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.
+37157 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:
+37157 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:
