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What is the equation of the line perpendicular to 2x – 3y = 13 that passes through the point (–6, 5)?

A. y = 2/3x + 9
B. y = -3/2x - 4
C. y = -3/2x - 13
D. y = 2/3x - 1

User Kina
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1 Answer

5 votes

Final answer:

The equation of the line perpendicular to 2x – 3y = 13 passing through the point (–6, 5) is found by taking the negative reciprocal of the original line's slope and using the point-slope form. It is option B: y = -\frac{3}{2}x - 4.

Step-by-step explanation:

The equation of a line that is perpendicular to another can be found by taking the negative reciprocal of the original line's slope. In this case, the given equation is 2x – 3y = 13. To find the slope of this line, we can rearrange it into slope-intercept form (y = mx + b), giving us y = 2/3x - 13/3 . The slope of this line is \frac{2}{3}, so the slope of the perpendicular line must be -3/2

Next, knowing that the line we're looking for passes through the point (–6, 5), we use the point-slope form: y - y_1 = m(x - x_1), where (x_1, y_1) is the point and m is the slope. Plugging in our values, we get y - 5 = -\frac{3}{2}(x + 6), which simplifies to y = -\frac{3}{2}x - 4. Therefore, the correct equation is option B: y = -3/2x - 4

User Tim Andersen
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