Question

A triangle $ABC$ lying in the first quadrant has two vertices as $A\left(1,2\right)$ and $B\left(3,1\right)$ If $BAC=90°$ and $ar\left(∆ABC\right)=5\sqrt{5}$ units, then the abscissa of the vertex $C$ is:

• A. $1+\sqrt{5}$
• B. $1+2\sqrt{5}$
• C. $2\sqrt{5}-1$
• D. $2+\sqrt{5}$

Answer 1

The correct option is B

$1+2\sqrt{5}$

Explanation for correct option:

Step-1: Draw the graph with the help of given data.

Given, two vertices as $A\left(1,2\right)$ and $B\left(3,1\right)$ If $BAC=90°$ and $ar\left(∆ABC\right)=5\sqrt{5}$

Step-2: Apply distance formula.

Given, vertices as $A\left(1,2\right)$ and $B\left(3,1\right)$

$\Rightarrow$ $AB=\sqrt{{\left(3-1\right)}^2+{\left(1-2\right)}^2}$$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{d}\mathbf{=}\sqrt{{\left(x_2-x_1\right)}^{\mathbf{2}}\mathbf{+}{\left(y_2-y_1\right)}^{\mathbf{2}}}$

$\Rightarrow$ $AB=\sqrt{5}$

Step-3: Apply formula for area of triangle.

Given, $ar\left(∆ABC\right)=5\sqrt{5}$

$\Rightarrow$ $ar\left(∆ABC\right)=\frac{1}{2}\times x\times \sqrt{5}$

$\Rightarrow$ $5\sqrt{5}=\frac{1}{2}\times x\times \sqrt{5}$

$\Rightarrow$ $x=10$

Step-4: Finding abscissa of point $C$.

Since, $BAC=90°$

Therefore, product of slope $AC$ and $AB$ is $-1$.

By using slope formula $m=\frac{y_2-y_1}{x_2-x_1}$.

$\Rightarrow$ $\frac{\left(b-2\right)}{\left(a-1\right)}\times \frac{\left(1-2\right)}{\left(3-1\right)}=-1$

$\Rightarrow$ $\frac{\left(b-2\right)}{\left(a-1\right)}=\frac{-2}{-1}$

$\Rightarrow$ $b=2a$

Step-5: Apply distance formula on side $AC$.

$\Rightarrow$ $\sqrt{{\left(a-1\right)}^2+{\left(b-2\right)}^2}=10$

$\Rightarrow$ ${\left(a-1\right)}^2+{\left(2a-2\right)}^2=100$

$\Rightarrow$ ${\left(a-1\right)}^{2}5=100$

$\Rightarrow$ $a=1+2\sqrt{5}$

Hence correct answer is option $\left(B\right)$