First slide

Equation of line in Straight lines

Question

The line x + 3y - 2 = 0 bisects the angle between a pair of straight lines of which one has equation x - 7y + 5 = 0.The equation of the other line is

Moderate

A

$3 x + 3 y - 1 = 0$
B

$x - 3 y + 2 = 0$
C

$5 x + 5 y - 3 = 0$
D

None of these

Solution

The family of line through the given lines is
$L \equiv x - 7 y + 5 + λ (x + 3 y - 2) = 0 = x (1 + λ) + y (- 7 + 3 λ) + 5 - 2 λ - - - - - (1)$
let (2, 0) be a point on x+3y-2=0 which is equidistant from x - 7y + 5 = 0 and the line L = 0
Therefore,
$|\frac{2 - 0 + 5}{\sqrt{1^{2} + 7^{2}}}| = |\frac{2 (1 + λ) + 0 (- 7 + 3 λ) + 5 - 2 λ}{\sqrt{{(1 + λ)}^{2} + {(- 7 + 3 λ)}^{2}}}|$
$|\frac{2 + 5}{\sqrt{50}}| = |\frac{2 + 2 λ + 5 - 2 λ}{\sqrt{(1 + λ)^{2} + (3 λ - 7)^{2}}}|$
$\begin{array}{l} or 10 λ^{2} - 40 λ = 0 \\ i.e., λ = 4 or 0 \\ Hence, L = 0, λ = 4 . \end{array}$
$Therefore, the required line is 5 x + 5 y - 3 = 0 .$

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