Question

Find the equation of a line perpendicular to3, x, plus, 2, y, equals, minus, 43x+2y=−4that passes through the point left bracket, 6, comma, minus, 8, right bracket(6,−8).

Answer
Multiple Choice Answers
minus, 2, x, plus, 3, y, equals, minus, 36−2x+3y=−36
y, equals, start fraction, 3, divided by, 2, end fraction, x, minus, 2y=
2
3
​
x−2
y, equals, minus, start fraction, 2, divided by, 3, end fraction, x, minus, 12y=−
3
2
​
x−12
3, x, plus, 2, y, equals, 23x+2y=2

Answer 1

The line perpendicular to 3x + 2y = -4 passing through (6, -8) is found by using the negative reciprocal of the given line's slope. This results in y = (2/3)x - 12.

Step-by-step explanation:

The question is asking us to find the equation of a line that is perpendicular to the given line 3x + 2y = -4 and passes through the point (6, -8).

To find the perpendicular line, we first need to determine the slope of the given line by rewriting it in slope-intercept form (y = mx + b), where m is the slope. The given equation can be transformed to y = -1.5x - 2, giving us a slope of -1.5. A line perpendicular to this will have a slope that is the negative reciprocal of -1.5, which is 2/3.

Using the point-slope form of a line's equation, y - y1 = m(x - x1), with m being 2/3 and the point (6, -8), we get:

y - (-8) = (2/3)(x - 6)

Expanding and simplifying, we obtain:

y + 8 = (2/3)x - 4

y = (2/3)x - 12

Therefore, the equation of the line perpendicular to 3x + 2y = -4 through (6, -8) is y = (2/3)x - 12.