# using y -3 = 3(x + 1) what is the equation in standard form of a perpendicular line that passes through (5,1)? What is the x interecept of the perpendicular line?

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The given equation of the line is `y-3=3(x + 1)` .

Rearranging it in the slope-intercept form i.e. `y=mx+b` where `m` is the slope of the line we get:

`y-3=3x+3`

`rArr y=3x+6`

Thus, slope of the line, `m=3` .

Again, the slope of the perpendicular line is the negative reciprocal of the slope of the original line.

So, slope of the perpendicular line=`-1/3`

Now, use the point-slope form of the equation of the straight line to find the equation of the perpendicular line.

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

Here, `(x_1,y_1)=(5,1)` and `m=-1/3` . Plug in these values in the above equation to get:

`y-1=-1/3(x-5)`

Multiply both sides by `3` :

`3y-3=-x+5`

`rArr x+3y=8`

**Therefore, the equation in standard form of a perpendicular line that passes through (5,1) is** **x+3y=8.**

To find the x intercept of the perpendicular line set y=0.

So, `x+3*0=8`

`rArr x=8`

**Hence, the x intercept of the perpendicular line is 8.**