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How do I find the equation of a perpendicular line given and equation and a point?

plz help!

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1) Put equation in slope-intercept form, y = mx + b
2) Get slope of line. It is m.
3) Make m into its negative reciprocal -1/m. This is slope of perpendicular line.
4) Take new slope and point and plug them into y = mx + b. Solve for b.
5) Take new slope, and b. Put them into y = mx + b for m and b. But leave
y as y and x as x.
Done.
( comment if you need more help )

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can you work out line 8x+6y=17 and passes threw the point (5,-3)
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I'll number the steps as I go along.
8x + 6y = 17 through (5,-3)
1. 8x + 6y = 17
-8x. -8x
6y = -8x + 17
/6. /6
y = -8/6x + 17
y = -4/3x + 17
2. m = -4/3
3. slope of perpendicular line = 3/4
4. -3 = 3/4(5) + b
-3 = 15/4 + b
-15/4. -15/4
-12/4 -15/4 = b
-27/4 = b
5. y = 3/4x - 27/14. ANSWER
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ok thanks that how i did it and i got the same :)
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Great!!!!
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