Find the value of y for which the points (5,-4),(3,-1) and (1,y)are collenior
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since the points are collinear then ,,the are of triangle formed by these three points will be be risk to zero ...
1/2[5(-1-y)+3(y+4)+1(-4+1)]=0
-5-5y+3y+12-4+1=0
2y=4
y=2 ......
i hope this will be helpful to you ....
1/2[5(-1-y)+3(y+4)+1(-4+1)]=0
-5-5y+3y+12-4+1=0
2y=4
y=2 ......
i hope this will be helpful to you ....
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