Q.
The equation of the locus of the foot of perpendicular drawn from (5,6) on the family of lines (x−2)+λ(y−3)=0 (where λ∈R ) is
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NTA AbhyasNTA Abhyas 2020Straight Lines
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Solution:
Let A=(5,6) and the point of concurrency of the family of lines (x−2)+λ(y−3)=0 is (2,3)=B and foot of the perpendicular from A to the family of lines is P=(h,k)
Now, PA is perpendicular to PB ⇒(slopeofPA)×(slopeofPB)=−1 ⇒h−5k−6×h−2k−3=−1⇒(k−6)(k−3)=−(h−5)(h−2) ⇒ locus is (x−2)(x−5)+(y−3)(y−6)=0