InterviewSolution
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InterviewSolution
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Find the values of k, if the points A (k+1,2k) ,B (3k,2k+3) and C (5k-1,5k) are collinear. |
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Answer» We know that , if three points are collinear , then the area of triangle formed by these points is zero. Since, the points A (k+1,2k) , B (3k,2k+3) and C (5k-1,5k) are collinear. Then , area of `triangle ABC =0` `rArr (1)/(2)[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2)]=0` Here , `x_(1)=k+1,x_(2)=3k,x_(3)=5k-1and y_(1)=2k+3,y_(3)=5k` `rArr(1)/(2)[(k+1)(-3k+3)+3k(3k)+(5k-1)(2k-3)]=0` `(1)/(2)[(k+1)(-3k+3)+3k(3k)+(5k-1)(2k-2k-3)]=0` `(1)/(2)[-3k^(2)+3k-3k+3+9k^(2)=15k+3]=0` `rArr(1)/(2)(6k^(2)-15k+6)=0` [multiply by 2] `rArr6k^(2)-15k+6=0` [by factorisation method] `2k^(2)-5k+2=0` [divide by 3] `rArr2k^(2)-4k-k+2=0` `2k(k-2)-(k-2)=0` `rArr(k-2)(2k-1)=0` If `k-2=0` , then `k=2` If `2k-1=0, then k=(1)/(2)` `:. k=2,(1)/(2)` Hence , the required values of k are and `(1)/(2)`. |
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