Question

What is the equation of the line in slope-intercept form that is perpendicular to the line y = x – 2 and passes through the point (–12, 10)?

y = x – 6
y = x + 6
y = x + 26
y = x + 10v

Answer 1

Answer:Slope intercept form : y=mx+b , where m=slope, y=y-intercept

Perpendicular to : y = 3/4x - 2 ; Passes through (-12, 10)

**Remember : When finding the slope of a perpendicular line, we must get the negative reciprocal **

Ex of negative reciprocal : 3/2 → -2/3

The original line has a slope of 3/4, therefore our perpendicular slope is -4/3

Now we have to use the point-slope formula to find our new equation.

y - y₁ = m(x - x₁)

Passes through (-12, 10) ; slope = -4/3

y₁ = 10 ; x₁ = -12

Simply plug in the numerals for y₁ and x₁ as well as m.

y - (10) = -4/3(x - (-12))

Simplify.

y - 10 = -4/3(x + 12)

Simplify.

y - 10 = -4/3x - 48/3

Simplify -48/3

-48/3 = -16

y - 10 = -4/3 - 16

Now, we need to put this into slope-intercept form.

**Remember : slope intercept form is y=mx+b where m=slope, y=y-intercept**

So, add 10 to both sides.

y = -4/3x - 16 + 10

Simplify.

y = -4/3x - 6

~hope I helped!~

Explanation:

Answer 2

The answer is:

y = -1/2x + 4

Work/explanation:

If one line is perpendicular to another, the slopes of the two lines are inverse reciprocals of each other.

The slope of y = x - 2 is 1; the inverse reciprocal of that is -1/2.

So I proceed to plug the data into point slope:

`\sf{y-y_1=m(x-x_1)}`

where m = slope.

Plug in the data:

`\sf{y-10=-(1)/(2)(x-(-12)}`

Simplify step by step

`\sf{y-10=-(1)/(2)(x+12)}`

`\sf{y-10=-(1)/(2)x-6}`

Add 10 on each side:

`\sf{y=-(1)/(2)x-6+10}`

`\sf{y=-(1)/(2)x+4}`

Hence, the equation is y=-1/2x + 4.