Question

How do you find the equation of the line perpendicular to y+5=3(x-2) that passes through the point (6, 2)?

Answer 1

Let,the equation of line be `y=mx+c`, where, `m` is its slope and `c` is the `Y` intercept.

Now,arranging the given equation in the above mentioned form to get its slope,

Given, `y+5=3(x-2)=y=3x-11`

So,its slope is `3`

Now,for two lines to be mutually perpendicular,their product of slope must be `-1`

So, `m*3=-1`

or. `m=-1/3`

So,our required line equation becomes, `y=-1/3 x+c`

Now,given,that the line passes through `(6,2)`,so putting the values in the equation to get `c`

So, `2=(-1/3)*6 +c`

or, `c=4`

So,the equation of the line is `y=-1/3 x+4`

or, `3y+x=12` graph{3y+x=12 [-10, 10, -5, 5]}

Answer 2

`y-2 = -1/3(x-6)`

Or `y = -1/3x +4`

Or `x+3y =12`

Or `x+3y-12=0`

Explanation:

There are several forms for the equation of a straight line:

`y = mx+c," and " ax +by +c =0" "` are well known.

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

where `(x_1, y_1)` is a point and `m` is the slope.

This is exactly the form we have for the given equation:

`y+5=color(blue)(3)(x-2)" "rarr :. color(blue)(m= 3)`

If lines are perpendicular, the product of their slopes is `-1`

One is the negative reciprocal of the other, so:

`m_1 = 3 " "rArr" "m_2 =-1/3`

Using `m= -1/3` and the point `(6,2)`

`y-y_1 = m(x-x_1)" "` gives`" "y-2 = -1/3(x-6)`

Or in another form:

`y= -1/3x+2+2" "rarr y = -1/3x +4`

Or even

`3y =-x+12" "rarr" "x+3y =12`