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  •  The point  divides the join of <img alt="\mathrm{P \equiv (-5, 1)\ and\ Q = (3, 5)}" src="h

 The point \mathrm{A} divides the join of \mathrm{P \equiv (-5, 1)\ and\ Q = (3, 5)} in the ratio \mathrm{k:1}. Find the two values of \mathrm{k} for which the areas of \mathrm{\bigtriangleup ABC} where \mathrm{B \equiv (1, 5), C \equiv (7, 2)} is equal to \mathrm{2} square units.

Option: 1

\mathrm{k=4, \frac{29}{7}}


Option: 2

\mathrm{k=7, \frac{32}{5}}


Option: 3

\mathrm{K=\frac{31}{9}, \frac{43}{7}}


Option: 4

\mathrm{k=7, \frac{31}{9}}


Answers (1)

best_answer

Co-ordinates of \mathrm{A}, dividing the join of \mathrm{P\equiv \left ( -5,1 \right )} and \mathrm{Q\equiv \left ( 3,5 \right )} in the ratio \mathrm{k : 1} are given by \mathrm{(3 \mathrm{k}-5 / \mathrm{k}+1,5 \mathrm{k}+1 / \mathrm{k}+1)}
Also, area of the \mathrm{ \triangle ABC} is given by \mathrm{ \triangle =\left|\frac{1}{2} \sum x_1\left(y_2-y_3\right)\right|}
\mathrm{\begin{aligned} & =1 / 2\left|\left[\mathrm{x}_1\left(\mathrm{y}_2-\mathrm{y}_3\right)+\mathrm{x}_2\left(\mathrm{y}_3-\mathrm{y}_1\right)+\mathrm{x}_3\left(\mathrm{y}_1-\mathrm{y}_2\right)\right]\right| \\ & =\left|\frac{1}{2}\left\{\frac{3 \mathrm{k}-5}{\mathrm{k}+1}(7)+\left(-2-\frac{5 \mathrm{k}+1}{\mathrm{k}+1}\right)+7\left(\frac{5 \mathrm{k}+1}{\mathrm{k}+1}-5\right)\right\}\right|=2 \end{aligned}}
\mathrm{\Rightarrow \left ( 1/2 \right )}
\mathrm{\left\{\frac{3 \mathrm{k}-5}{\mathrm{k}+1}(7)+\left(-2-\frac{5 \mathrm{k}+1}{\mathrm{k}+1}\right)+7\left(\frac{5 \mathrm{k}+1}{\mathrm{k}+1}-5\right)\right\}= \pm 2}
\mathrm{\begin{aligned} & \Rightarrow 14 \mathrm{k}-66=4 \mathrm{k}+4, \Rightarrow 10 \mathrm{k}=70, \Rightarrow \mathrm{k}=7 \\ & 14 \mathrm{k}-66=-4 \mathrm{k}-4, \Rightarrow 18 \mathrm{k}=62, \Rightarrow \mathrm{k}=(31 / 9). \end{aligned}}
Therefore value of the \mathrm{k = 7, 31/9}

Posted by

sudhir kumar

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