Question

The points $A(2,9)$, $B(a,5)$, $C(5,5)$ are the vertices of a triangle ABC right angled at B. find the value of ${}^{\prime }{a}^{\prime }$ and hence the area of $\Delta ABC$.

Answer 1

Given the points $A(2,9),B(a,5),C(5,5)$are the vertices of right angle triangle whose angle $B$ is right angle.

Then distance

$AB=\sqrt{{(2-a)}^2+{(9-5)}^2}$

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

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

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

$=\sqrt{a^2+25-10a}$

$CA=\sqrt{{(5-2)}^2+{(5-9)}^2}$

$=\sqrt{9+16}$

$=\sqrt{25}=5$

We know that, In a right angle triangle

By phythagoras theorm,

$A{B}^2+B{C}^2=C{A}^2$

${\left(\sqrt{a^2-4a+20}\right)}^2+{\left(\sqrt{a^2+25-10a}\right)}^2=5^2$

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

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

$\Rightarrow 2(a^2-7a+10)=0$

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

By factorize this equation

$\Rightarrow a=2,5$

Put first $a=2$, In $AB,BCandCA$

Then,

$AB=\sqrt{2^2-4\times 2+20}$

$=\sqrt{4-8+20}$

$=\sqrt{16}=4$

$BC=\sqrt{a^2-10a+25}$

$=\sqrt{2^2-10\times 2+25}$

$=\sqrt{9}=3$

Now, put $a=5$ and we get,

$AB=\sqrt{5^2-4\times 5+20}$

$=\sqrt{25-20+20}$

$=\sqrt{25}=5$

$BC=\sqrt{a^2-10a+25}$

$=\sqrt{5^2-10\times 5+25}$

$=\sqrt{0}=0$

Then $a=5$ is not possible

Hence $a=2$ is possible for all value of right angle triangle.