If the points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a ΔABC right angled at B, then find the values of a and hence the area of ΔABC.
Given,
The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ΔABC right angled at B.
By using Pythagoras theorem,
AC2 = AB2 + BC2 ……(i)
Now by using distance formula;
As distance between two points (x1, y1) and (x2, y2);
So,
AB =
BC = 
Put the values of AB, BC and AC in Eq. (i),
We get;
25 = a2 – 4a + 20 + 25 + a2 – 10a
2a2 – 14a + 20 = 0
a2 – 7a + 10 = 0
a2 – 2a – 5a + 10 = 0
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 ∆ABC = 1/2 [x1 (y2 – y3 ) + x2 (y3 - y1 ) + x3 (y1 - y2 )]
[2(5 - 5) + 2(5 - 9) + 5(9 - 5)]
= 
[2 × 0 + 2( - 4) + 5(4)]
= 
(0 – 8 + 20) = 
× 12 = 6 
Hence, the required area of ΔABC is 6 sq. units.