Coordinate Geometry : Exercise - 7.1 (Mathematics NCERT Class 10th)

Q.1     Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)       (ii) (–5, 7), (–1, 3)       (iii) (a, b),(–a,–b)
Sol.      (i) Let P(2, 3) and Q (4, 1) be the given points.
Here, ${x_1} = 2,{y_1} = 3$ and ${x_2} = 4,{y_2} = 1$
Therefore, $PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

$\Rightarrow PQ = \sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {1 - 3} \right)}^2}}$ $= \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}}$
$\Rightarrow PQ = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2$

(ii) Let P(– 5, 7) and Q(–1, 3) be the given points.
Here ${x_1} = - 5,{y_1} = 7$ and ${x_2} = - 1,{y_2} = 3$
Therefore, $PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
$\Rightarrow PQ = \sqrt {{{\left( { - 1 + 5} \right)}^2} + {{\left( {3 - 7} \right)}^2}}$ $= \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}}$
$\Rightarrow PQ = \sqrt {16 + 16} = \sqrt {32} = \sqrt {16 \times 2} = 4\sqrt 2$

(iii) Let P(a, b) and Q(–a, –b) be the given points.
Here, ${x_1} = a,{y_1} = b$ and ${x_2} = - a,{y_2} = - b$
Therefore, $PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
$\Rightarrow PQ = \sqrt {{{\left( { - a - a} \right)}^2} + {{\left( {- b - b} \right)}^2}}$ $= \sqrt {{{\left( { - 2a} \right)}^2} + {{\left( { - 2b} \right)}^2}}$
$\Rightarrow PQ = \sqrt {4{a^2} + 4{b^2}} = 2\sqrt {{a^2} + {b^2}}$

Q.2     Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
Sol.     Let P(0, 0) and Q(36, 15) be the given points.
Here ${x_1} = 0,{y_1} = 0\,$ and ${x_2} = 36,{y_2} = 15$
Therefore, $PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
$\Rightarrow PQ = \sqrt {{{\left( {36 - 0} \right)}^2} + {{\left( {15 - 0} \right)}^2}}$ $= \sqrt {1296 + 225} = \sqrt {1521} = 39$
In fact, the positions of towns A and B are given by (0, 0) and (36, 15) respectively and so the distance between them = 39 km as calculated above.

Q.3     Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Sol.       Let A (1, 5), B(2, 3) and C(–2, –11) be the given points. Then, we have
$AB = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {3 - 5} \right)}^2}}$$= \sqrt {{1^2} + {{\left( { - 2} \right)}^2}}$ $= \sqrt {1 + 4} = \sqrt 5$
$BC = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( { - 11 - 3} \right)}^2}}$
$= \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 14} \right)}^2}}$ $= \sqrt {16 + 196}$
$= \sqrt {212} = \sqrt {4 \times 53} = 2\sqrt {53}$
and, $AC = \sqrt {{{\left( { - 2 - 1} \right)}^2} + {{\left( { - 11 - 5} \right)}^2}}$
$= \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 16} \right)}^2}}$
$= \sqrt {9 + 256} = \sqrt {265}$

Clearly, $BC \ne AB + AC,AB \ne BC + AC$ and  $AC \ne BC$
Hence, A,B and C are not collinear.

Q.4     Check whether (5, –2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Sol.      Let A(5, –2), B(6, 4) and C(7, –2) are the given point. Then,
$AB = \sqrt {{{\left( {6 - 5} \right)}^2} + {{\left( {4 + 2} \right)}^2}}$
$= \sqrt {1 + 36} = \sqrt {37}$
$BC = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( {-2 - 4} \right)}^2}}$
$= \sqrt {1 + 36} = \sqrt {37}$
Clearly, AB = BC
Therefore, $\Delta ABC$ is an isosceles triangle.

Q.5     In a classroom, four friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square ?" Chameli disagrees. Using distance formula, find which of  them is correct, and why? Sol.
Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).

By using distance formula, we get
$AB = \sqrt {{{\left( {6 - 3} \right)}^2} + {{\left( {7 - 4} \right)}^2}}$
$= \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$
$BC = \sqrt {{{\left( {9 - 6} \right)}^2} + {{\left( {4 - 7} \right)}^2}}$
$= \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$
$CD = \sqrt {{{\left( {6 - 9} \right)}^2} + {{\left( {1 - 4} \right)}^2}}$
$= \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$

and, $DA = \sqrt {{{\left( {3 - 6} \right)}^2} + {{\left( {4 - 1} \right)}^2}}$
$= \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$
$\Rightarrow AB = BC = CD = DA = 3\sqrt 2$
Also, $AC = \sqrt {{{\left( {9 - 3} \right)}^2} + {{\left( {4 - 4} \right)}^2}}$
$= \sqrt {36 + 0} = \sqrt {36} = 6$
and, $BD = \sqrt {{{\left( {6 - 6} \right)}^2} + {{\left( {1 - 7} \right)}^2}}$
$= \sqrt {0 + 36} = \sqrt {36} = 6$
$\Rightarrow AC = BD = 6$
Thus, the four sides are equal and the diagonals are also equal. Therefore, ABCD is a square.
Hence, Champa is correct.

Q.6     Name the type of quadrilateral formed, if any, the following points, and give reasons for your answer:
(i) (–1,–2), (1, 0), (–1, 2), (–3,0)     (ii) (–3, 5), (3, 1), (0, 3), (–1,– 4)     (iii) (4, 5), (7, 6), (4, 3), (1, 2)
Sol.     (i) Let A(–1,–2), B(1, 0), C(–1, 2) and D(–3, 0) be the given points. Then
$AB = \sqrt {{{\left( {1 + 1} \right)}^2} + {{\left( {0 + 2} \right)}^2}} = \sqrt {4 + 4} = \sqrt 8$
$BC = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {2 - 0} \right)}^2}} = \sqrt {4 + 4} = \sqrt 8$
$CD = \sqrt {{{\left( { - 3 + 1} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = \sqrt {4 + 4} = \sqrt 8$
$DA = \sqrt {{{\left( { - 1 + 3} \right)}^2} + {{\left( { - 2 - 0} \right)}^2}} = \sqrt {4 + 4} = \sqrt 8$
$AC = \sqrt {{{\left( { - 1 + 1} \right)}^2} + {{\left( {2 + 2} \right)}^2}} = \sqrt {0 + 16} = 4$
and $BD = \sqrt {{{\left( { - 3 - 1} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = \sqrt {16 + 0} = 4$
Clearly, four sides AB, BC, CD and DA are equal. Also, diagonals AC and BD are equal.
Therefore, the quadrilateral ABCD is a square.

(ii) Let A(–3, 5), B(3, 1), C(0, 3) and D(–1,– 4) are the given points. Plot these points as shown. Clearly, the points A, C and B are collinear. So, no quadrilateral is formed by these points.

(iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) be the given points. Then,
$AB = \sqrt {{{\left( {7 - 4} \right)}^2} + {{\left( {6 - 5} \right)}^2}} = \sqrt {9 + 1} = \sqrt {10}$
$BC = \sqrt {{{\left( {4 - 7} \right)}^2} + {{\left( {3 - 6} \right)}^2}} = \sqrt {9 + 9} = \sqrt {18}$
$CD = \sqrt {{{\left( {1 - 4} \right)}^2} + {{\left( {2 - 3} \right)}^2}} = \sqrt {9 + 1} = \sqrt {10}$
$DA = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( {5 - 2} \right)}^2}} = \sqrt {9 + 9} = \sqrt {18}$
$AC = \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( {3 - 5} \right)}^2}} = \sqrt {0 + 4} = \sqrt 4 = 2$
and, $BD = \sqrt {{{\left( {1 - 7} \right)}^2} + {{\left( {2 - 6} \right)}^2}} = \sqrt {36 + 16} = \sqrt {52}$
Clearly, AB = CD, BC = DA and $AC \ne BD$
Therefore, the quadrilateral ABCD is a parallelogram.

Q.7     Find the point on the x-axis which is equidistant from (2, – 5) and (– 2, 9).
Sol.       Since the point on x - axis have its ordinate = 0, so (x, 0) is any point on the x - axis.
Since P(x, 0) is equidistant from A(2, –5) and B(–2, 9)
PA = PB $\Rightarrow P{A^2} = P{B^2}$
$\Rightarrow {\left( {x - 2} \right)^2} + {\left( {0 + 5} \right)^2} = {\left( {x + 2} \right)^2} + {\left( {0 - 9} \right)^2}$
$\Rightarrow {x^2}-4x + 4 + 25 = {x^2} + 4x + 4 + 81$
$\Rightarrow -\,4x - 4x = 81 - 25$
$\Rightarrow -8x = 56$
$\Rightarrow x = {{56} \over { - 8}} = - 7$
Therefore, the point equidistant from given points on the axis is (–7, 0).

Q.8     Find the value of y for which the distance between the points P(2, –3) and Q (10, y) is 10 units.
Sol.      P(2, –3) and Q(10, y) are given points such that PQ = 10 units.
But, $PQ = \sqrt {{{\left( {10 - 2} \right)}^2} + {{\left( {y + 3} \right)}^2}}$
$\Rightarrow 10 = \sqrt {64 + {y^2} + 6y + 9}$
$\Rightarrow 100 = 73 + {y^2} + 6y$
$\Rightarrow {y^2} + 6y - 27 = 0 \Rightarrow (y + 9)(y - 3) = 0$
$\Rightarrow$ $y = - 9$ or 3
Thus, the possible values of y are – 9 or 3.

Q.9    If Q (0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Sol.      Since the point Q(0, 1) is equidistant from P(5, –3) and R(x, 6) therefore,
QP = QR $\Rightarrow Q{P^2} = Q{R^2}$
$\Rightarrow {\left( {5 - 0} \right)^2} + {\left( { - 3 - 1} \right)^2} = {\left( {x - 0} \right)^2} + {\left( {6 - 1} \right)^2}$
$\Rightarrow 25 + 16 = {x^2} + 25$
$\Rightarrow {x^2} = 16 \Rightarrow x = \pm 4$
Thus, R is (4, 6) or (– 4, 6).
Now, QR = Distance between Q (0, 1) and R (4, 6)
$= \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {6 - 1} \right)}^2}} = \sqrt {16 + 25} = \sqrt 41$
Also, QR = Distance between Q(0, 1) and R(– 4, 6)
$= \sqrt {{{\left( {- 4 - 0} \right)}^2} + {{\left( {6 - 1} \right)}^2}} = \sqrt {16 + 25} = \sqrt {41}$
and, PR = Distance between P(5, –3) and R( 4, 6)
$= \sqrt {{{\left( {4 - 5} \right)}^2} + {{\left( {6 + 3} \right)}^2}} = \sqrt {1 + 81} = \sqrt {82}$
Also, PR = Distance between P(5, –3) and R(– 4, 6)
$= \sqrt {{{\left( { 4 - 5} \right)}^2} + {{\left( {6 + 3} \right)}^2}} = \sqrt {81 + 81} = 9\sqrt 2$

Q.10   Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (–3, 4).
Sol.      Let the point P(x, y) be equidistant from the points A(3, 6) and B(–3, 4)

i.e., PA = PB
$\Rightarrow P{A^2} = P{B^2}$
$\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x + 3} \right)^2} + {\left( {y - 4} \right)^2}$
$\Rightarrow {x^2} - 6x + 9 + {y^2} - 12y + 36 = {x^2} + 6x + 9 + {y^2} - 8y + 16$
$\Rightarrow - 6x - 6x - 12y + 8y + 36 - 16 = 0$
$\Rightarrow - 12x - 4y + 20 = 0$

$\Rightarrow 3x + y - 5 = 0$
This is the required relation.

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