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Misc-II · Q130

Q.Find the co-ordinates of the foot of the perpendicular drawn from the point P(−1,3) to the line 3x − 4y − 16 = 0.

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The line 3x−4y−16=03x-4y-16=0 has slope 3/43/4, so the perpendicular from P(−1,3)P(-1,3) has slope −4/3-4/3: y−3=−43(x+1)⇒3y−9=−4x−4⇒4x+3y−5=0y-3=-\dfrac43(x+1) \Rightarrow 3y-9=-4x-4 \Rightarrow 4x+3y-5=0. Solving with 3x−4y−16=03x-4y-16=0: scale the first by 4 (16x+12y−20=016x+12y-20=0) and the second by 3 (9x−12y−48=09x-12y-48=0); adding gives 25x−68=0⇒x=682525x-68=0 \Rightarrow x=\dfrac{68}{25}. From 4x+3y=54x+3y=5 …

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