Question

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

• A. $a=-1$ , Area $=$ $15$ sq. unit
• B. $a=2$ , Area $=$ $6$ sq. unit
• C. $a=0$ , Area $=$ $8$ sq. unit
• D. $a=1$ , Area $=$ $12$ sq. unit

Answer 1

The correct option is B $a=2$ , Area $=$ $6$ sq. unit

Given that the vertices of the $\Delta ABC$ are $A(x_1,y_1)=(2,9),B(x_2,y_2)=(a,5)$ and $C(x_3,y_3)=(5,5)and\angle B={90}^o$.

So, $AC$ is the hypotenuse.

$\therefore$ by Pythagoras theorem, we have

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

Now, by distance formula

$d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

So, $AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(2-a)}^2+{(9-5)}^2}=\sqrt{a^2-4a+20}$,

$BC=\sqrt{{(x_3-x_2)}^2+{(y_3-y_2)}^2}=\sqrt{{(5-a)}^2+{(5-5)}^2}=\sqrt{a^2-10a+25}$ and

$AC=\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2}=\sqrt{{(5-2)}^2+{(5-9)}^2}3$ units $=\sqrt{25}3$ units.

$\therefore$ by (i), we get

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

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

$\Rightarrow a=5,2$

We reject $a=5$, since $B$ and $C$ will coincide in that case and $\Delta ABC$ will collapse.

So, $a=2.i.e.AB=\sqrt{a^2-4a+20}=\sqrt{4-8+20}3$ units $=43$ units

$BC=\sqrt{a^2-10a+25}=\sqrt{4-20+25}3$ units $=3$ units.

Now,

We know that the area of the triangle whose vertices are $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ is $\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Therefore, required area is $6$ square units.