Given:Area of triangle formed by ( 5 , 6 ) , ( 3 , 2 ) and ( α , β ) = 12 s q u a r e u n i t s \left(5, 6\right), \left(3, 2\right) \text{ and } \left(\alpha, \beta\right) = 12squareunits ( 5 , 6 ) , ( 3 , 2 ) and ( α , β ) = 12 s q u a re u ni t s
∴ \dfrac{1}{2}\left| \begin{cases}\alpha & \beta & 1 \\ 5 & 6 & 1 \\ 3 & 2 & 1\end{cases}\right| = 12
⇒ \left| \begin{cases}\alpha & \beta & 1 \\ 5 & 6 & 1 \\ 3 & 2 & 1\end{cases}\right| = 24
⇒ α ( 6 − 2 ) − β ( 5 − 3 ) + 1 ( 10 − 18 ) = ± 24 ⇒ \alpha\left(6 - 2\right) - \beta\left(5 - 3\right) + 1\left(10 - 18\right) = \pm 24 ⇒ α ( 6 − 2 ) − β ( 5 − 3 ) + 1 ( 10 − 18 ) = ± 24
⇒ 4 α − 2 β − 8 = ± 24 ⇒ 4\alpha - 2\beta - 8 = \pm 24 ⇒ 4 α − 2 β − 8 = ± 24
⇒ 4 α − 2 β = 32 , 4 α − 2 β + 16 = 0 ⇒ 4\alpha - 2\beta = 32, 4\alpha - 2\beta + 16 = 0 ⇒ 4 α − 2 β = 32 , 4 α − 2 β + 16 = 0
⇒ 2 α − β − 16 = 0 , 2 α − β + 8 = 0 ⇒ 2\alpha - \beta - 16 = 0, 2\alpha - \beta + 8 = 0 ⇒ 2 α − β − 16 = 0 , 2 α − β + 8 = 0
So point will be either \left(\alpha, 2\alpha + 8\right), \left(\alpha, 2\alpha - 16\right)< / p > < p > F o r t h e p o i n t ( α , 2 α + 8 ) </p><p>For the point \left(\alpha, 2\alpha + 8\right) < / p >< p > F or t h e p o in t ( α , 2 α + 8 )
Distance from originD = α 2 + ( 2 α + 8 ) 2 = 5 α 2 + 32 α + 64 ( ∵ β = 2 α + 8 ) D = \sqrt{{\alpha}^{2} + ( 2\alpha + 8{)}^{2}} = \sqrt{5{\alpha}^{2} + 32\alpha + 64}\left(∵ \beta = 2\alpha + 8\right) D = α 2 + ( 2 α + 8 ) 2 = 5 α 2 + 32 α + 64 ( ∵ β = 2 α + 8 )
⇒ D 2 = 5 α 2 + 32 α + 64 ⇒ {D}^{2} = 5{\alpha}^{2} + 32\alpha + 64 ⇒ D 2 = 5 α 2 + 32 α + 64
For maximum or minimum length,
d ( D 2 ) d α = 10 α + 32 = 0 \dfrac{d\left({D}^{2}\right)}{d\alpha} = 10\alpha + 32 = 0 d α d ( D 2 ) = 10 α + 32 = 0
⇒ α = − 16 5 ⇒ \alpha = - \dfrac{16}{5} ⇒ α = − 5 16
∴ β = − 32 5 + 8 = 8 5 ∴ \beta = \dfrac{- 32}{5} + 8 = \dfrac{8}{5} ∴ β = 5 − 32 + 8 = 5 8
∴ D = ( − 16 5 ) 2 + ( 8 5 ) 2 = 8 5 5 = 8 5 ∴ D = \sqrt{{\left(- \dfrac{16}{5}\right)}^{2} + {\left(\dfrac{8}{5}\right)}^{2}} = \dfrac{8}{5}\sqrt{5} = \dfrac{8}{\sqrt{5}} ∴ D = ( − 5 16 ) 2 + ( 5 8 ) 2 = 5 8 5 = 5 8
Similarly if β = 2 α − 16 , D = α 2 + ( 2 α − 16 ) 2 = 5 α 2 − 64 α + 256 \beta = 2\alpha - 16, D = \sqrt{{\alpha}^{2} + {\left(2\alpha - 16\right)}^{2}} = \sqrt{5{\alpha}^{2} - 64\alpha + 256} β = 2 α − 16 , D = α 2 + ( 2 α − 16 ) 2 = 5 α 2 − 64 α + 256
∴ D 2 = 5 α 2 − 64 α + 256 ∴ {D}^{2} = 5{\alpha}^{2} - 64\alpha + 256 ∴ D 2 = 5 α 2 − 64 α + 256
For maximum or minimum length,
d ( D 2 ) d α = 0 \dfrac{d\left({D}^{2}\right)}{d\alpha} = 0 d α d ( D 2 ) = 0
⇒ 10 α − 64 = 0 ⇒ 10\alpha - 64 = 0 ⇒ 10 α − 64 = 0
∴ α = 32 5 ∴ \alpha = \dfrac{32}{5} ∴ α = 5 32
∴ D = ( 32 5 ) 2 + ( 64 5 − 16 ) 2 = 16 5 ∴ D = \sqrt{{\left(\dfrac{32}{5}\right)}^{2} + {\left(\dfrac{64}{5} - 16\right)}^{2}} = \dfrac{16}{\sqrt{5}} ∴ D = ( 5 32 ) 2 + ( 5 64 − 16 ) 2 = 5 16 α = 32 5 \alpha = \dfrac{32}{5} α = 5 32 8 5 \dfrac{8}{\sqrt{5}} 5 8
