Area of Triangle with Coordinates of Vertices Given
Let two point...
Question
Let two points be A(1,−1) and B(0,2). If a point P(x′,y′) be such that the area of ΔPAB=5 sq. units and it lies on the line, 3x+y−4λ=0, then the value of λ is:
A
4
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B
1
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C
−3
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D
3
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Solution
The correct option is D3 Area of triangle is A=12∣∣
∣∣1−11021x′y′1∣∣
∣∣=±5
⇒(2−y′)−x′−2x′=±10 ⇒−3x′−y′+2=±10 ⇒3x′+y′=12 or 3x′+y′=−8 ⇒λ=3 or λ=−2