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Question
For what values of k are the points (8, 1), (3, –2k) and (k, –5) collinear ?
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Solution
\[\therefore \frac{1}{2}\left[ 8\left( - 2k + 5 \right) + 3\left( - 5 - 1 \right) + k\left( 1 + 2k \right) \right] = 0\]
\[ \Rightarrow - 16k + 40 - 18 + k + 2 k^2 = 0\]
\[ \Rightarrow 2 k^2 - 15k + 22 = 0\]
\[ \Rightarrow 2 k^2 - 11k - 4k + 22 = 0\]
\[ \Rightarrow k\left( 2k - 11 \right) - 2\left( 2k - 11 \right) = 0\]
\[ \Rightarrow \left( k - 2 \right)\left( 2k - 11 \right) = 0\]
\[ \Rightarrow k - 2 = 0 or 2k - 11 = 0\]
\[ \Rightarrow k = 2 or k = \frac{11}{2}\]
Thus, the values of k are 2 and
\[\frac{11}{2}\]
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Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
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If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by ______.
The distance between the points A(0, 6) and B(0, –2) is ______.
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
