SOLUTION: write the equation of the line that contains the point (8,-3) and is perpindicular to 4x-3y=10 graph the equation. write it in standard form. please help me:/

Algebra ->  Graphs -> SOLUTION: write the equation of the line that contains the point (8,-3) and is perpindicular to 4x-3y=10 graph the equation. write it in standard form. please help me:/      Log On


   

Question 983884: write the equation of the line that contains the point (8,-3) and is perpindicular to 4x-3y=10 graph the equation. write it in standard form. please help me:/
Found 2 solutions by josmiceli, josgarithmetic:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+4x+-+3y+=+10+
+3y+=+4x+-+10+
+y+=+%284%2F3%29%2Ax+-+10%2F3+
----------------------
The line is now in the form:
+y+=+m%2Ax+%2B+b+ where
+m+ = slope
For this line, +m+=+4%2F3+
-------------------------
ANY line perpendicular to this line
will have slope =
+m%5B1%5D+=+-1%2Fm+
+m%5B1%5D+=+-1%2F%284%2F3%29+
+m%5B1%5D+=+-3%2F4+
-------------------------
The required line goes through: (8,-3)
You can use the point-slope formula:
+%28+y+-+%28-3%29+%29+%2F+%28+x+-+8+%29+=+-3%2F4+
+y+%2B+3+=+%28+-3%2F4+%29%2A%28+x+-+8+%29+
+y+%2B+3+=+%28-3%2F4%29%2Ax+%2B+6+
+y+=+%28-3%2F4%29%2Ax+%2B+3+ answer
----------------------------
check:
Does it go through (8,-3) ?
+y+=+%28-3%2F4%29%2Ax+%2B+3+
+-3+=+%28-3%2F4%29%2A8+%2B+3+
+-3+=+-6+%2B+3+
+-3+=+-3+
OK
Here are the plots of the lines:

Hope this helps

Answer by josgarithmetic(39617) About Me  (Show Source):