NCERT Solutions for Class 9th Maths Chapter 1 In-text Questions — Distance Between Two Points in the 2-D Plane
Book page 11 Updated on2026-09-08
Q1.
What if, x₁, x₂, y₁, y₂ take negative values? In Fig. 1.9, triangle AMD is reflected in the y-axis. What are the coordinates of the images of points A, M, and D?
Answer
Reflection in the y-axis sends (x, y) to (−x, y).
A (3, 4) → A′ = (−3, 4) M (9, 6) → M′ = (−9, 6) D (7, 1) → D′ = (−7, 1) (and the helper point C (3, 1) → C′ = (−3, 1))
∆ADM in Quadrant I and its mirror image ∆A′D′M′ in Quadrant II. Each point keeps its height and swaps sides.
Why the x-coordinate changes sign and the y-coordinate does not: The mirror is the y-axis. Reflection keeps a point at the same perpendicular distance from the mirror but on the opposite side. That distance is measured along the x-direction, so its size stays the same and its sign flips. Nothing moves the point up or down, so the y-coordinate is untouched.
The side lengths are unchanged, and the arithmetic shows why:
A′D′: shifts |−3 − (−7)| = 4 and |4 − 1| = 3 → √(16 + 9) = 5 units
D′M′: shifts |−9 − (−7)| = 2 and |6 − 1| = 5 → √(4 + 25) = √29 units
M′A′: shifts |−9 − (−3)| = 6 and |6 − 4| = 2 → √(36 + 4) = √40 units
Why negatives cause no trouble in the distance formula: Only the differences x₂ − x₁ and y₂ − y₁ enter the formula, and they are then squared. Changing the sign of both x-coordinates changes the sign of their difference but not its square, so the distance is unaffected. The formula was never restricted to the first quadrant.