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Chapter 7

Coordinate Geometry

NCERT solutions and explanations for Class 10 Mathematics Chapter 7 Coordinate Geometry — covering Distance formula, section formula, and applications in the coordinate plane.

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Questions

6
Q1

State the distance formula between two points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2).

The distance is d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
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Q2

Find the distance between (2,3)(2,3) and (6,6)(6,6).

Distance =(6−2)2+(6−3)2=16+9=5=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{16+9}=5.
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Q3

How can the distance formula be used to test whether three points are collinear?

Find the three pairwise distances. If the sum of two smaller distances equals the largest distance, the points are collinear.
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Q4

State the section formula for internal division.

If P(x,y)P(x,y) divides A(x1,y1)A(x_1,y_1) and B(x2,y2)B(x_2,y_2) internally in the ratio m:nm:n, then x=mx2+nx1m+nx=\frac{mx_2+nx_1}{m+n} and y=my2+ny1m+ny=\frac{my_2+ny_1}{m+n}.
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Q5

Derive the midpoint formula from the section formula.

For midpoint, ratio is 1:11:1. Substituting m=n=1m=n=1 gives P(x1+x22,y1+y22)P\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).
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Q6

Find the point dividing (1,2)(1,2) and (7,8)(7,8) internally in the ratio 2:12:1.

Using section formula, x=2(7)+1(1)3=5x=\frac{2(7)+1(1)}{3}=5, y=2(8)+1(2)3=6y=\frac{2(8)+1(2)}{3}=6. The point is (5,6)(5,6).
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