# CBSE 2017 Board Paper

10th standard Math - Set 3 Q21. Coordinate Geometry

#### 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

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