What is the equation of the line which passes through the point of intersection of the lines y=x and x+y=6 and which is perpendicular to the line with equation 3x+6y=12?

1 Answer
Jan 27, 2018

The line is y=2x-3y=2x3.

Explanation:

First, find the intersection point of y=xy=x and x+y=6x+y=6 using a system of equations:

y+x=6y+x=6
=>y=6-xy=6x

y=xy=x
=>6-x=x6x=x
=>6=2x6=2x
=>x=3x=3
and since y=xy=x:
=>y=3y=3

The lines' intersection point is (3,3)(3,3).

Now we need to find a line that goes through the point (3,3)(3,3) and is perpendicular to the line 3x+6y=123x+6y=12.

To find the slope of the line 3x+6y=123x+6y=12, convert it to slope-intercept form:

3x+6y=123x+6y=12

6y=-3x+126y=3x+12

y=-1/2x+2y=12x+2

So the slope is -1/212. The slopes of perpendicular lines are opposite reciprocals, so that means the slope of the line we're trying to find is -(-2/1)(21) or 22.

We can now use point-slope form to make an equation for our line from the point and slope that we found before:

y-y_1=m(x-x_1)yy1=m(xx1)

=>y-3=2(x-3)y3=2(x3)

=>y-3=2x-6y3=2x6

=>y=2x-3y=2x3

The line is y=2x-3y=2x3.