SOLUTION: Find an equation of the line L that passes through the point (−2, 6)and satisfies the given condition. (Let x be the independent variable and y be the dependent variable.) L

Algebra ->  Test -> SOLUTION: Find an equation of the line L that passes through the point (−2, 6)and satisfies the given condition. (Let x be the independent variable and y be the dependent variable.) L      Log On


   

Question 973834: Find an equation of the line L that passes through the point (−2, 6)and satisfies the given condition. (Let x be the independent variable and y be the dependent variable.)
L is perpendicular to the line 4x + 3y = 6
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+4x+%2B+3y+=+6+
+3y+=+-4x+%2B+6+
+y+=+-%284%2F3%29%2Ax+%2B+2+
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Any line perpendicular to this line
will have slope = +-1%2Fm+
where +m+=+-4%2F3+
+-1%2Fm+=+3%2F4+
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The required line goes through ( -2,6 )
Now use point-slope formula
+%28+y+-+6+%29+%2F+%28+x+-%28-2%29+%29+=+3%2F4+
+y+-+6+=+%283%2F4%29%2A%28+x%2B2+%29+
+4y+-+24+=+3x+%2B+6+
+4y+=+3x+%2B+30+
+y+=+%283%2F4%29%2Ax+%2B+15%2F2+ answer
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check:
does it go through ( -2,6 )?
+y+=+%283%2F4%29%2Ax+%2B+15%2F2+
+6+=+%283%2F4%29%2A%28-2%29+%2B+15%2F2+
+6+=+-3%2F2+%2B+15%2F2+
+6+=+12%2F2+
+6+=+6+
OK
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Here are plots of the 2 lines: