Question

Write an equation in standard form for the line that passes through (5, -2) and is perpendicular to 3x-4y=12

Answer 1

The equation in standard form for the line that passes through (5, -2) and is perpendicular to 3x - 4y = 12 is 4x + 3y = 14

Solution:

Given that line that passes through (5, -2) and is perpendicular to 3x - 4y = 12

We have to find the equation of line

The slope intercept form is given as:

y = mx + c  ------ eqn 1

Where "m" is the slope of line and "c" is the y-intercept

Let us first find the slope of line

Given equation of line is 3x - 4y = 12

On rearranging the above equation to slope intercept form,

3x - 4y = 12

4y = 3x - 12

$y = \frac{3}{4}x - 3$

On comparing the above equation with slope intercept form,

$m = \frac{3}{4}$

Thus the slope of line is $m = \frac{3}{4}$

We know that product of slope of given line and slope of line perpendicular to given line is always -1

slope of given line x slope of line perpendicular to given line = -1

$\begin{array}{l}{\frac{3}{4} \times \text { slope of line perpendicular to given line }=-1} \\\\ {\text {slope of line perpendicular to given line }=\frac{-4}{3}}\end{array}$

Now we have to find the equation of line with slope $m = \frac{-4}{3}$ and passes through (5, -2)

Substitute $m = \frac{-4}{3}$ and (x, y) = (5, -2) in eqn 1

$-2 = \frac{-4}{3}(5) + c\\\\-2 = \frac{-20}{3} + c\\\\-6 = -20 + 3c\\\\3c = 14\\\\c = \frac{14}{3}$

The required equation of line is:

Now substitute $m = \frac{-4}{3}$ and $c = \frac{14}{3}$

$y = \frac{-4}{3}x + \frac{14}{3}$

The standard form of an equation is Ax + By = C

x and y are variables and A, B, and C are integers

Rewriting the above equation,

$y = \frac{-4}{3}x + \frac{14}{3}$

3y = -4x + 14

4x + 3y = 14

Thus the equation of line in standard form is found out