#### 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 (x_{1}, y_{1}) (x_{2}, y_{2}) and (x_{3}, y_{3})
= \\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}) {x_{1}(y_{2} – y_{3}) – x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})}

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

= -3k^{2} + 3k – 3k + 3 + 9k^{2} – 15k + 3 = 0

6k^{2} – 15k + 6 = 0

2k^{2} – 5k + 2 = 0

2k^{2} – 4k – k + 2 = 0

2k(k – 2) – 1(k – 2) = 0

(2k – 1)(k – 2) = 0

k = \\frac{1}{2}) or k = 2