Question

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} $

Answer 1

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.

Complete step-by-step answer:
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| $