Question

If the points (4,0), (7,3) and (,5) lies on the same line, what is the value of 'k-7'?​

Answer 1

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Answer 2

Given,

$\[\left( {4,0} \right),\left( {7,3} \right),\left( {k - 3,5} \right)\]$

Solution,

If points $\[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\]$ lies on a line then area of triangle will be zero.

i.e. $\[\frac{1}{2}\left\{ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right\} = 0\]$

Put the points coordinate in the above given condition,

$\[\begin{array}{l}\frac{1}{2}\left\{ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right\} = 0\\ \Rightarrow \frac{1}{2}\left\{ {4\left( {3 - 5} \right) + 7\left( {5 - 0} \right) + \left( {k - 3} \right)\left( {0 - 3} \right)} \right\} = 0\\ \Rightarrow \left\{ { - 8 + 35 - 3k + 9} \right\} = 0\\ \Rightarrow - 3k + 36 = 0\\ \Rightarrow k = 12\end{array}\]$

$\[\begin{array}{l} = k - 7\\ = 12 - 7\\ = 5\end{array}\]$

Hence the value of $\[k - 7\]$ is 5.