Question

Find the value of k if the points A(2, 3), B(4, k) and C(6, -3) are collinear​

Answer 1

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.