Answer:
Equation of required line is: [tex]\mathbf{x+3y=-9 }[/tex]
Step-by-step explanation:
We need to find equation for a line that's perpendicular to [tex]6x-2y=8[/tex] when it passes through (6,-5)
We will use the slope-intercept form: [tex]y=mx+b[/tex] where m is slope and b is y-intercept.
Finding Slope m:
When two lines are perpendicular there slopes are [tex]m=-\frac{1}{m}[/tex]
Finding slope of given equation first: [tex]6x-2y=8[/tex]
Writing in slope-intercept form:
[tex]6x-2y=8\\-2y=-6x+8\\y=\frac{-6x}{-2}+\frac{8}{-2}\\y=3x-4[/tex]
Comparing it with slope intercept form: [tex]y=mx+b[/tex] , m= 3
So, slope of given equation is 3
Now, the slope of required equation will be [tex]m=-\frac{1}{m}[/tex]
So, Slope m=-1/3
Finding y-intercept
Using slope m=-1/3 and point (6,-5) we can find y-intercept
[tex]y=mx+b\\-5=\frac{-1}{3}(6)+b\\-5=-2+b\\b=-5+2\\b=-3[/tex]
So, y-intercept is b=-3
Now, finding the equation of required line having Slope m=-1/3 and y-intercept b= -3
[tex]y=mx+b\\y=-\frac{1}{3}x-3[/tex]
Writing in standard form:
[tex]y=-\frac{1}{3}x-3\\y=\frac{-x-9}{3}\\3y=-x-9\\x+3y=-9[/tex]
So, equation of required line is: [tex]\mathbf{x+3y=-9 }[/tex]
