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What is an equation of the line that passes through the point (-2,5) and is perpendicular
to the line whose equation is y=-x+ 5?
O y = 2x+9
Oy=-2x+1
Oy= 2x+1
Oy=-2x-9

A What is an equation of the line that passes through the point (-2,5) and is perpendicular-example-1

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Answer: The given line has a slope of -1, since its equation is y = -x + 5. The line that is perpendicular to this line will have a slope that is the negative reciprocal of -1, which is 1. So, we know that the equation of the line we're looking for will have a slope of 1.

To find the equation of this line, we need to use the point-slope form of the equation of a line:

y - y1 = m(x - x1)

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

We know that the point (-2, 5) is on the line we're looking for, and we know that the slope of the line is 1. So we can substitute these values into the point-slope form:

y - 5 = 1(x - (-2))

Simplifying, we get:

y - 5 = x + 2

Adding 5 to both sides, we get:

y = x + 7

Therefore, the equation of the line that passes through the point (-2, 5) and is perpendicular to the line y = -x + 5 is y = x + 7.

Explanation:

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