Question

The line $3x+2y=24$ meets $y$-axis at $A$ and $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $\left(0,-1\right)$ parallel to $x$-axis at $C$. The area of the triangle $ABC$ is

• A. $182$ sq. units
• B. $91$ sq. units
• C. $48$ sq. units
• D. None of these

Answer 1

The correct option is B

$91$ sq. units

Step 1: Calculate the co-ordinates of the triangle.

The given equation of line is $3x+2y=24$…….$\left(i\right)$

When $x=0,y=\frac{24}{2}=12$

So, coordinates of $A=\left(0,12\right)$.

When $y=0,x=\frac{24}{3}=8$

So, coordinates of $B=\left(8,0\right)$.

Let $P$ be the midpoint of $AB$.

So coordinates of

$\begin{array}{rcl}P& =& \left(\frac{0+8}{2},\frac{12+0}{2}\right)\\ & =& \left(4,6\right)\end{array}$

Equation of line through $\left(0,-1\right)$parallel to $x$-axis is

$y=-1....\left(ii\right)$

Slope of equation $\left(i\right)=-\frac{3}{2}$

Slope of perpendicular bisector of $AB=\frac{2}{3}$

Equation of perpendicular bisector of $AB$ passing through $\left(4,6\right)$ is

$$\begin{gathered} \Rightarrow y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-6=\frac{2}{3}\left(x-4\right) \\ \Rightarrow 3y-18=2x-8 \\ \Rightarrow 3y-2x-10=0....\left(iii\right) \end{gathered}$$

Put $y=-1$ in equation $\left(iii\right)$,

$\begin{array}{rcl}& \Rightarrow & -3-2x-10=0\\ & \Rightarrow & x=-\frac{13}{2}\end{array}$

So the coordinates of $C=\left(-\frac{13}{2},1\right)$.

Step 2: Calculate the area of triangle.

Area of a triangle is

$\begin{array}{rcl}& =& \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\\ & =& \frac{1}{2}\left|\begin{array}{ccc}0& 12& 1\\ 8& 0& 1\\ -\frac{13}{2}& -1& 1\end{array}\right|\\ & =& \frac{1}{2}\left|-174-8\right|\\ & =& 91\end{array}$

Hence, option $\left(B\right)$ is correct.