/3, q). Find the values of p and q.
[Foreign 2013]

Sol.             Given that line segment joining the points A(3, –4) and B(1, 2) is trisected at the points P(p, –2)
and Q $\left( {{5 \over 3},q} \right)$. AP = PQ = QB (given)
P is mid-point of AQ so, $p = {{3 + {5 \over 3}} \over 2}$
$\Rightarrow 2p = {{9 + 5} \over 3}$
$\Rightarrow 6p = 14$
$\Rightarrow p = {7 \over 3}$
$- 2 = {{ - 4 + q} \over 2}$
$\Rightarrow - 2 \times 2 =- 4 + q$
$\Rightarrow - 4 =- 4 + q$
$\Rightarrow - 4 + 4 = q$
$\Rightarrow - 4 + 4 = q$
$\Rightarrow$ $q = 0$
Hence, the value of $p = {7 \over 3}$ and q = 0.

Long answer type questions : -

Q.1           The three vertices of parallelogram ABCD are A(3, –4), B(–1, –3) and C(–6, 2). Find the coordinates of vertex D and find the area of ABCD.
[Delhi 2013]
Sol.           A(3, –4), B(–1, –3), C(–6,2) and D are the given points of parallelogram ABCD.
Let D(x,y) be the fourth vertex of parallelogram ABCD.

${{x_1 + x_3 } \over 2} = {{x_2 + x_4 } \over 2}$ .........................(1)
put $x_1 = 3,x_2 =- 1,x_3 =- 6\,and\,x_4 = x$ in equation (1)
${{3 + ( - 6)} \over 2} = {{ - 1 + x} \over 2}$
$\Rightarrow 3 - 6 = - 1 + x$
$\Rightarrow - 3 + 1 = x$
$\Rightarrow x = - 2$
${{y_1 + y_3 } \over 2} = {{y_2 + y_4 } \over 2}$..........................(2)

${{ - 4 + 2} \over 2} = {{ - 3 + y} \over 2}$
$\Rightarrow - 2 = - 3 + y$
$\Rightarrow - 2 + 3 = y$
$\Rightarrow y = 1$
So, the vertex D(–2, 1).
Area of parallelogram ABCD = 2 × Area of $\Delta$ABC

Area of $\Delta$ABC = ${1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$................................(3)
Put values of $x_1 ,x_2 ,x_3 ,y_1 ,y_2 \,and\,y_3$ in equation (3) -
= ${1 \over 2}\left| {3\left( { - 3 - 2} \right) + \left( { - 1} \right)\left\{ {2 - \left( { - 4} \right)} \right\} + \left( { - 6} \right)\left\{ { - 4 - \left( { - 3} \right)} \right\}} \right|$
$= {1 \over 2}\left| {3\left( {- 5} \right) + - 1\left( {2 + 4} \right) + \left( { - 6} \right)\left( { - 4 + 3} \right)} \right|$
$= {1 \over 2}\left| { - 15 - 1 \times 6 - 6\left( { - 1} \right)} \right|$
$= {1 \over 2}\left| { - 15 - 6 + 6} \right|$
$= {{15} \over 2}$ sq. units
Area of parallelogram ABCD = $\left( {2 \times {{15} \over 2}} \right)$ sq. units

= 15 sq. units
Hence, the fourth vertex of parallelogram is D(–2, 1) and area of parallelogram ABCD = 15 Sq. units.

Q.2           If the points A(1, –2), B(2, 3), C(–3, 2) and D(–4, –3) are the vertices of parallelogram ABCD, then taking AB as the base, find the height of the parallelogram.
[AI 2013]

Sol. Points A(1, –2), B(2, 3), C(–3, 2) and D(–4, –3) are the given points of
parallelogram ABCD where AB is base
Let h be the height of parallelogram ABCD
In $\Delta ABC$
$\Rightarrow$ $AB = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( { - 2 - 3} \right)}^2}}$ ( Using distance formula )
$\Rightarrow$ $AB = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2}}$
$\Rightarrow$ $AB = \sqrt {1 + 25}$
$\Rightarrow$ $AB = \sqrt {26}$ Units
Area of $\Delta ABC$
$= {1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$
$= {1 \over 2}\left| {1\left( {3 - 2} \right) + 2\left\{ {2 - \left( { - 2} \right)} \right\} + \left( { - 3} \right)\left( { - 2 - 3} \right)} \right|$
$= {1 \over 2}\left| {1\left( 1 \right) + 2\left( 4 \right) - 3\left( { - 5} \right)} \right|$
$= {1 \over 2}\left| {1 + 8 + 15} \right|$
$= {1 \over 2} \times 24$
$= 12$ sq. units
Area of ||gm ABCD = Base x Height

$\Rightarrow \,12 = AB \times h$
$\Rightarrow \,12 = \sqrt {26}\times h$
$\Rightarrow \,h = {{12} \over {\sqrt {26} }}$
$\Rightarrow {\rm{h = }}{{{\rm{12}}} \over {\sqrt {26} }} \times {{\sqrt {26} } \over {\sqrt {26} }}$
$\Rightarrow {\mkern 1mu} h = {{12} \over {26}} \times \sqrt {26}$
$\Rightarrow {\mkern 1mu} h = {6 \over {13}}\sqrt {26}$ units
Hence , the height of parallelogram ABCD is ${6 \over {13}}\sqrt {26}$ units.

Q.3           If the area of the triangle formed by points A(x, y), B(1, 2) and C(2, 1) is 6 square units, then show that x + y = 15.

[Foreign 2013].
Sol.           Given that A(x, y), B(1, 2) and C(2, 1) are the given points of $\Delta ABC$.

To prove – x + y = 15
Area of $\Delta ABC$ = 6 sq. units (given)
Area of $\Delta ABC = {1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {y3 - {y_1}} \right) + {x_3}\left( {y - {y_2}} \right)} \right|$.......................(1)
Put $x_1 = x,x_2 = 1,x_3 = 2,y_1 = y,y_2 = 2\,and\,y_3 = 1$ in (1)
$\Rightarrow 6 = {1 \over 2}\left| {x\left( {2 - 1} \right) + 1\left( {1 - y} \right) + 2\left( {y - 2} \right)} \right|$
$\Rightarrow 6 \times 2 = \left| {x\left( 1 \right) + 1 - y + 2y - 4} \right|$
$\Rightarrow 12 = \left| {x + 1 - y + 2y - 4} \right|$
$\Rightarrow 12 = \left| {x - 3 + y} \right|$
$\Rightarrow 12 + 3 = x + y$
$\Rightarrow 15 = x + y$
$\Rightarrow x + y = 15$
“Hence proved”

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