Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Write the slope-intercept equation for the line that passes through (-5, -8) and is perpendicular to 10x – 6y = -11 Please show all of your work

OpenStudy (mysesshou):

Eep. . ok. perpendicular means \[m2 = \frac{-1}{m1}\] so you start with m1 which is in your original equation, then solve for m2 which is the slope for the equation that is perpendicular.

OpenStudy (mysesshou):

first, find the slope of this 10x – 6y = -11 change to the format of y=m*x + b so, subtract x from both sides: – 6y = -11 -10x then divide whole equation by -6 y = (-11/-6) -(10/-6)x rearrange it for proper format: y = -(10/-6)x +(-11/-6) simplify the - and +: y = (10/6)x +(11/6) simplify the fraction: y = (5/3)x +(11/6) so, slope m = 5/3 and y intercept b=11/6 ok?

OpenStudy (mysesshou):

so for your original equation, the slope is m1 = 5/3 to find the perpendicular slope, \[m2 = \frac{-1}{5/3} = \frac{-3}{5}\] use this new slope and that point you were given to fine the new equation... m2=(-3/5)

OpenStudy (mysesshou):

Point: (-5, -8) slope: m= -3/5 put into equation format: (y-y1) = m * (x-x1) x1 = -5 y1 = -8 m = -3/5 (y-(-8)) = (-3/5) * (x-(-5)) minus a negative is + y+8 = (-3/5) * (x+5) multiply through by the -3/5 y+8 = (-3/5) * x + (-3/5) *5 simplify: y+8 = (-3/5) * x -3 subtract both sides 8: y = (-3/5) * x -11 therefore, slope intercept form is y = (-3/5) * x -11 slope is -3/5 y intercept is -11

OpenStudy (mysesshou):

ok?

OpenStudy (anonymous):

got you

OpenStudy (mysesshou):

Good luck !

OpenStudy (anonymous):

thanks soooooo much

OpenStudy (mysesshou):

you're welcome ! hope you get a good score and understood well enough!

OpenStudy (anonymous):

me too lol

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!