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 values of a and hence the area of ∆ABC.

Answer 1

Given that, the points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a ΔABC right angled at B.

By Pythagoras theorem,

AC² = AB² + BC²    

Now, by distance formula,

AB = `sqrt((a - 2)^2 + (5 - 9)^2)`   ...(i) `[∵ "Distance between two point"  (x_1, y_1)  "and"  (x_2, y_2) = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`

= `sqrt(a^2 + 4 - 4a + 16)`

= `sqrt(a^2 - 4a + 20)`

BC = `sqrt((5 - a)^2 + (5 - 5)^2`

= `sqrt((5 - a)^2 + 0)`

= 5 – a

And AC = `sqrt((2 - 5)^2 + (9 - 5)^2`

= `sqrt((-3)^2 + (4)^2`

= `sqrt(9 + 16)`

= `sqrt(25)`

= 5

Put the values of AB, BC and AC in equation (i), we get

(5)² = `(sqrt(a^2 - 4a + 20))^2 + (5 - a)^2`

⇒ 25 = a² – 4a + 20 + 25 + a² – 10a

⇒ 2a² – 14a + 20 = 0

⇒ a² – 7a + 10 = 0

⇒ a² – 2a – 5a + 10 = 0  ...[By factorisation method]

⇒ a(a – 2) – 5(a – 2) = 0

⇒ (a – 2)(a – 5) = 0

∴ a = 2, 5

Here, a ≠ 5, since at a = 5, the length of BC = 0.

It is not possible because the sides AB, BC and CA form a right angled triangle.

So, a = 2

Now, the coordinate of A, B and C becomes (2, 9), (2, 5) and (5, 5), respectively.

∵ Area of ΔABC = `1/2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]`

∴ Δ = `1/2[2(5 - 5) + 2(5 - 9) + 5(9 - 5)]`

= `1/2[2 xx 0 + 2(-4) + 5(4)]`

= `1/2(0 - 8 + 20)`

= `1/2 xx 12`

= 6

Hence, the required area of ΔABC is 6 sq units.