First slide

Equation of line in Straight lines

Question

Let $2 a + 3 b + c = 0$ . If $x + 5 y + 7 = 0$ and L = 0 are angle bisectors of two lines $L_{1} = 0, L_{2} = 0 a n d L = 0$ is a member in the family of lines $a x + b y + c = 0$ then area of triangle formed by L = 0 with coordinate axes is (in sq.units)

Moderate

A

49
B

$\frac{49}{2}$
C

$\frac{49}{5}$
D

$\frac{49}{10}$

Solution

Point of concurrence of $a x + b y + c = 0$ is (2, 3)
Slope of L is $\frac{- 1}{(- \frac{1}{5})} = 5$
Equation of line L = 0 is y-3 = 5 (x-2)
$\Rightarrow 5 x - y - 7 = 0$
Area of the triangle $= \frac{{(7)}^{2}}{2 | 5 (- 1) |} = \frac{49}{10} s q . u n i t s$

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