Question

Can I please have help with this one

Answer 1

The second line cuts the x-axis at x = -5/2.

Explanation:

The given line:

`3x-2y=12`

Intersects a second line at x = 2.

This second line is perpendicular to the first.

And we want to find the second line's x-intercept.

First, rearrange the first line into slope intercept form:

`-2y=-3x+12`

So:

`\displaystyle y=(3)/(2)x-6`

We can see that the slope of the first line is 3/2.

It intersects the second line at x = 2.

So, the y-value of both the first and second line at x = 2 is the same.

Find the y-value by using the first line:

`\displaystyle y=(3)/(2)(2)-6=3-6=-3`

So, a point on both lines is (2, -3).

Since the second line is perpendicular to the first, its slope it the first's negative reciprocal.

The negative reciprocal of 3/2 is -2/3.

So, the slope of the second line is -2/3.

And we determined that it passes through (2, -3).

Then by the point-slope form:

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

Substitute:

`\displaystyle y-(-3)=-(2)/(3)(x-2)`

Distribute:

`\displaystyle y+3=-(2)/(3)x+(4)/(3)`

So:

`\displaystyle y=-(2)/(3)x-(5)/(3)`

The second line will cut the x-axis when y = 0.

So:

`\displaystyle 0 =-(2)/(3)x-(5)/(3)`

Multiply both sides by -3:

`0=2x+5`

Solve for x:

`\displaystyle x=-(5)/(2)`

The second line cuts the x-axis at x = -5/2.