Create a line perpendicular to the line that goes between the points (-7,0) and (0,13)

Create a line perpendicular to the line that goes between the points (-7,0) and (0,13)

It is asking where the perpendicular line can bisect in a triangle.. The given options are inside the triangle, on the triangle, or outside of the triangle.

Exampl; (2,-1) y=-2x+1

I need the answer to this problem.

i dont understand this question can you guys please help me.

Find the value of c for which the lines 5x+cy=9 and x-3y=9 are perpendicular

Please explain how you did it.

write the equation of the line through the given points perpendicular to the given line

A line contains the points (-1,1) and (5,4) The slope is 1/2 The equation of the line in point-slope form is: (y-1) = 1/2(x- -1) The equation of the line in slope-intercept form is: y...

Through: (4, 2), perp. to y = -4/3x – 4

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i got the answer y=1/5x+3 but it was supposedly incorrect . Can you please help me figure this out ?

I know I have to do the point slop form, however got a weird answer every time I used that equation.

Either be r=4.5, r=9, r=-3 or r=1.5 Please help.

Which equation describes the line that passes through the point (-9,9) and is perpendicular to the graph of y=-8/9x-4/9? a. y=9/8x + 153/8 b. y= -9/8x - 9/8 c. y=9/8x -9/8 d...

find the equation of the line that is perpendicular to 4y+x+40=O that passes through the point (1/2,0)

How do I solve

I need help with this. I have to give the equation in slope-intercept form and standard form. I know perpendicular lines are a negative reciprocal.

Determine an equation of a line that passess through the point (8,6) and is perpendicular to the graph of y=-7/3x+8/3

What is an equation of a line that passes through the point (-5,-2) and is perpendicular to the graph of y=3/4x+1/8

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