Answer:
The equation in slope-intercept form perpendicular to the given equation [tex]y = -3x + 4[/tex] through the given point (6,-2) is [tex]\mathbf{y=\frac{1}{3}x}[/tex]
Step-by-step explanation:
We need to write an equation in slope-intercept form perpendicular to the given equation through the given point
y = -3x + 4
(6,-2)
The general equation in slope intercept form is: [tex]y=mx+b[/tex] where m is slope and b is y-intercept.
We need to find slope and y-intercept.
Finding slope:
The given line is perpendicular to required line, so there slopes are opposite
The slope of given line [tex]y = -3x + 4[/tex] is m = -3, (By comparing it with general equation y=mx+b, we get m = -3)
The slope of required line is: [tex]m = \frac{1}{3}[/tex]
Finding y-intercept:
y-intercept can be found by using slope [tex]m = \frac{1}{3}[/tex] and the point(6,-2)
[tex]y=mx+b\\-2=\frac{1}{3}(-6)+b\\-2=-2+b\\b=-2+2\\b=0[/tex]
So, we get y-intercept b =0
Equation of required line:
So, the equation of required line having slope [tex]m = \frac{1}{3}[/tex] and y-intercept b =0 is:
[tex]y=mx+b\\y=\frac{1}{3}x+0\\y=\frac{1}{3}x[/tex]
So, the equation in slope-intercept form perpendicular to the given equation [tex]y = -3x + 4[/tex] through the given point (6,-2) is [tex]\mathbf{y=\frac{1}{3}x}[/tex]
