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
[tex]y = \frac{3}{4}x - 3[/tex]
On comparing the above equation with slope intercept form,
[tex]m = \frac{3}{4}[/tex]
Thus the slope of line is [tex]m = \frac{3}{4}[/tex]
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
[tex]\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}[/tex]
Now we have to find the equation of line with slope [tex]m = \frac{-4}{3}[/tex] and passes through (5, -2)
Substitute [tex]m = \frac{-4}{3}[/tex] and (x, y) = (5, -2) in eqn 1
[tex]-2 = \frac{-4}{3}(5) + c\\\\-2 = \frac{-20}{3} + c\\\\-6 = -20 + 3c\\\\3c = 14\\\\c = \frac{14}{3}[/tex]
The required equation of line is:
Now substitute [tex]m = \frac{-4}{3}[/tex] and [tex]c = \frac{14}{3}[/tex]
[tex]y = \frac{-4}{3}x + \frac{14}{3}[/tex]
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,
[tex]y = \frac{-4}{3}x + \frac{14}{3}[/tex]
3y = -4x + 14
4x + 3y = 14
Thus the equation of line in standard form is found out
