Find the equation of the line passing through | vedclass

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

Question

(C) The equation of a line passing through the intersection of $L_1: x + y - 3 = 0$ and $L_2: 2x - y + 1 = 0$ is given by $(x + y - 3) + \lambda(2x -

Vedclass · Accepted Answer

$4x + y - 5 = 0$

Vedclass · Answer

(C) The equation of a line passing through the intersection of $L_1: x + y - 3 = 0$ and $L_2: 2x - y + 1 = 0$ is given by $(x + y - 3) + \lambda(2x - y + 1) = 0$.Since this line passes through the point $(2, -3)$,we substitute $x = 2$ and $y = -3$ into the equation:$(2 - 3 - 3) + \lambda(2(2) - (-3) + 1) = 0$$-4 + \lambda(4 + 3 + 1) = 0$$-4 + 8\lambda = 0$$8\lambda = 4 \implies \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ back into the family of lines equation:$(x + y - 3) + \frac{1}{2}(2x - y + 1) = 0$Multiply by $2$ to clear the fraction:$2(x + y - 3) + (2x - y + 1) = 0$$2x + 2y - 6 + 2x - y + 1 = 0$$4x + y - 5 = 0$.

Find the equation of the line passing through the intersection of the lines $x + y - 3 = 0$ and $2x - y + 1 = 0$ and the point $(2, -3)$.

Similar Questions

For the family of lines given by the equation $(2 + k)x + (1 + k)y = 5 + 7k$,which of the following statements is true for different values of $k$?

If the straight lines $2x - y + 1 = 0$,$4x + y + 2 = 0$,and $x + y - k = 0$ are concurrent,then $k$ equals

The equation of a line passing through the point of intersection of the lines $x + 5y + 7 = 0$ and $3x + 2y - 5 = 0$,and perpendicular to the line $7x + 2y - 5 = 0$,is given by

For $a \neq b \neq c$,if the lines $x+2ay+a=0$,$x+3by+b=0$ and $x+4cy+c=0$ are concurrent,then $a, b, c$ are in

If the lines $ax + by + c = 0$,$bx + cy + a = 0$,and $cx + ay + b = 0$ are concurrent,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module