please.help. im clueless
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
Through: (-3, -1), perp. To y = -8x + 4
y = -5/2x
idk this is hard i need help pls help
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: (-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.