SOLUTION: Write an equation of a line in standard form that passes through (6, -5) perpendicular to the line whose equation is 3x-1/5y=3. Can you help me solve?

Algebra ->  Graphs -> SOLUTION: Write an equation of a line in standard form that passes through (6, -5) perpendicular to the line whose equation is 3x-1/5y=3. Can you help me solve?      Log On


   

Question 391825: Write an equation of a line in standard form that passes through (6, -5) perpendicular to the line whose equation is 3x-1/5y=3.
Can you help me solve?
Answer by jjordan95(63) About Me  (Show Source):
You can put this solution on YOUR website!
In order to find a perpendicular line, you need the equation in the form: y=mx%2Bb
So convert your function into the form mentioned above:
3x-%281%2F5%29y=3
=%28-1%2F5%29y=3-3x
=y=15x-15
Next, we take the inverse reciprocal of the slope (m is the slope in the equation y=mx+b.
The reciprocal of 15 is 1/15, to invert the number, simply make it the opposite sign that it currently is (positive turns to negative, and vice-versa). Therefore, the slope of the line perpendicular to y=15x-15, is -1/15.
The perpendicular line, now has the equation: y=%28-1%2F15%29x%2Bb
Now, we must solve for b. To solve for b, we simply plug in the x and y values from the point above. (x,y), (6,-5).
-5=%28-1%2F15%29%286%29%2Bb
-5=-6%2F15%2Bb
-5%2B6%2F15=b
-75%2F15%2B6%2F15=b
-69%2F15=b
Now plug b back into the equation and you have your answer:
highlight%28y=%28-1%2F15%29x-69%2F15%29