# The points $\mathrm{A}(2,9), \mathrm{B}(a, 5)$ and $\mathrm{C}(5,5)$ are the vertices of a triangle $\mathrm{ABC}$ right angled at $\mathrm{B}$. Find the values of $a$ and hence the area of $\triangle \mathrm{ABC}$.

Given:

The points $A (2, 9), B (a, 5)$ and $C (5, 5)$ are the vertices of a triangle ABC right angled at B.

To do:

We have to find the values of $a$ and hence the area of $\triangle ABC$.

Solution:

We know that,

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

The triangle is right-angled at B.

This implies, by Pythagoras theorem,

$AB^2+BC^2=AC^2$.......(i)

Therefore,

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

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

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

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

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

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

$=\sqrt{(-3)^{2}+(4)^{2}}$

$=\sqrt{9+16}$

$=\sqrt{25}=5$

Substituting the values of $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ in (i), we get,

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

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

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

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

$\Rightarrow a^{2}-2 a-5 a+10=0$

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

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

$\Rightarrow a=2$ or $a=5$

If $a=5,$ the length of $\mathrm{BC}=0$ which is not possible because the sides $\mathrm{AB}, \mathrm{BC}$ and CA from a right angled triangle.

Therefore, $a=2$

The coordinates of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $(2,9),(2,5)$ and $(5,5),$ respectively.

Area of $\Delta \mathrm{ABC}=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

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

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

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

$=\frac{1}{2} \times 12=6$

Hence, the required area of $\Delta \mathrm{ABC}$ is 6 sq. units.

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Updated on: 10-Oct-2022

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