Find the value of k if the points A(2, 3), B(4, k) and C(6, -3) are collinear
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Step-by-step explanation:
distance between A and B=√{(x2-x1)²+(y2-y1)²}
AB= √{(4-2)²+(k-3)²}
AB=√{2²+(k+3)²}
AB=√{4+k²+9+6k}
AB=√(k²+6k+13)
distance between B and C= √{(x2-x1)²+(y2-y1)²}
BC= √{(6-2)²+(-3-k)²}
BC=√{(4)²+(k+3)²}
BC=√(16+k²+9+6k)
BC=√(k²+6k+25)
distance between A and = √{(x2-x1)²+(y2-y1)²}
AC=√{(6-2)²+(-3-3)²}
AC=√(4²+6²)
AC=√(16+36)
AC=√52
A,B and C are collinear so,
AB+ BC = AC
√(k²+6k+13) + √(k²+6k+25)=√52
now solve To get the value of k.
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