L_1 equiv ax-3y+5=0 and L_2 equiv 4x-6y+8=0 a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Since $L_1$ and $L_2$ are parallel,their slopes are equal.For $L_2 \equiv 4x-6y+8=0$,the slope is $m_2 = -\frac{4}{-6} = \frac{2}{3}$.Since $L_1$

Vedclass · Accepted Answer

$2x+3y=0$

Vedclass · Answer

(B) Since $L_1$ and $L_2$ are parallel,their slopes are equal.For $L_2 \equiv 4x-6y+8=0$,the slope is $m_2 = -\frac{4}{-6} = \frac{2}{3}$.Since $L_1$ is parallel to $L_2$,the slope of $L_1 \equiv ax-3y+5=0$ is $\frac{a}{3} = \frac{2}{3}$,which gives $a=2$.For $L_1: 2x-3y+5=0$,the $X$-intercept $p$ is found by setting $y=0$,so $2p+5=0 \implies p = -\frac{5}{2}$.The $Y$-intercept $q$ is found by setting $x=0$,so $-3q+5=0 \implies q = \frac{5}{3}$. Point $(p, q) = (-\frac{5}{2}, \frac{5}{3})$.For $L_2: 4x-6y+8=0$,the $X$-intercept $m$ is found by setting $y=0$,so $4m+8=0 \implies m = -2$.The $Y$-intercept $n$ is found by setting $x=0$,so $-6n+8=0 \implies n = \frac{4}{3}$. Point $(m, n) = (-2, \frac{4}{3})$.The slope of the line passing through $(p, q)$ and $(m, n)$ is $M = \frac{\frac{4}{3} - \frac{5}{3}}{-2 - (-\frac{5}{2})} = \frac{-\frac{1}{3}}{\frac{1}{2}} = -\frac{2}{3}$.The equation of the line is $y - \frac{4}{3} = -\frac{2}{3}(x + 2) \implies 3y - 4 = -2x - 4 \implies 2x + 3y = 0$.

$L_1 \equiv ax-3y+5=0$ and $L_2 \equiv 4x-6y+8=0$ 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

Similar Questions

If the straight line $L \equiv 3x+4y-k=0$ cuts the line segment joining the points $P(2,-1)$ and $Q(1,1)$ in the ratio $4:1$,then the equation of the line parallel to the line $y=x$ and concurrent with the lines $PQ$ and $L=0$ is

The equation of a straight line passing through $(3, 2)$ and cutting an intercept of $2 \text{ units}$ between the lines $3x + 4y = 11$ and $3x + 4y = 1$ is:

Straight lines $2x + y = 5$ and $x - 2y = 3$ intersect at the point $A$. Points $B$ and $C$ are chosen on these two lines such that $AB = AC$. Then the equation of a line $BC$ passing through the point $(2, 3)$ is

The equation of one of the sides of an isosceles right-angled triangle,whose hypotenuse is $3x + 4y = 4$ and the vertex opposite to the hypotenuse is $(2, 2)$,is:

One side of a rectangle lies along the line $4x + 7y + 5 = 0.$ Two of its vertices are $(-3, 1)$ and $(1, 1).$ Then the equations of the other three sides are

Vedclass Test Series

Exam Paper Generator

Online Exam Module