Question

Find the equation of the line joining \[A\left( {1,3} \right)\] and \[B\left( {0,0} \right)\] using determinants and fink k if D \[\left( {k,0} \right)\] is a point such that area of triangle ABD is 3sq unit.

Answer 1

Hint:
First, we have to put the area ABP as 0. Then using the distance formula, we have to find the equation of line AB. Then find the value of k by putting triangle ABD as 3.

Complete step by step solution:
Let us assume $P(x,y)$ be any point on the line AB.
Then $area\left( {\Delta ABP} \right) = 0$
Now, using determinant,
 \[ \Rightarrow \dfrac{1}{2}\left| {\begin{array}{*{20}{l}}
  {131} \\
  {001} \\
  {xy1}
\end{array}} \right| = 0\]
 $ \Rightarrow \dfrac{1}{2}\left\{ {1\left( {0 - y} \right) - 3\left( {0 - x} \right) + 1\left( {0 - 0} \right)} \right\} = 0$
 $ \Rightarrow 3x - y = 0$, which is the required equation of line AB
Now, same for $area\left( {\Delta ABD} \right) = 3$ sq unit
 \[ \Rightarrow \dfrac{1}{2}\left| {\begin{array}{*{20}{l}}
  {131} \\
  {001} \\
  {k01}
\end{array}} \right| = \pm 3\]
 \[ \Rightarrow \left| {\begin{array}{*{20}{l}}
  {131} \\
  {001} \\
  {k01}
\end{array}} \right| = \pm 6\]
 $ \Rightarrow 1\left( {0 - 0} \right) - 3\left( {0 - k} \right) + 1\left( {0 - 0} \right) = \pm 6$
 $ \Rightarrow 3k = \pm 6$
 $ \Rightarrow k = \pm 2$

$\therefore $ The value of k is $ \pm 2$.

Note:
Alternate Method:
Using determinant, the line joining \[A\left( {1,3} \right)\] and \[B\left( {0,0} \right)\] is given by \[\left| {\begin{array}{*{20}{c}}
  {xy1} \\
  {131} \\
  {001}
\end{array}} \right| = 0\]
When we expanding along ${R_3}$ we get $1\left( {3x - y} \right) = 0$ or $y = 3x$
Now, D(k,0) is appointing such that area $\Delta ABD = 3$ sq units
 \[ \Rightarrow \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {131} \\
  {001} \\
  {k01}
\end{array}} \right| = \left| 3 \right|\]
Expanding along ${R_2}$
 $\left( {0 - 3k} \right) = \pm 6$
 $ \Rightarrow k = \pm 2$