$L_1 = ax - 3y + 5 = 0$ and $L_2 = 4x - 6y + 8 = 0$ are two parallel lines. If $p, q$ are the intercepts made by $L_1 = 0$ and $m, n$ are the intercepts made by $L_2 = 0$ on the X and Y coordinate axes, respectively, then the equation of the line passing through the points $(p, q)$ and $(m, n)$ is
Show Hint
- $3x + 3y + 2 = 0$
- $2x + 3y = 0$
- $6x + 6y + 5 = 0$
- $x + 3y = 2$
The Correct Option is B
Solution and Explanation
For $L_1: ax - 3y + 5 = 0$: - X-intercept ($p$):
Set $y = 0$, $ax + 5 = 0$, $p = -\frac{5}{a}$.
- Y-intercept ($q$): Set $x = 0$, $-3y + 5 = 0$, $q = \frac{5}{3}$. For $L_2: 4x - 6y + 8 = 0$ or $2x - 3y + 4 = 0$:
- X-intercept ($m$): Set $y = 0$, $2x + 4 = 0$, $m = -2$.
- Y-intercept ($n$): Set $x = 0$, $-3y + 4 = 0$, $n = \frac{4}{3}$.
Since $L_1$ and $L_2$ are parallel, their slopes are equal.
Slope of $L_1$: $\frac{a}{3}$. Slope of $L_2$: $\frac{2}{3}$.
Thus: \[ \frac{a}{3} = \frac{2}{3} \implies a = 2 \] So, $p = -\frac{5}{2}$, $q = \frac{5}{3}$.
Points are $\left( -\frac{5}{2}, \frac{5}{3} \right)$ and $\left( -2, \frac{4}{3} \right)$.
Slope of the line through these points: \[ m = \frac{\frac{4}{3} - \frac{5}{3}}{-2 - \left( -\frac{5}{2} \right)} = \frac{-\frac{1}{3}}{\frac{1}{2}} = -\frac{2}{3} \] Equation using point $\left( -2, \frac{4}{3} \right)$: \[ y - \frac{4}{3} = -\frac{2}{3} (x + 2) \] \[ 3y - 4 = -2x - 4 \implies 2x + 3y = 0 \] Option (2) is correct. Options (1), (3), and (4) do not satisfy the points or slope.
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