Question

Identify an equation in point-slope form for the line perpendicular to y = -1/3x - 6 that passes through (-1, 5).

A. y + 1 = 3(x - 5)
B. y + 5 = 1/3(x - 1)
C. y - 5 = -1/3(x + 1)
D. y - 5 =3(x + 1)

Answer 1

D. y - 5 =3(x + 1)

Explanation:

Given equation of line:

`y = -(1)/(3)x-6`

Comparing it with the standard form of equation of line

y=mx+b

m = -1/3

Let m1 be the slope of line perpendicular to the given line.

We know that the product of slopes of two perpendicular lines is -1.

`m*m_1 = -1\\-(1)/(3) * m_1 = -1\\ m_1 = -1 * -(3)/(1)\\ m_1 = 3`

The equation of line in point slope form is:

`y-y_1 = m(x-x_1)`

where x_1 and y_1 is the point from which the line passes.

So, putting the values of slope and point,

`y-5 = 3[x-(-1)]\\y-5=3(x+1)`

Option D is correct ..

Answer 2

D. y - 5 =3(x + 1)

Explanation:

y = -1/3x -6

The slope is -1/3

Take the negative reciprocal to find the slope of the line that is perpendicular

- (-3) = 3

The slope would be 3

We have the slope and a point

Using point slope form

y-y1 = m(x-x21)

y-5 = 3(x--1)

y-5 = 3(x+1)