What is an equation of the line that is perpendicular to y- 4 = 2(x-6) and
passes through the point (-3,-5)?
o a. y + 5 = 2(x+3)
o b. y-5=-2(x-3)
o c. v-5-x-3)
od. y +5 - -}(x+3)

D. Y+5=-(1/2)*(x+3)

Step-by-step explanation:

Perpendicular Lines are those with the following condition:

y=a*x+b (1)

y=c*x+d (2)

Where 'a' and 'c' are the respective slope

If These two lines are perpendicular, then

a=- 1/c

Equation (1) for our case is written as y=2x-8, meaning that a=2 and b = -8

Using those principles we have that the slope for our needed line ('c') has to be -(1/2).

Now we most use the given point to find the remaining term of the equation (d) so, evaluate (-3,-5) in eq (2) to have this:

-5=(-1/2)*(-3)+d

resulting that d=-5-(3/2)

Eq (2) is written now as the following: y= (-1/2)*x - (5+3/2)

Rearranging terms, we have the following:

y+5=(-1/2)*x-(3/2)

where you can obtain a more pretty expression:

y+5=(-1/2)*(x+3)

its an acute right triangle

step-by-step explanation: