182k views
1 vote
Find the equation of a line perpendicular to y+7=-4/3x that passes through the point (-8,-1)

1 Answer

3 votes

Answer:

y = (3/4)x + 5 or x = y - (20/3) or 4y = 3x + 20

Step-by-step explanation:

To find the equation of a line that is perpendicular to the given line y + 7 = -4/3x and passes through the point (-8, -1), you can use the following steps:

First, determine the slope of the given line. The given line is in the slope-intercept form (y = mx + b), where m is the slope. In this case, the slope is -4/3.

The slope of a line perpendicular to the given line is the negative reciprocal of the slope of the given line. So, the slope of the perpendicular line is 3/4.

Use the point-slope form of the equation of a line to find the equation of the perpendicular line. The point-slope form is:

y - y1 = m(x - x1),

where (x1, y1) is the point through which the line passes, and m is the slope.

In this case, (x1, y1) = (-8, -1), and m = 3/4. So, the equation of the perpendicular line is:

y - (-1) = (3/4)(x - (-8)).

Simplify this equation:

y + 1 = (3/4)(x + 8).

To put it in standard form, you can multiply both sides of the equation by 4 to eliminate the fraction:

4(y + 1) = 3(x + 8).

Now, distribute on both sides:

4y + 4 = 3x + 24.

To isolate y, subtract 4 from both sides:

4y = 3x + 20.

Finally, divide by 4 to solve for y:

y = (3/4)x + 5.

So, the equation of the line perpendicular to y + 7 = -4/3x and passing through the point (-8, -1) is y = (3/4)x + 5.

User MrJre
by
7.9k points