The point A divides the join of $P (-5, 1)$ and $Q (3, 5)$ in the ratio $k : 1$. Find the two values of $k$ for which the area of $\triangle ABC$ where $B$ is $(1, 5)$ and $C (7, -2)$ is equal to 2 units.

Given:

The point A divides the join of $P (-5, 1)$ and $Q (3, 5)$ in the ratio $k : 1$.

The area of $\triangle ABC$ where $B$ is $(1, 5)$ and $C (7, -2)$ is equal to 2 units.

To do:

We have to find the two values of $k$.

Solution:

Let the coordinates of $A$ be $(x, y)$ which divides the join of $P (-5, 1)$ and $Q (3, 5)$ in the ratio $k:1$. 

The coordinates of $A$ are \( \left(\frac{k \times 3+1 \times(-5)}{k+1}, \frac{k(5)+1(1)}{k+1}\right) \)

\( =\left(\frac{3 k-5}{k+1}, \frac{5 k+1}{k+1}\right) \)

$A(\frac{3 k-5}{k+1}, \frac{5 k+1}{k+1}), B(1, 5)$ and $C(7, -2)$ are the vertices of $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC = 2 \) sq. units

\( \Rightarrow 2=\frac{1}{2}\left[\frac{3 k-5}{k+1}(5+2)+1\left(-2-\frac{5 k+1}{k+1}\right)+7\left(\frac{5 k+1}{k+1}-5\right)\right. \)

\( \Rightarrow 2=\frac{1}{2}[\frac{3 k-5}{k+1} \times 7-2-\frac{5 k+1}{k+1}+7(\frac{5 k+1}{(k+1)})-35] \)

\( \Rightarrow 4=\left[\frac{21 k-35}{k+1}-\frac{5 k+1}{k+1}+\frac{35 k+7}{k+1}-37\right] \)

\( \Rightarrow 4=\left[\frac{21 k-35-5 k+1+35 k+7-37 k-37}{k+1}\right] \)

\( \Rightarrow\left|\frac{14 k-66}{k+1}\right|=\pm 4 \)

This implies,

\( \frac{14 k-66}{k+1}=4 \)

\( \Rightarrow 14 k-66=4 k+4 \)

\( \Rightarrow 14 k-4 k=66+4 \)

\( \Rightarrow 10 k=70 \)

\( \Rightarrow k=\frac{70}{10}=7 \)

And,

\( \frac{14 k-66}{k+1}=-4 \)

\( \Rightarrow 14 k-66=-4 k-4 \)

\( \Rightarrow 14 k+4 k=-4+66 \)

\( \Rightarrow 18 k=62 \)

\( \Rightarrow k=\frac{62}{18} \)

\( =\frac{31}{9} \)

The two values of $k$ are $7$ and $\frac{31}{9}$.

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Updated on: 10-Oct-2022

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