What is the equation that is perpendicular to the line y=2x-3 and passes through the point (-6,5)? Show all of your work. y = 2x + 17 y = 1/2x + 8 y = -2x - 7 y = -1/2x + 2

y = -1/2x + 2

Step-by-step explanation:

If two lines are perpendicular then the product of their slopes are -1

m1*m2 = -1

where m1 amd m2 is slope of two lines.

Given line

y = 2x-3\

it is in form of y = mx+c (slope intercept form)

where m is slope and c is y intercept

Thus, slope of line  y = 2x-3 is 2

Now we know that product of two lines are -1

let m be slope of other line

then m*2 = -1

m = -1/2

thus, slope of required line is -1/2

let the line be y = mx+c but m = -1/2

thus

y = -1/2 x + c

given that point(-6,5) lies on it

we use -6 in plca of x and 5 in place of y

5 = -1/2 *-6 + c

=>5 = 3 +c

=> c = 5-3 = 2

Thus, equation that is perpendicular to the line y=2x-3 and passes through the point (-6,5) is y = -1/2x + 2

you already gave the answer

step-by-step explanation:

step-by-step explanation:

the answer is 128