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write an equation for a line that is perpendicular to the line y-4=2/3(x-9) that goes through the point (6,-5)

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2 votes

Answer:

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

Explanation:

y - 4 = (2/3)(x - 9)

y = (2/3)x - (2/3)(9) + 4

y = (2/3)x - 2

The slope of this line is 2/3.

A line that is perpendicular to this line will have a slope that is the negative reciprocal of 2/3. To find the negative reciprocal, we flip the fraction and change the sign:

slope of perpendicular line = -3/2

Now we can use the point-slope form of the equation of a line, which is:

y - y1 = m(x - x1)

where m is the slope of the line, and (x1, y1) is a point on the line.

Substituting the given point (6,-5) and the slope -3/2, we get:

y - (-5) = (-3/2)(x - 6)

y + 5 = (-3/2)x + 9

y = (-3/2)x + 4

Therefore, the equation of the line that is perpendicular to y - 4 = (2/3)(x - 9) and passes through the point (6,-5) is y = (-3/2)x + 4.

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User MEvans
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