First slide

Equation of line in Straight lines

Question

Equation of the line passing through the point of intersection of $x + y - 4 = 0$ and $2 x + 3 y - 10 = 0$ , parallel to the line $5 x - 7 y + 5 = 0$

Moderate

A

$7 x + 5 y + 3 = 0$
B

$5 x - 7 y + 10 = 0$
C

$5 x - 7 y + 6 = 0$
D

$5 x - 7 y + 4 = 0$

Solution

$The equation of any line passing through the point of intersection of two lines x + y - 4 = 0$
$and 2 x + 3 y - 10 = 0 can be taken as$
$\begin{matrix} x + y - 4 + λ (2 x + 3 y - 10) = 0 \\ x (1 + 2 λ) + y (1 + 3 λ) - 4 - 10 λ = 0 \end{matrix}$
$Since the line x (1 + 2 λ) + y (1 + 3 λ) - 4 - 10 λ = 0 is parallel to the line 5 x - 7 y + 5 = 0,$
$\frac{(1 + 2 λ)}{5} = \frac{(1 + 3 λ)}{- 7}$
Cross multiply and then simplify
$\begin{matrix} - 7 (1 + 2 λ) = 5 (1 + 3 λ) \\ - 7 - 14 λ = 5 + 15 λ \\ 29 λ = - 12 \\ λ = - \frac{12}{29} \end{matrix}$
$Substitute λ = - \frac{12}{29} in the equation x + y - 4 + λ (2 x + 3 y - 10) = 0$
It implies that
$\begin{array}{l} x + y - 4 - \frac{12}{29} (2 x + 3 y - 10) = 0 \\ 29 x + 29 y - 116 - 24 x - 36 y + 120 = 0 \\ 5 x - 7 y + 4 = 0 \end{array}$
$Therefore, the equation of the required line is 5 x - 7 y + 4 = 0$

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