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3The point (2, 1) is translated parallel to the lineL: x -y 4 by 2/3 units. If the new point Qlies in the third quadrant, then, the equation ofthe line passing through Q and perpendicular toL is[JEE Main Online - 2016](A) x +y 2-6 (B)x+y 3-36(C) x+y 3-2/6 (D) 2x + 2y 1-6

Answer»

Slopes ofx−y=4

=>tanθ=1

=>(sinθ=1/√2,cosθ=1√2)or(sinθ=−1/√2,cosθ=1/√2)(sin⁡θ=−12,cos⁡θ=12)

Q is(2+2√3–√(1/√2),1+2√3(-1/√2))

=(2−√6,1-√6)

equation of required line isx+y=3−2√6



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