Question

Find the value of $\lambda$ such that points $P(1,2),Q(-2,3)$ and $R(\lambda +1,\lambda )$ are not forming a triangle?

Answer 1

$P-\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}$

$Q--2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}$

$R-(\lambda +1)\stackrel{\wedge }{i}+\lambda \stackrel{\wedge }{j}$

$\overline{PQ}=(-2-1)\stackrel{\wedge }{i}+(3-2)\stackrel{\wedge }{j}$

$=-3\stackrel{\wedge }{i}+\stackrel{\wedge }{j}$

$\overline{QR}=(\lambda +1+2)\stackrel{\wedge }{i}+(\lambda -3)\stackrel{\wedge }{j}$

$=(\lambda +3)\stackrel{\wedge }{i}+(\lambda -3)\stackrel{\wedge }{j}$

If $PQR$ form a triangle,

$\overline{PQ}+\overline{QR}=\overline{PR}$

$\overline{PR}=(\lambda +1-1)\stackrel{\wedge }{i}+(\lambda -2)\stackrel{\wedge }{j}$

$=\lambda \stackrel{\wedge }{i}+(\lambda -2)\stackrel{\wedge }{j}$

They don't form a triangle.

$Hence,\overline{PQ}+\overline{QR}\ne \overline{PR}$

$⟹\stackrel{\wedge }{i}(\lambda +3-3)+\stackrel{\wedge }{j}(\lambda -3+1)\ne \stackrel{\wedge }{i}\left(\lambda \right)+\stackrel{\wedge }{j}(\lambda -2)$

$⟹\stackrel{\wedge }{i}\left(\lambda \right)+\stackrel{\wedge }{j}(\lambda -3+1)\ne \stackrel{\wedge }{i}\left(\lambda \right)+\stackrel{\wedge }{j}(\lambda -2)$

$⟹\stackrel{\wedge }{i}\left(\lambda \right)+\stackrel{\wedge }{j}(\lambda -2)\ne \stackrel{\wedge }{i}\left(\lambda \right)+\stackrel{\wedge }{j}(\lambda -2)$

$\because$ L.H.S and R.H.S can't be equal,

$\therefore$ Both have to be zero,then$\overline{PR}=\overline{0},$

hence no side and hence,no triangle will be possible.$

$⟹\lambda =0,2.$

But sine $\lambda$ can't be equal to $0$ and $2$ at the same time.

Hence,there is no value of $\lambda$ for which the given points can't form a triangle.