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# Need help

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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.

Feb 27, 2018

#1
+18106
+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:

Feb 27, 2018

#1
+18106
+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