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,

AC^{2} = AB^{2} + BC^{2} ……(i)

Now by using distance formula;

As distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2});

So,

AB =

BC =

Put the values of AB, BC and AC in Eq. (i),

We get;

25 = a^{2} – 4a + 20 + 25 + a^{2} – 10a

2a^{2} – 14a + 20 = 0

a^{2} – 7a + 10 = 0

a^{2} – 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 [x_{1} (y_{2} – y_{3} ) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{1} - y_{2} )]

[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.

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