Question

A line 3x-4y+12=0 meets the x axis at the point p , fimd the equation of the line through p , perpendicular to the line 3x+5y-15=0
Plzzzz solve this ... tomorrow is my exam

Answer 1

Answer:

$5x-3y+20=0$

Step-by-step explanation:

Given the equation of the line

$L_1: 3x-4y+12=0$

Putting y = 0 we get x = -4

Thus, the line $L_1$ meets the x-axis at $(-4,0)$

Line $L_2$ is

$L_2: 3x+5y-15=0$

or, $5y=-3x+15$

or, $y=\frac{-3}{5}x+3$

Comparing the above equation with the equation $y=mx+c$ we get

$m=-\frac{3}{5}$

Thus the slope of the line $L_2$ is $m_1=-\frac{3}{5}$

Let the slope of the line perpendicular to the line $L_2$ be $m_2$

Then

$m_1m_2=-1$

$\implies -\frac{3}{5}\times m_2=-1$

$\implies m_2=\frac{5}{3}$

Equation of the line whose slope is $\frac{5}{3}$

$y=\frac{5}{3}x+c$

Since this line passes through the point $(-4,0)$ therefore

$0=\frac{5}{3}\times (-4)+c$

$\implies c=\frac{20}{3}$

Therefore, the equation of the line will be

$y=\frac{5}{3}x+\frac{20}{3}$

or, $3y=5x+20$

or, $5x-3y+20=0$

Hope this helps.