# Coordinate Geometry : Exercise - 7.4 (Mathematics NCERT Class 10th)

Q.1     Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the
points A(2, –2) and B(3, 7).

Sol.       Suppose the line 2x + y – 4 = 0 divides the line segment joining A(2, –2) and B(3, 7) in the ratio k : 1 at
point C. Then, the coordinates of C are $\left( {{{3k + 2} \over {k + 1}},{{7k - 2} \over {k + 1}}} \right)$

But C lies on 2x + y – 4 = 0 therefore,
$2\left( {{{3k + 2} \over {k + 1}}} \right) + \left( {{{7k - 2} \over {k + 1}}} \right) - 4 = 0$
$\Rightarrow 6k + 4 + 7k - 2 - 4k - 4 = 0$
$\Rightarrow$ 9k – 2 = 0
$\Rightarrow$ 9k = 2         $\Rightarrow$ $k = {2 \over 9}$
So, the required ratio is 2 : 9 internally.

Q.2      Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Sol.        The points A(x, y), B(1, 2) and C(7, 0) will be are collinear.
x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
$\Rightarrow$ 2x – y + 7y – 14 = 0
$\Rightarrow$ 2x + 6y – 14 = 0
$\Rightarrow$ x + 3y – 7 = 0
which is the relation between x and y.

Q.3      Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).
Sol.        Let P(x, y) be the centre of the circle passing through the points A(6, –6) B(3, –7) and C(3, 3).
Then, AP = BP = CP.

Now, AP = BP
$\Rightarrow A{P^2} = B{P^2}$

$\Rightarrow {\left( {x - 6} \right)^2} + {\left( {y + 6} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y + 7} \right)^2}$
$\Rightarrow {x^2} - 12x + 36 + {y^2} + 12y + 36$  $= {x^2} - 6x + 9 + {y^2} + 14y + 49$
$\Rightarrow - 12x + 6x + 12y - 14y + 72 - 58$ = 0
$\Rightarrow - 6x - 2y + 14 = 0$
$\Rightarrow 3x + y - 7 = 0$                                ....(1)
and, BP = CP
$\Rightarrow B{P^2} = C{P^2}$
$\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y + 7} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y - 3} \right)^2}$
$\Rightarrow {x^2} - 6x + 9 + {y^2} + 14y + 49$  $= {x^2} - 6x + 9 + {y^2} - 6y + 9$
$\Rightarrow - 6x + 6x + 14y + 6y + 58 - 18$ = 0
$\Rightarrow 20y + 40 = 0$
$\Rightarrow y = {{ - 40} \over {20}} = - 2$         ...(2)
Putting y = – 2 in (1), we get
3x – 2 – 7 = 0
$\Rightarrow$ 3x = 9
$\Rightarrow$ x = 3
Thus, the centre of the circle is (3, –2).

Q.4     The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
Sol.     Let ABCD be a square and let A(–1, 2) and C(3, 2) be the given angular points. Let B(x, y) be the unknown vertex.

Then,    AB = BC
$\Rightarrow A{B^2} = B{C^2}$
$\Rightarrow {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2}$
$\Rightarrow {x^2} + 2x + 1 + {y^2} - 4y + 4 = {x^2} - 6x + 9 + {y^2} - 4y + 4$
$\Rightarrow 2x + 1 = - 6x + 9$
$\Rightarrow$ 8x = 8
$\Rightarrow$ x = 1            ...(1)
In $\Delta$ ABC, we have
$A{B^2} + B{C^2} = A{C^2}$
$\Rightarrow$ ${\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2}+{\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {3 + 1} \right)^2} + {\left( {2 - 2} \right)^2}$
$\Rightarrow 2{x^2} + 2{y^2} + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16$
$\Rightarrow 2{x^2} + 2{y^2} - 4x - 8y + 2 = 0$
$\Rightarrow {x^2} + {y^2} - 2x - 4y + 1 = 0$           ...(2)
Substituting the value of x from (1) into (2), we get
$1 + {y^2} - 2 - 4y + 1 = 0$
$\Rightarrow {y^2} - 4y = 0 \Rightarrow y\left( {y - 4} \right) = 0$
$\Rightarrow y = 0$ or 4
Hence, the required vertices of the square are (1, 0) and (1, 4).

Q.5     The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their  gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of $\Delta PQR$ if C is the origin ? Also calculate the area of the triangle in these cases. What do you observe ?
Sol.      (i) Taking A as the origin, AD and AB as the coordinate axes. Clearly, the points P,Q and R are (4, 6),
(3, 2) and (6, 5) respectively.

(ii) Taking C as the origin, CB and CD as the coordinate axes. Clearly, the points P, Q and R are
given by (12, 2), (13, 6) and
(10, 3) respectively,
We know that the area of the triangle whose vertices are $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ is given by
= ${1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
Therefore, area of $\Delta$ PQR in the 1st case
${1 \over 2}\left[ {4\left( {2 - 5} \right) + 3\left( {5 - 6} \right) + 6\left( {6 - 2} \right)} \right]$
$= {1 \over 2}\left( {4 \times - 3 + 3 \times - 1 + 6 \times 4} \right)$
$= {1 \over 2}\left( { - 12 - 3 + 24} \right) = {9 \over 2}$ sq. units
and, area of $\Delta PQR$ in the 2nd case
$= {1 \over 2}\left[ {12\left( {6 - 3} \right) + 13\left( {3 - 2} \right) + 10\left( {2 - 6} \right)} \right]$
$= {1 \over 2}\left( {12 \times 3 + 13 \times 1 + 10 \times - 4} \right)$
$= {1 \over 2}\left( {36 + 13 - 40} \right) = {9 \over 2}$ sq. units
Thus, we observe that the areas are the same in both the cases.

Q.6      The vertices of a $\Delta ABC$ are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that ${{AD} \over {AB}} = {{AE} \over {AC}} = {1 \over 4}$. Calculate the area of $\Delta ADE$ and compare it with the area of $\Delta ABC$.
Sol.       We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to
the third side.

Therefore, DE || BC
Clearly, $\Delta ADE \sim \Delta ABC$ (AAA similarity)
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any
two corresponding sides.

Therefore, ${{Area\left( {\Delta ADE} \right)} \over {Area\left( {\Delta ABC} \right)}} = {{A{D^2}} \over {A{B^2}}}$
$= {\left( {{{AD} \over {AB}}} \right)^2} = {\left( {{1 \over 4}} \right)^2} = {1 \over {16}}$          ...(1)
Now, Area $\left( {\Delta ABC} \right)$  $= {1 \over 2}\left[ {4\left( {5 - 2} \right) + 1\left( {2 - 6} \right) + 7\left( {6 - 5} \right)} \right]$
$= {1 \over 2}\left( {12 - 4 + 7} \right)$
$= {{15} \over 2}$ sq. units            ...(2)
From (1) and (2),
Area $\left( {\Delta ADE} \right) = {1 \over 16} \times Area\left( {\Delta ABC} \right)$
= $\left( {{1 \over {16}} \times {{15} \over 2}} \right)$ sq. units
${ = {{15} \over {32}}}$ sq. units
Also, form (1), area $\left( {\Delta ADE} \right)$ : area $\left( {\Delta ABC} \right)$ = 1 : 16.
ALITER : (Using only Coordinate Geometry)
Since ${{AD} \over {AB}} = {1 \over 4}$
$\Rightarrow {{AB} \over {AD}} = {4 \over 1} \Rightarrow {{AD + DB} \over {AD}} = {{1 + 3} \over 1}$
$\Rightarrow 1 + {{DB} \over {AD}} = 1 + {3 \over 1}$
$\Rightarrow {{DB} \over {AD}} = {3 \over 1} \Rightarrow {{AD} \over {DB}} = {1 \over 3}$
Since D divides AB in the ratio 1 : 3
Therefore, coordinates of D are
$\left( {{{1 \times 1 + 3 \times 4} \over {1 + 3}},{{1 \times 5 + 3 \times 6} \over 4}} \right)$, i.e., $\left( {{{13} \over 4},{{23} \over 4}} \right)$
From ${{AE} \over {AC}} = {1 \over 4}$, we find that
${{AE} \over {EC}} = {1 \over 3}$
Since E divides AC in the ratio 1 : 3

Therefore, coordinates of E are

$\left( {{{1 \times 7 + 3 \times 4} \over {1 + 3}},{{1 \times 2 + 3 \times 6} \over {1 + 3}}} \right)$,i.e., $\left( {{{19} \over 4},5} \right)$
Now, Area $\left( {\Delta ADE} \right)$
= $\left[ {4\left( {{{23} \over 4} - 5} \right) + {{13} \over 4}\left( {5 - 6} \right) + {{19} \over 4}\left( {6 - {{23} \over 4}} \right)} \right]$
$= {1 \over 2}\left[ {4 \times {3 \over 4} - {{13} \over 4} + {{19} \over 4} \times {1 \over 4}} \right]$

$= {1 \over {32}}[48 - 52 + 19]$
$= {{15} \over {32}}$ sq. units
Also find the area $\left( {\Delta ABC} \right)$
Area ${\Delta ABC = {{15} \over 2}}$ sq. units
Area $\left( {\Delta ADE} \right)$ : Area (${\Delta ABC}$)
$= {{15} \over {32}}:{{15} \over 2} = {1 \over {32}}:{1 \over 2} = 1:16$

Q.7     Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ${\Delta ABC}$.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that
BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe ?
(Note : The point which is common to all the three medians is called centroid and this point divides each median in the ratio 2 : 1).
(v) If $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$ are the vertices of ${\Delta ABC}$, find the coordinates of the centroid of the triangle.
Sol.      Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ${\Delta ABC}$.

(i) Since AD is the median of ${\Delta ABC}$, Therefore, D is the mid-point of BC.
Its coordinates are $\left( {{{6 + 1} \over 2},{{5 + 4} \over 2}} \right)$,i.e., $\left( {{7 \over 2},{9 \over 2}} \right)$.
(ii) Since P divides AD in the ratio 2 : 1 , so its coordinates are $P\left( {{{2 \times {7 \over 2} + 1 \times 4} \over {2 + 1}},{{2 \times {9 \over 2} + 1 \times 2} \over {2 + 1}}} \right)$
or $P\left( {{{7 + 4} \over 3},{{9 + 2} \over 3}} \right)$  i.e., $P\left( {{{11} \over 3},{{11} \over 3}} \right)$

(iii) Since BE is the median of ${\Delta ABC}$, so E is the mid-point of AC and its coordinates are $E\left( {{{4 + 1} \over 2},{{2 + 4} \over 2}} \right)$  i.e., $E\left( {{5 \over 2},3} \right)$

Since Q divides BE in the ratio 2 : 1, so its coordinates are
$Q\left( {{{2 \times {5 \over 2} + 1 \times 6} \over {2 + 1}},{{2 \times 3 + 1 \times 5} \over {2 + 1}}} \right)$ or $Q\left( {{{5 + 6} \over 3},{{6 + 5} \over 3}} \right)$, i.e., $Q\left( {{{11} \over 3},{{11} \over 3}} \right)$

Since CF is the median of ${\Delta ABC}$, so F is the mid-point of AB. Therefore, its coordinates are
$F\left( {{{4 + 6} \over 2},{{2 + 5} \over 2}} \right)$
i.e., $F\left( {5,{7 \over 2}} \right)$.
Since R divides CF in the ratio 2 : 1, so its coordinates are
$R\left( {{{2 \times 5 + 1 \times 1} \over {2 + 1}},{{2 \times {7 \over 2} + 1 \times 4} \over {2 + 1}}} \right)$ or $R\left( {{{10 + 1} \over 3},{{7 + 4} \over 3}} \right)$ or $R\left( {{{11} \over 3},{{11} \over 3}} \right)$

(iv) We observe that the points P,Q and R coincide, the medians AD, BE and CF are concurrent at the point
$\left( {{{11} \over 3},{{11} \over 3}} \right)$
This point is known as the centroid of the triangle.

(v) Let $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$ be the vertices of ${\Delta ABC}$ whose medians are AD, BE, and CF respectively. So, D,E and F are respectively the mid-points of BC,CA and AB.

Coordinates of D are $\left( {{{{x_2} + {x_3}} \over 2},{{{y_2} + {y_3}} \over 2}} \right)$
Coordinates of a point dividing AD in the ratio 2 : 1 are
$\left( {{{1.{x_1} + 2\left( {{{{x_2} + {x_3}} \over 2}} \right)} \over {1 + 2}}{{1.{y_1} + 2\left( {{{{y_2} + {y_3}} \over 2}} \right)} \over {1 + 2}}} \right)$ $= \left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$

The coordinates of E are $\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)$.The coordinates of a point dividing BE in the ratio 2 : 1 are
$\left( {{{1.{x_2} + 2\left( {{{{x_1} + {x_3}} \over 2}} \right)} \over {1 + 2}}, 1.{y_2} + 2\left( {{{{y_1} + {y_3}} \over 2}} \right)} \right)$
$= \left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$
Similarly, the coordinates of a point dividing CF in the ratio 2 : 1 are
$\left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$
Thus, the point $\left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$ is common to AD,BE and CF and divides them in the ratio 2 : 1
Therefore, the medians of a triangle are concurrent and the coordinates of the centroid are
$\left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$

Q.8     ABCD is a rectangle formed by joining the points A(– 1,–1), B(– 1, 4), C(5, 4) and D(5,–1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle ? or a rhombus ? Justify your answer.
Sol.      Various points are marked in the adjoining figure.

$PQ = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {4 - {3 \over 2}} \right)}^2}}$ = $\sqrt {{3^2} + {{\left( {{5 \over 2}} \right)}^2}}$
= $\sqrt {9 + {{25} \over 4}}$ $= \sqrt {{{36 + 25} \over 4}} = \sqrt {{{61} \over 4}}$
$QR = \sqrt {{{\left( {5 - 2} \right)}^2} + {{\left( {{3 \over 2} - 4} \right)}^2}}$
= $\sqrt {{3^2} + {{\left( {{{ - 5} \over 2}} \right)}^2}} = \sqrt {9 + {{25} \over 4}}$
$= \sqrt {{{36 + 25} \over 4}} = \sqrt {{{61} \over 4}}$
$RS = \sqrt {{{\left( {2 - 5} \right)}^2} + {{\left( { - 1 - {3 \over 2}} \right)}^2}}$
= $\sqrt {\left( { - {3^2}} \right) + {{\left( {{{ - 5} \over 2}} \right)}^2}}$
$= \sqrt {9 + {{25} \over 4}} = \sqrt {{{36 + 25} \over 4}} = \sqrt {{{61} \over 4}}$
and, $SP = \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( {{3 \over 2} + 1} \right)}^2}}$
$= \sqrt {{{\left( { - 3} \right)}^2} + {{\left( {{5 \over 2}} \right)}^2}}$
$= \sqrt {9 + {{25} \over 4}} = \sqrt {{{36 + 25} \over 4}} = \sqrt {{{61} \over 4}}$
$\Rightarrow$ PQ = QR = RS = SP
Now, $PR = \sqrt {{{\left( {5 + 1} \right)}^2} + {{\left( {{3 \over 2} - {3 \over 2}} \right)}^2}}$ = $\sqrt {36} = 6$
and $SQ = \sqrt {{{\left( {2 - 2} \right)}^2} + {{\left( {4 + 1} \right)}^2}}$  $= \sqrt {25} = 5$
$\Rightarrow PR \ne SQ$
Since all the sides are equal but the diagonals are not equal,
therefore, PQRS is a rhombus.

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