JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 14)
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 :
4
1
-3
3
Explanation
Point P(x', y') lies on the line 3x + y – 4$$\lambda $$ = 0
$$ \therefore $$ 3x' + y' – 4$$\lambda $$ = 0
Area of $$\Delta $$PAB = 5
$$ \Rightarrow $$ $${1 \over 2}\left| {\matrix{ 1 & 1 & { - 1} \cr 1 & 0 & 2 \cr 1 & {x'} & {y'} \cr } } \right|$$ = $$ \pm $$ 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
$$ \therefore $$ 3x' + y' – 4$$\lambda $$ = 0
Area of $$\Delta $$PAB = 5
$$ \Rightarrow $$ $${1 \over 2}\left| {\matrix{ 1 & 1 & { - 1} \cr 1 & 0 & 2 \cr 1 & {x'} & {y'} \cr } } \right|$$ = $$ \pm $$ 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
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