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.

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.

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

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

Please explain how you did it.

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

none

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

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.

Through: (-3, -1), perp. To y = -8x + 4

y = -5/2x

idk this is hard i need help pls help

Through: (-1, -1), perpendicular to y = -1/5x + 4

Through: (3, 2), perpendicular to y = -x - 5

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...

y = 2x - 1

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