Question

Write an equation of the line passing through the point (-8, -4) that is perpendicular to the line given by y= 1/6 x+3.

Answer 1

Answer:

y+4=-6(x+8)  point-slope form

y=-6x-52 slope-intercept form

6x+y=-52 standard form

Step-by-step explanation:

Slope-intercept form of a line is y=mx+b where m is the slope and b is the y-intercept.

Lines that are perpendicular have opposite reciprocal slopes.

So the slope of y=(1/6)x+3 is 1/6.

The opposite reciprocal of (1/6) is -6.

So the equation for the line we are looking for is in the form:

y=-6x+b         (Since the slope of our new line is -6)

Now we want our line to go through (-8,-4).

So plug that in:

-4=-6(-8)+b

-4=48+b

Subtract 48 on both sides:

-52=b

The equation for the line we are looking for is

y=-6x-52.

Now you could do other forms.

Another one is point-slope form.

We already know it goes through (-8,-4) and a slope of -6.

Point slope form is: y-y1=m(x-x1) where m is the slope and (x1,y1) is a point on the line.

Plug in the information to get:

y-(-4)=-6(x-(-8))

y+4=-6(x+8)

I'm going to do one more form.

Standard form is ax+by=c where a,b,c are integers.

y=-6x-52

Add 6x on both sides:

6x+y=-52