Answer:
An equation of the line that is perpendicular to y-4=2(x-6) and passes thru the point (-3,-5) will be:
Step-by-step explanation:
We know the point-slope form of the line equation is
[tex]y-y_1=m\left(x-x_1\right)[/tex]
where
Given the line
y-4 = 2(x-6)
comparing with the point-slope form of the line equation
The slope = m = 2
We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:
slope = m = 2
so, the slope of the line perpendicular to y-4 = 2(x-6) will be:
– 1/m = -1/2 = -1/2
substituting the values of the slope = -1/2 and the point (-3, -5)
[tex]y-y_1=m\left(x-x_1\right)[/tex]
[tex]y-\left(-5\right)=-\frac{1}{2}\left(x-\left(-3\right)\right)[/tex]
[tex]y+5=-\frac{1}{2}\left(x-\left(-3\right)\right)[/tex]
[tex]y+5=-\frac{1}{2}\left(x+3\right)[/tex]
Subtract 5 from both sides
[tex]y+5=-\frac{1}{2}\left(x+3\right)[/tex]
Simplify
[tex]y=-\frac{1}{2}x-\frac{13}{2}[/tex]
Therefore, an equation of the line that is perpendicular to y-4=2(x-6) and passes thru the point (-3,-5) will be:
