If the points A (2, 9), B (a, 5) and C (5, 5)

Question:

If the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ABC right angled at B, then find the values of a and hence the area of ΔABC.

Solution:

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     ,,.(i)

$\left[\because\right.$ distance between two points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right)=\sqrt{\left(x_{2}-x_{1}^{2}\right)+\left(y_{2}-y_{1}\right)^{2}}\right]$

$=\sqrt{a^{2}+4-4 a+16}=\sqrt{a^{2}-4 a+20}$

$B C=\sqrt{(5-a)^{2}+(5-5)^{2}}$

$=\sqrt{(5-a)^{2}+0}=5-a$

and $A C=\sqrt{(2-5)^{2}+(9-5)^{2}}$

$=\sqrt{(-3)^{2}+(4)^{2}}=\sqrt{9+16}=\sqrt{25}=5$

Put the values of $A B, B C$ and $A C$ in Eq. (i), we get

$(5)^{2}=\left(\sqrt{a^{2}-4 a+20}\right)^{2}+(5-a)^{2}$

$\Rightarrow \quad 25=a^{2}-4 a+20+25+a^{2}-10 a$

$\Rightarrow \quad 2 a^{2}-14 a+20=0$

$\Rightarrow \quad a^{2}-7 a+10=0$

$\Rightarrow \quad a^{2}-2 a-5 a+10=0 \quad$ [by factorisation method]

$\Rightarrow \quad a(a-2)-5(a-2)=0$

$\Rightarrow \quad(a-2)(a-5)=0$

$\therefore \quad a=2,5$

Here, $a \neq 5$, since at $a=5$, the length of $B C=0$. It is not possible because the sides $A B, B C$ and $C A$ form a right angled triangle.

so,

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

$\because \quad$ Area of $\Delta A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\therefore$ $\Delta=\frac{1}{2}[2(5-5)+2(5-9)+5(9-5)]$

$=\frac{1}{2}[2 \times 0+2(-4)+5(4)]$

$=\frac{1}{2}(0-8+20)=\frac{1}{2} \times 12=6$

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