Question

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the value(s) of a and hence find area of ∆ABC.

Answer 1

2+9+2+9+5+5+5 =

=37 is. the answer

Answer 2

a = 2

Area of ∆ABC = 6 sq. units

Explanation:

Given: The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B.

Since ABC is a right-angled triangle, we have:

AB^2 + BC^2 = CA^2

Please refer to the attached picture for the diagram.

Using distance formula for coordinates, we get:

[(2 - a)^2 + (9 - 5)^2] + [(a - 5)^2 + (5 - 5)^2] = (2 - 5)^2 + (9 - 5)^2

4 + a^2 - 4a + 16 + a^2 + 25 - 10a = 9 + 16

2a^2 - 14a + 20 = 0

a^2 - 7a + 10 = 0

(a - 5) (a - 2) = 0

a = 5 or 2

But a cannot be 5 as then the points B and C will coincide.

Hence solved that a = 2.

Area = 1/2 [2(5-5) + 2(5-9) + 5(9-5)]

= 1/2 (0 - 8 + 20)

= 12/ 2

= 6 sq. units