The equation of the line passing through the | vedclass

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

Question

(C) Step $1$: Find the point of intersection of the lines $3x - 2y - 1 = 0$ and $x - 4y + 3 = 0$.Multiply the second equation by $3$: $3x - 12y + 9 =

Vedclass · Accepted Answer

$x - y = \pi(1 - y)$

Vedclass · Answer

(C) Step $1$: Find the point of intersection of the lines $3x - 2y - 1 = 0$ and $x - 4y + 3 = 0$.Multiply the second equation by $3$: $3x - 12y + 9 = 0$.Subtract this from the first equation: $(3x - 2y - 1) - (3x - 12y + 9) = 0$ $\Rightarrow 10y - 10 = 0$ $\Rightarrow y = 1$.Substitute $y = 1$ into $x - 4y + 3 = 0$: $x - 4(1) + 3 = 0$ $\Rightarrow x - 1 = 0$ $\Rightarrow x = 1$.The point of intersection is $(1, 1)$.Step $2$: Find the equation of the line passing through $(1, 1)$ and $(\pi, 0)$.The slope $m = \frac{0 - 1}{\pi - 1} = \frac{-1}{\pi - 1}$.The equation is $y - 0 = \frac{-1}{\pi - 1}(x - \pi)$.$(\pi - 1)y = -x + \pi$ $\Rightarrow x + (\pi - 1)y = \pi$ $\Rightarrow x - \pi = -(\pi - 1)y$ $\Rightarrow x - \pi = -\pi y + y$ $\Rightarrow x - y = \pi - \pi y$ $\Rightarrow x - y = \pi(1 - y)$.

The equation of the line passing through the point of intersection of the lines $3x - 2y - 1 = 0$ and $x - 4y + 3 = 0$ and the point $(\pi, 0)$ is:

Similar Questions

$A$ line passes through the point of intersection of $2x + y = 5$ and $x + 3y + 8 = 0$ and is parallel to the line $3x + 4y = 7$. Find the equation of this line.

The points $(3a, 0)$,$(0, 3b)$,and $(a, 2b)$ are:

The equation of the line parallel to the $y$-axis and passing through the point of intersection of the lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ is:

The lines $15x - 18y + 1 = 0$,$12x + 10y - 3 = 0$ and $6x + 66y - 11 = 0$ are

Which of the following lines is concurrent with the lines $3x + 4y + 6 = 0$ and $6x + 5y + 9 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module