find the value of k for which A(1,2),(4,-11), (5,-k)are collinear
Answers
Answered by
0
x1 = 1, x2= 4 ,x3=5 y1 = 2 , y2 = -11 , y3= -k
=> 1/2 × [x1(y2 - y3) + x2(y3 - y1) + x3( y1 - y2) ]=0
=> 1/2 × [1( -11+k) + 4(-k-2) + 5(2+11)] =0
=> 1/2 × [ -11+k - 4k - 8 + 65] = 0
=> 1/2 × [ -3k + 46] = 0
=> -3k + 46 = 0
=> -3k = -46
=> 3k = 46
=> k = 46/3
=> 1/2 × [x1(y2 - y3) + x2(y3 - y1) + x3( y1 - y2) ]=0
=> 1/2 × [1( -11+k) + 4(-k-2) + 5(2+11)] =0
=> 1/2 × [ -11+k - 4k - 8 + 65] = 0
=> 1/2 × [ -3k + 46] = 0
=> -3k + 46 = 0
=> -3k = -46
=> 3k = 46
=> k = 46/3
Similar questions