Which equation represents the line that passes through the point (-2,5) and is perpendicular to the graph of x-6y =4?

y=6x+7

Step-by-step explanation:

First we need to convert x-6y=4 into slop-intercept form.  which is y=mx+b.

to do so we nee y isolated on one side. so we need to move 6 and x.

To love numbers or variables we need to use the inverse operations.

x is being added to -6y , so to move it we need to subtract it on both sides.

once we do so we get

-6y=4-x

now we need to move -6. -6 is being multiplied to y so to move it we need to divided it by -6. when we do so we get

y=4-x/-6

we are still not done

we need to divide -6 with both numbers....

4/6 and x/-6

since 4 cant be divide into 6 evenly we leave it as a fraction. But we do need to simplify the fraction.

4/2=2

-6/2=3

so simplest form of 4/-6 is -2/3

now x/-6 cant be solved so we need to put it as a fraction

since x is = to 1

we write -1/6x

so the slope-intercept form is

y=-1/6x - 2/3

Now to find a equation that is perpendicular to line y=-1/6x-2/3

we remember that any two perpendicular lines are negative reciprocals of each other. (basically means opposite... one is positive one is negative and the numbers flip)

so the slope for the perpendicular line ( also known as mx in the formula y=mx+b) is Postive 6/1x. ths can also be writen as 6x

now we need to find the y-intercept( also known as B in the  formula y=mx+b)

to find b we need to use (-2,5) and a formula

y-y1=m(x-x1)

so plug in the numbers we know

y+5=(6)(x-(-2))

ima do tis part on paper since a gets a bit trick the steps are on te pic below.

17.08

correct me if i am wrong.