Find the equation of A line.

The line has the smae x-intercept as 3x-2y+12=0 and passes through (-4,3)

The line is paralle to x= -4 and passes through (-2,5)

The line is perpendicular to y = 5 and passes through the point (6,0)

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(1) The equation of a line with the same x-intercept as 3x-2y+12=0 and contains ((-4,3).

The intercept of the given line is found by setting y=0 and solving for x. Thus 3x+12=0 ==> x=-4.

We now have two points on the line; (-4,3) and (-4,0). This is a vertical line (the x values are the same) **so the line is x=-4.**

(2) The equation of a line parallel to x=-4 containing the point (-2,5).

Again, since x=-4 is a vertical line, the line we seek is a vertical line. If the line contains the point (-2,5) then every point on the line has an x-coordinate of -2. **Thus the line is x=-2.**

(3) The equation of a line perpendicular to y=5 containing the point (6,0).

The line y=5 is horizontal, so the line perpendicular to it is vertical and of the form x=k. Since the line contains (6,0) every x-coordinate is 6 and **the line is x=6.**

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