Answer: [tex]\frac{4}{3}x+y=\frac{14}{3}[/tex]
Step-by-step explanation:
The Slope-Intercept form of an equation of the line is:
[tex]y=mx+b[/tex]
Where "m" is the slope and "b" is the y-intercept.
The equation of the line in Standard form is:
[tex]Ax + By = C[/tex]
Where "A" is a positive integer, and "B" and "C" are integers.
Given the equation:
[tex]3x-4y=12[/tex]
Solve for "y" in order to write it in Slope-Intercept form:
[tex]3x-4y=12\\\\-4y=-3x+12\\\\y=\frac{3}{4}x-3[/tex]
Notice that:
[tex]m=\frac{3}{4}[/tex]
Since the slopes of perpendicular lines are negative reciprocals, the slope of the other line is:
[tex]m=-\frac{4}{3}[/tex]
Knowing tha it passes through the point (5, -2), you can substistute the slope and the coordinates of that point into [tex]y=mx+b[/tex] and solve for "b":
[tex]-2=-\frac{4}{3}(5)+b\\\\-2=-\frac{20}{3}+b\\\\-2+\frac{20}{3}=b\\\\b=\frac{14}{3}[/tex]
Then, the equation of this line in Slope-Intercept form is:
[tex]y=-\frac{4}{3}x+\frac{14}{3}[/tex]
In order to write it in Standard form, add [tex]-\frac{4}{3}x[/tex] to both sides of the equation. Then:
[tex]\frac{4}{3}x+y=-\frac{4}{3}x+\frac{14}{3}+\frac{4}{3}x\\\\\frac{4}{3}x+y=\frac{14}{3}[/tex]
