The line $L_1$ passes through the points $(-1, 2)$ and $(3, 6)$. If $L_1$ divides the line $L_2$ (which passes through $(3, -1)$) in the ratio $1:3$ internally, and the point of intersection lies on $L_1$, find the equation of $L_2$ given that it is perpendicular to $L_1$.
- A$4x - 3y - 15 = 0$
- B$x + y - 2 = 0$
- C$x + y + 2 = 0$
- D$x - y - 4 = 0$
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View SolutionMatch the items given in List-$I$ to the items given in List-$II$. List-$I$ List-$II$ $A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$ $1$. $2x - 5y + 4 = 0$ $B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes $2$. $3x + 5y = 3$ $C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is $3$. $10x - 15y + 4 = 0$ $D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is $4$. $10x - 15y = 4$ $5$. $5x - 2y - 20 = 0$
| List-$I$ | List-$II$ |
| $A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$ | $1$. $2x - 5y + 4 = 0$ |
| $B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes | $2$. $3x + 5y = 3$ |
| $C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is | $3$. $10x - 15y + 4 = 0$ |
| $D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is | $4$. $10x - 15y = 4$ |
| $5$. $5x - 2y - 20 = 0$ |
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