Straight LineHardBloom L3

Question

In a $\bigtriangleup ABC$, the equations of right bisectors of sides $AB$ and $CA$ are $3x + 4y = 20$ and $8x + 6y = 65$ respectively. If the vertex $A$ be $(10,10)$, then the area of $\bigtriangleup ABC$ will be :

Options

A.14
B.21
C.42
D.63

Solution

Sol. [Key Concept: $A$ reflected across each right bisector gives $B$ and $C$; then area via the coordinate determinant.]

Let perpendicular bisector of $AB$ is $3x + 4y - 20 = 0$ and perpendicular bisector of $AC$ is $8x + 6y - 65 = 0$.

⇒ Image of $A$ w.r.t. $3x + 4y - 20 = 0$ is $B$ and image of $A$ w.r.t. $8x + 6y - 65 = 0$ is $C$.

For B , $\frac{x - 10}{3} = \frac{y - 10}{4} = - 2\left( \frac{30 + 40 - 20}{25} \right)$ $\Rightarrow \ B \equiv ( - 2, - 6)$

For C, $\frac{x - 10}{8} = \frac{y - 10}{6} = - 2\left( \frac{80 + 60 - 65}{100} \right)$ $\Rightarrow \ C \equiv ( - 2,1)$

Area of $\bigtriangleup ABC = \frac{1}{2}(10 + 2)(1 + 6) = 42$

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