Answer:
The second line cuts the x-axis at x = -5/2.
Step-by-step explanation:
The given line:
[tex]3x-2y=12[/tex]
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:
[tex]-2y=-3x+12[/tex]
So:
[tex]\displaystyle y=\frac{3}{2}x-6[/tex]
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:
[tex]\displaystyle y=\frac{3}{2}(2)-6=3-6=-3[/tex]
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:
[tex]y-y_1=m(x-x_1)[/tex]
Substitute:
[tex]\displaystyle y-(-3)=-\frac{2}{3}(x-2)[/tex]
Distribute:
[tex]\displaystyle y+3=-\frac{2}{3}x+\frac{4}{3}[/tex]
So:
[tex]\displaystyle y=-\frac{2}{3}x-\frac{5}{3}[/tex]
The second line will cut the x-axis when y = 0.
So:
[tex]\displaystyle 0 =-\frac{2}{3}x-\frac{5}{3}[/tex]
Multiply both sides by -3:
[tex]0=2x+5[/tex]
Solve for x:
[tex]\displaystyle x=-\frac{5}{2}[/tex]
The second line cuts the x-axis at x = -5/2.
