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

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,

{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$