Let two points be A(1, --1) and B(0, 2).

Question

Let two points be A(1, --1) and B(0, 2). If a point
P(x', y') be such that the area of $$\Delta $$PAB = 5 sq.
units and it lies on the line, 3x + y -- 4$$\lambda $$ = 0,
then a value of $$\lambda $$ is :

Accepted Answer

Point P(x', y') lies on the line 3x + y -- 4$$\lambda $$ = 0

$$ \therefore $$ 3x' + y' -- 4$$\lambda $$ = 0

Area of $$\Delta $$PAB = 5

$$ \Rightarrow $$ $${1 \over 2}\left| {\begin{pmatrix} 1 & 1 & { - 1}  \
   1 & 0 & 2  \
   1 & {x'} & {y'}  \

\end{pmatrix} } \right|$$ = $$ \pm $$ 5

$$ \Rightarrow $$ [(--2x') -- (y' -- 2) -- (x')] = $$ \pm $$ 10

$$ \Rightarrow $$ -- 3x' -- y' + 2 = $$ \pm $$ 10

$$ \Rightarrow $$ 2 - 4$$\lambda $$ = $$ \pm $$ 10

$$ \Rightarrow $$ $$\lambda $$ = -2 or $$\lambda $$ = 3

Solutions · 1