Question If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, then find the value of k.
Video Explanation
Explanatory Answer
A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
If the points are collinear, the area of the triangle formed by joining these points will be ‘0’.
Area of triangle, coordinates of whose vertices are (x1, y1) (x2, y2) and (x3, y3) = \\frac{1}{2}{\begin{vmatrix} { x }_{ 1 } & { x }_{ 2 } & { x }_{ 3 } \\ { y }_{ 1 } & { y }_{ 2 } & { y }_{ 3 } \\ 1 & 1 & 1 \end{vmatrix}})
= \\frac{1}{2}) {x1(y2 – y3) – x2(y3 – y1) + x3(y1 – y2)}
Because the points are collinear, \\frac{1}{2}{\begin{vmatrix} ({k + 1 }) & 3k & ({5k - 1}) \\ 2k & ({ 2k + 3}) & 5k \\ 1 & 1 & 1 \end{vmatrix}}) = 0
= \\frac{1}{2}) {(k + 1)(2k + 3 – 5k) – 3k(2k – 5k) + (5k – 1)(2k – (2k + 3))} = 0
= (k + 1)(-3k + 3) – 3k(-3k) + (5k - 1)(-3) = 0
= -3k2 + 3k – 3k + 3 + 9k2 – 15k + 3 = 0
6k2 – 15k + 6 = 0
2k2 – 5k + 2 = 0
2k2 – 4k – k + 2 = 0
2k(k – 2) – 1(k – 2) = 0
(2k – 1)(k – 2) = 0
k = \\frac{1}{2}) or k = 2