Find the equation of a line perpendicular of 3x-12y+16=0 that has the same y intercept as the line 14x-13y-52=0.

Guest Feb 27, 2018

#1**+2 **

Let's re-arrange the line equations to y=mx+b form for easier work

3x-12y+ 16 = 0 is the same as

y = 1/4 x + 1 1/4 Slope = 1/4 perpindicular slope is -4

14x-13y-52=0 is the same as

y=4/13x-4 y intercept (put x=0 into the equation) = -4

So now we have the slope of our line is -4 and a point 0,-4

y = -4x + b Sustitute 0, -4 to find b

-4 = -4(0) + b

b = -4

Line then becomes y = -4x - 4

Here is a graph:

ElectricPavlov
Feb 27, 2018

#1**+2 **

Best Answer

Let's re-arrange the line equations to y=mx+b form for easier work

3x-12y+ 16 = 0 is the same as

y = 1/4 x + 1 1/4 Slope = 1/4 perpindicular slope is -4

14x-13y-52=0 is the same as

y=4/13x-4 y intercept (put x=0 into the equation) = -4

So now we have the slope of our line is -4 and a point 0,-4

y = -4x + b Sustitute 0, -4 to find b

-4 = -4(0) + b

b = -4

Line then becomes y = -4x - 4

Here is a graph:

ElectricPavlov
Feb 27, 2018