Straight LineHardBloom L3
Question
In a triangle $ABC$, if $A(2, - 1)$ and $7x - 10y + 1 = 0$ and $3x - 2y + 5 = 0$ are equations of an altitude and an angle bisector respectively drawn from $B$, then equation of $BC$ is :
Options
A.$x + y + 1 = 0$
B.$5x + y + 17 = 0$
C.$4x + 9y + 30 = 0$
D.$x - 5y - 7 = 0$
Solution
Sol. [Key Concept: $A$ given with the altitude and angle bisector from $B$ ⇒ reflect / intersect to obtain $BC$.]
Image of $A(2, - 1)$ w.r.t.
$3x - 2y + 5 = 0,A^{'}$ is given by
$$\begin{matrix} & \frac{x - 2}{3} = \frac{y + 1}{- 2} = - 2 \cdot \frac{(6 + 2 + 5)}{13} = - 2 \\ & A^{'} \equiv ( - 4,3) \end{matrix}$$
Equation of $BC$ is
$$\begin{matrix} & y - 3 = \frac{3 + 2}{- 4 + 3}(x + 4) \\ & 5x + y + 17 = 0 \end{matrix}$$
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