Class 10 Coordinate Geometry — practice questions with answers

Distance = $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.

Sides: (0,4)-(0,0) = 4; (0,0)-(3,0) = 3; (3,0)-(0,4) = $\sqrt{9+16}$ = 5. Perimeter = 4 + 3 + 5 = 12 units.

AB = $\sqrt{(6-3{)}^2+(4-0{)}^2}=\sqrt{25}=5$. CA = $\sqrt{(-1-3{)}^2+(3-0{)}^2}=\sqrt{25}=5$. BC = $\sqrt{(-1-6{)}^2+(3-4{)}^2}=\sqrt{50}$. AB = CA (isosceles) and AB² + CA² = 25 + 25 = 50 = BC² (right angle at A). So it is an isosceles right triangle.

A point on the y-axis is (0, y). Equate squared distances: (0 − 3)² + (y − 6)² = (0 + 1)² + (y − 4)² → 9 + y² − 12y + 36 = 1 + y² − 8y + 16 → 45 − 12y = 17 − 8y → 28 = 4y → y = 7. The mast goes at (0, 7), and both distances are $\sqrt{10}$ km. Averaging the two y-coordinates to get (0, 5) is the common error — that is the midpoint of PQ, which is not on the y-axis. (7, 0) has the axes swapped.

Suppose Q divides AB in the ratio k : 1. Q lies on the x-axis, so its y-coordinate must be 0: $\frac{k(-3)+1(6)}{k+1}=0$ → −3k + 6 = 0 → k = 2, i.e. the ratio 2 : 1. Now verify the x-coordinate with that ratio: $\frac{2(6)+1(-3)}{3}=\frac{9}{3}=3$ ✓, matching Q(3, 0), so Q genuinely lies on AB. Using the ratio the wrong way round, 1 : 2, would place the point at $(\frac{6+2(-3)}{3},\frac{-3+12}{3})=(0,3)$ — on the y-axis, not at Q.