NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1#
Exercise 7.1 focuses on the Distance Formula — finding the distance between two points given their coordinates, and using it to determine properties of geometric figures (collinearity, type of triangle/quadrilateral).
Distance Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is:
$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$Distance from Origin: Distance of point $P(x, y)$ from origin $= \sqrt{x^2 + y^2}$
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)
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. #
Answer
Yes, the distance between the two towns A and B = 39 km (if the coordinates represent distances in km).
3. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear. #
Answer
Let $A(1, 5)$, $B(2, 3)$, $C(-2, -11)$.
$$AB = \sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1 + 4} = \sqrt{5}$$$$BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53}$$$$CA = \sqrt{(1-(-2))^2 + (5-(-11))^2} = \sqrt{9 + 256} = \sqrt{265}$$Check: $AB + BC = \sqrt{5} + 2\sqrt{53} \neq \sqrt{265} = CA$
Since $AB + BC \neq CA$, the points are not collinear.
4. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle. #
Answer
Let $A(5, -2)$, $B(6, 4)$, $C(7, -2)$.
$$AB = \sqrt{(6-5)^2 + (4-(-2))^2} = \sqrt{1 + 36} = \sqrt{37}$$$$BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1 + 36} = \sqrt{37}$$$$CA = \sqrt{(5-7)^2 + (-2-(-2))^2} = \sqrt{4 + 0} = 2$$Since $AB = BC = \sqrt{37}$ (two sides are equal), the triangle is isosceles.
5. In a classroom, 4 friends are seated at the points A, B, C and D as given in the 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. #
Answer
From the figure, the coordinates are: $A(3, 4)$, $B(6, 7)$, $C(9, 4)$, $D(6, 1)$.
$$AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{9+9} = 3\sqrt{2}$$$$BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{9+9} = 3\sqrt{2}$$$$CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{9+9} = 3\sqrt{2}$$$$DA = \sqrt{(3-6)^2 + (4-1)^2} = \sqrt{9+9} = 3\sqrt{2}$$All sides are equal. Now check diagonals:
$$AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{36} = 6$$$$BD = \sqrt{(6-6)^2 + (1-7)^2} = \sqrt{36} = 6$$Since all sides are equal and diagonals are equal, ABCD is a square. Champa is correct.
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: #
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
Let $A(-1,-2)$, $B(1,0)$, $C(-1,2)$, $D(-3,0)$.
$$AB = \sqrt{4+4} = 2\sqrt{2}$$$$BC = \sqrt{4+4} = 2\sqrt{2}$$$$CD = \sqrt{4+4} = 2\sqrt{2}$$$$DA = \sqrt{4+4} = 2\sqrt{2}$$Diagonals:
$$AC = \sqrt{0+16} = 4$$$$BD = \sqrt{16+0} = 4$$All sides equal, diagonals equal → Square.
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
Let $A(-3,5)$, $B(3,1)$, $C(0,3)$, $D(-1,-4)$.
$$AB = \sqrt{36+16} = \sqrt{52}$$$$BC = \sqrt{9+4} = \sqrt{13}$$$$CD = \sqrt{1+49} = \sqrt{50}$$$$DA = \sqrt{4+81} = \sqrt{85}$$All sides are different, and checking if any are collinear:
Check if A, B, C are collinear: slope AB = $\dfrac{1-5}{3-(-3)} = \dfrac{-4}{6} = -\dfrac{2}{3}$; slope BC = $\dfrac{3-1}{0-3} = \dfrac{2}{-3} = -\dfrac{2}{3}$.
Since slope AB = slope BC and B is common, A, B, C are collinear. Therefore these points do not form a quadrilateral.
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Let $A(4,5)$, $B(7,6)$, $C(4,3)$, $D(1,2)$.
$$AB = \sqrt{9+1} = \sqrt{10}$$$$BC = \sqrt{9+9} = 3\sqrt{2}$$$$CD = \sqrt{9+1} = \sqrt{10}$$$$DA = \sqrt{9+9} = 3\sqrt{2}$$Opposite sides equal: $AB = CD$, $BC = DA$.
Diagonals:
$$AC = \sqrt{0+4} = 2$$$$BD = \sqrt{36+16} = \sqrt{52}$$Diagonals are not equal. → Parallelogram (opposite sides equal, diagonals unequal).
7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). #
Answer
Let the point on x-axis be $P(x, 0)$.
Given: $PA = PB$ where $A(2,-5)$ and $B(-2,9)$.
$$PA^2 = (x-2)^2 + 25$$$$PB^2 = (x+2)^2 + 81$$Setting $PA^2 = PB^2$:
$$(x-2)^2 + 25 = (x+2)^2 + 81$$$$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$$$$-4x + 29 = 4x + 85$$$$-8x = 56$$$$x = -7$$∴ The required point is $\mathbf{(-7, 0)}$.
8. Find the values of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units. #
Answer
$y = 3$ or $y = -9$
∴ $y = 3$ or $y = -9$.
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. #
Answer
Given: $QP = QR$
$$QP = \sqrt{(5-0)^2 + (-3-1)^2} = \sqrt{25+16} = \sqrt{41}$$$$QR = \sqrt{x^2 + (6-1)^2} = \sqrt{x^2+25}$$Setting $QP = QR$:
$$x^2 + 25 = 41$$$$x^2 = 16$$$$x = \pm 4$$Case 1: $x = 4$, so $R(4, 6)$
$$QR = \sqrt{16+25} = \sqrt{41}$$$$PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1+81} = \sqrt{82}$$Case 2: $x = -4$, so $R(-4, 6)$
$$QR = \sqrt{16+25} = \sqrt{41}$$$$PR = \sqrt{(5+4)^2+(-3-6)^2} = \sqrt{81+81} = 9\sqrt{2}$$∴ $x = 4$ or $x = -4$; $QR = \sqrt{41}$; $PR = \sqrt{82}$ or $PR = 9\sqrt{2}$.
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4). #
Answer
Let $P(x, y)$, $A(3, 6)$, $B(-3, 4)$.
Given: $PA = PB$
$$PA^2 = PB^2$$$$(x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2$$$$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$$$$-6x + 9 - 12y + 36 = 6x + 9 - 8y + 16$$$$-6x - 12y + 45 = 6x - 8y + 25$$$$-12x - 4y + 20 = 0$$$$3x + y = 5$$∴ The required relation is $3x + y - 5 = 0$ (or $3x + y = 5$).