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Coordinate Geometry

All 6 textbook questions with direct answer previews. Open any question for simple explanations and exam-ready answers.

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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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