Triangle ADM is an acute angled triangle in the first quadrant. How do we find the lengths of its sides AD, DM and MA?

Question

Accepted Answer

None of AD, DM, MA is parallel to an axis, so we cannot subtract coordinates directly. Instead, make each side the hypotenuse of a right triangle whose legs are parallel to the axes.

Dropping A straight down to C (3, 1) makes ∆ACD right-angled at C, with legs parallel to the axes.

A (3, 4), D (7, 1), M (9, 6)

AD: horizontal shift = |7 − 3| = 4, vertical shift = |4 − 1| = 3
AD = √(4² + 3²) = √25 = 5 units

DM: horizontal shift = |9 − 7| = 2, vertical shift = |6 − 1| = 5
DM = √(2² + 5²) = √29 ≈ 5.39 units

MA: horizontal shift = |9 − 3| = 6, vertical shift = |6 − 4| = 2
MA = √(6² + 2²) = √40 = 2√10 ≈ 6.32 units

Why it happens: Take any two points. Travel from the first to the second in two stages — first horizontally, then vertically. The corner where you turn is the third vertex, C (3, 1) for the side AD, and since one stage is along a horizontal line and the other along a vertical line, the angle there is a right angle. The Baudhāyana–Pythagoras Theorem then gives the direct distance from the two stages. This works for any pair of points, which is why it becomes the general distance formula √((x₂−x₁)² + (y₂−y₁)²).

Check it yourself: The book calls ∆ADM acute-angled. Verify it. The longest side is MA, with MA² = 40. Now AD² + DM² = 25 + 29 = 54, and 54 > 40, so the angle opposite MA (the angle at D) is acute. The other two angles are opposite shorter sides, so they are acute as well. All three angles acute — the claim checks out, and it was settled by arithmetic alone.

NCERT Solutions for Class 9th Maths Chapter 1 In-text Questions — Distance Between Two Points in the 2-D Plane