If the straight line L equiv 3x+4y-k=0 cuts t | vedclass

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

Question

(C) The point of intersection $R$ of the line segment $PQ$ dividing it in the ratio $4:1$ is given by the section formula:$R = \left(\frac{4(1)+1(2)}{

Vedclass · Accepted Answer

$5x-5y-3=0$

Vedclass · Answer

(C) The point of intersection $R$ of the line segment $PQ$ dividing it in the ratio $4:1$ is given by the section formula:$R = \left(\frac{4(1)+1(2)}{4+1}, \frac{4(1)+1(-1)}{4+1}\right) = \left(\frac{6}{5}, \frac{3}{5}\right)$Since $R$ lies on the line $L \equiv 3x+4y-k=0$,we have:$3\left(\frac{6}{5}\right) + 4\left(\frac{3}{5}\right) - k = 0$ $\Rightarrow \frac{18+12}{5} = k$ $\Rightarrow k=6$Thus,$L \equiv 3x+4y-6=0$.The equation of the line $PQ$ passing through $(2,-1)$ and $(1,1)$ is:$y - 1 = \frac{1-(-1)}{1-2}(x-1)$ $\Rightarrow y-1 = -2(x-1)$ $\Rightarrow 2x+y-3=0$The family of lines passing through the intersection of $L=0$ and $PQ=0$ is given by:$(3x+4y-6) + \lambda(2x+y-3) = 0$$(3+2\lambda)x + (4+\lambda)y - (6+3\lambda) = 0$Since this line is parallel to $y=x$ (slope $m=1$),its slope must be $1$:$-\frac{3+2\lambda}{4+\lambda} = 1$ $\Rightarrow -3-2\lambda = 4+\lambda$ $\Rightarrow 3\lambda = -7$ $\Rightarrow \lambda = -\frac{7}{3}$Substituting $\lambda = -\frac{7}{3}$ into the equation:$(3+2(-\frac{7}{3}))x + (4-\frac{7}{3})y - (6+3(-\frac{7}{3})) = 0$$(3-\frac{14}{3})x + (\frac{12-7}{3})y - (6-7) = 0$$-\frac{5}{3}x + \frac{5}{3}y + 1 = 0$ $\Rightarrow -5x+5y+3=0$ $\Rightarrow 5x-5y-3=0$

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

Similar Questions

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:

$A$ straight line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$ square units. The equation of the line $L$ can be

The point on the line $3x + 4y = 5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is

The coordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC$ $(AC = BC)$ are $(-2, 3)$ and $(2, 0)$ respectively. $A$ line parallel to $AB$ and having a $y$-intercept equal to $\frac{43}{12}$ passes through $C$. Then the coordinates of $C$ are:

$P$ is a point on either of the two lines $y - \sqrt{3}|x| = 2$ at a distance of $5 \ units$ from their point of intersection. The coordinates of the foot of the perpendicular from $P$ on the bisector of the angle between them are

Vedclass Test Series

Exam Paper Generator

Online Exam Module