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How do you write the equation in point slope form given (5,-6) and perpendicular to the line whose equation is x-7y=3?

Answer 1

Julia Adams

#=>y+6=-7(x-5)#

Point slope format : #y-y_1=m(x-x_1)#

Given:#" "x-7y=3#

In standard form this is #y=1/7x-3/7#

So the gradient of the perpendicular line is of form: # (-1)xx1/m#

#=>" required gradient "= -7#

So standard form of the new equation is

#y=-7x+c#

We are given the point #(x,y)->(5,-6)#

So in slope point form we have

#y-(-6)=-7(x-5)#

#=>y+6=-7(x-5)#

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Answer 2

Kennedy Adams

First, determine the slope of the line (x - 7y = 3) by rearranging the equation into slope-intercept form (y = mx + b), where (m) is the slope. Then, since the line we want to find is perpendicular to this line, the slope of the new line will be the negative reciprocal of the slope of the given line. Once you have the slope, you can use the point-slope form of a line (y - y_1 = m(x - x_1)), where ((x_1, y_1)) is the given point, to write the equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.