Question

Write the point-stope form of the linear equation that passes through the point (2, - 3) and is perpendicular to 3x - y = 4 .

Answer 1

Answer:

Equation of a line that is perpendicular to y = 3x-4 and that passes through the point (2 -3) is  $y=-\frac{1}{3} x-\frac{7}{3}$

Solution:

The slope - intercept form equation of line is given as

y = mx+c    ----- (1)

Where m is the slope of the line. The coefficient of “x” is the value of slope of the line.

Given that

3x  -y = 4          

Converting above equation in slope intercept form,

y = 3x - 4 ----  (2)

On comparing equation (1) and (2) we get slope of equation (2) is m=3

Consider equation of the line which is perpendicular to equation (2) is $y=m_{1} x+b$ --- eqn 3

If two lines having slope m1 and m1 are perpendicular then relation between their slope is $m_{1} m_{2}=-1$

That is if slope of the line (2) is 3 then slope of equation (3) is  $m_{1}=-\frac{1}{3}$

On substituting value of m1 in equation (3), we get

$y=-\frac{1}{3} x+b$ --- eqn 4

Given that equation (4) passes through (2, -3) that is x = 2 and y = -3. So on substituting value of x and y in equation (4),

$-3=-\frac{2}{3}+b$

On simplifying above equation,

$\begin{aligned} b &=\frac{2}{3}-3 \\ b &=\frac{2-9}{3} \\ b &=-\frac{7}{3} \end{aligned}$

On substituting value of b in equation (4),

$y=-\frac{1}{3} x-\frac{7}{3}$

Hence equation of a line that is perpendicular to y=3x-4 and that passes through the point (2 -3)  is $y=-\frac{1}{3} x-\frac{7}{3}$