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# The point A divides the join of $P\left( { - 5,1} \right)$ and $Q\left( {3,5} \right)$ in the ratio .What is the value of for which the area of the $\Delta ABC$ where $B(1,5)$ , $C\left( {7, - 2} \right)$ is 2 square units.?A. $7,\dfrac{{31}}{9}$ B. $- 7,\dfrac{{31}}{9}$ C. $7, - \dfrac{{31}}{9}$ D. $- 7, - \dfrac{{31}}{9}$

Last updated date: 15th Sep 2024
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Hint: The coordinates of point is to be calculated using the section formula. The coordinates of point which divides the join of two points internally in the ratio $m:n$ , $D\left( {{x_1},{y_1}} \right)$ and $E\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$ . Then calculate the area of the triangle using the coordinates of the points A,B and C.

Given information
Point A divides the join of points $P\left( { - 5,1} \right)$ and $Q\left( {3,5} \right)$ in the ratio of $k:1$ .

Let the coordinates of point be $A\left( {x,y} \right)$ which divides the join of two points internally in the ratio is given by,
$\Rightarrow \left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$
Here $m = k$ , $n = 1$ , ${x_1} = - 5$ , ${y_1} = 1$ , ${x_2} = 3$ , and ${y_2} = 5$ . Using it, calculate the value of coordinates of point A.
The x-coordinate of point A is given by,
$x = \dfrac{{k \times 3 + 1 \times \left( { - 5} \right)}}{{k + 1}} \\ x = \dfrac{{3k - 5}}{{k + 1}} \\$

The x-coordinate of point B is given by,
$y = \dfrac{{k \times 5 + 1 \times \left( 1 \right)}}{{k + 1}} \\ y = \dfrac{{5k + 1}}{{k + 1}} \\$
The coordinates of point is $A\left( {\dfrac{{3k - 5}}{{k + 1}},\dfrac{{5k + 1}}{{k + 1}}} \right)$ .
The area of the triangle whose coordinates is $\left( {{x_1},y{}_1} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ is given by,
$\Rightarrow {A_r} = \dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right| \cdots \left( 1 \right)$
The coordinates of the vertices of the triangle are $A\left( {\dfrac{{3k - 5}}{{k + 1}},\dfrac{{5k + 1}}{{k + 1}}} \right)$ , $B(1,5)$ , and $C\left( {7, - 2} \right)$ .Substitute the value A,B and C in equation (1), we get
$\Rightarrow {A_r} = \dfrac{1}{2}\left| {\dfrac{{3k - 5}}{{k + 1}}\left( {5 - \left( { - 2} \right)} \right) + 1\left( { - 2 - \left( {\dfrac{{5k + 1}}{{k + 1}}} \right)} \right) + 7\left( {\left( {\dfrac{{5k + 1}}{{k + 1}}} \right) - 5} \right)} \right| \cdots \left( 2 \right)$
Substitute the value of in equation (2) and solve the equation further to obtain the value of k.
$\Rightarrow 2 = \dfrac{1}{2}\left| {\dfrac{{3k - 5}}{{k + 1}}\left( 7 \right) + 1\left( {\dfrac{{ - 2\left( {k + 1} \right) - \left( {5k + 1} \right)}}{{k + 1}}} \right) + 7\left( {\dfrac{{5k + 1 - 5\left( {k + 1} \right)}}{{k + 1}}} \right)} \right| \\ 2 \times 2 = \left| {\dfrac{{21k - 35}}{{k + 1}} + \dfrac{{ - 2k - 2 - 5k - 1}}{{k + 1}} + \dfrac{{35k + 7 - 35k - 35}}{{k + 1}}} \right| \\ \Rightarrow 4 = \left| {\dfrac{{21k - 35 - 2k - 2 - 5k - 1 + 35k + 7 - 35k - 35}}{{k + 1}}} \right| \\ \Rightarrow 4 = \left| {\dfrac{{14k - 66}}{{k + 1}}} \right| \cdots \left( 3 \right) \\$
Equation (3) can have 2 values of k
When modulus opens with a positive sign
$\Rightarrow 4 = \dfrac{{14k - 66}}{{k + 1}}$
$\Rightarrow 4k + 4 = 14k - 66 \\ \Rightarrow 10k = 70 \\ \Rightarrow k = 7 \\$

When modulus opens with a negative sign
$\Rightarrow 4 = - \dfrac{{14k - 66}}{{k + 1}}$
$\Rightarrow 4k + 4 = - 14k + 66 \\ \Rightarrow 18k = 62 \\ \Rightarrow k = \dfrac{{62}}{{18}} \\ \Rightarrow k = \dfrac{{31}}{9} \\$
Hence, the two values of k are $k = 7,\dfrac{{31}}{9}$

So, the correct answer is “Option A”.

Note: The important steps and formulae in the question are,
The use of section formula, when point $A\left( {x,y} \right)$ divides the join of $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ in the ratio of $m:n$ internally,
$\left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$
Also when it divides externally it is given by
$\left( {x = \dfrac{{m{x_2} - n{x_1}}}{{m - n}},y = \dfrac{{m{y_2} - n{y_1}}}{{m - n}}} \right)$
The area of the triangle whose coordinates are $\left( {{x_1},y{}_1} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ is,
${A_r} = \dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$