First slide

Equation of line in Straight lines

Question

$In △ A B C if A (0, 0), internal angle bisector through vertex B is x + 2 y - 5 = 0 and$ $perpendicular bisector of the side A C is 3 x - y - 10 = 0, then the absolute value of slope of$ $line B C is$

Moderate

A

$\frac{1}{2}$
B

2
C

$- \frac{3}{2}$
D

4

Solution

$Point C will be image of A in 3 x - y - 10 = 0$
$\frac{x - 0}{3} = \frac{y - 0}{- 1} = - 2 [- \frac{10}{10}]$
$C = (x, y) = (6, - 2)$
$Image of A in x + 2 y - 5 = 0 lie on B C$
$\frac{x - 0}{1} = \frac{y - 0}{2} = - 2 [- \frac{5}{5}] (∵ Im age f o r m u l a \frac{h - x_{1}}{a} = \frac{k - y_{1}}{b} = \frac{- 2 (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}})$
$(x, y) = (2, 4)$
$∴ Slope of B C, m_{BC} = \frac{6}{- 4} = - \frac{3}{2}$

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