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Home / Doubts and Solutions/JEE/Mathematics

A Straight line passing through the point A(2,3) has inclination 45 degree and intersects the line       2x-3y+9=0 at point P.Find the distance AP.

Asked by neeraj.unnao 5th November 2016, 11:34 PM
Answered by Expert
Answer:
begin mathsize 16px style The space equation space of space straight a space line space passing space through space straight A left parenthesis 2 comma 3 right parenthesis space that space has space an space inclination space of space 45 degree space is fraction numerator straight x space minus space straight x subscript 1 over denominator cos space straight theta end fraction equals fraction numerator straight y space minus space straight y subscript 1 over denominator sin space straight theta end fraction equals straight r space where space straight r space represents space the space distance space of space any space point space on space this space line space from space the space given space point space left parenthesis space straight x subscript 1 comma space straight y subscript 1 right parenthesis that space is comma space fraction numerator straight x space minus space 2 over denominator cos space 45 end fraction equals fraction numerator straight y space minus space 3 over denominator sin space 45 end fraction equals AP space rightwards double arrow space fraction numerator straight x space minus space 2 over denominator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction equals fraction numerator straight y space minus space 3 over denominator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction equals AP equals straight r left parenthesis say right parenthesis The space coordinates space of space any space point space straight P space on space this space line space are space open parentheses 2 plus straight r fraction numerator 1 over denominator square root of 2 end fraction comma space space 3 plus straight r fraction numerator 1 over denominator square root of 2 end fraction close parentheses rightwards double arrow space open parentheses 2 plus fraction numerator straight r over denominator square root of 2 end fraction comma space space 3 plus fraction numerator straight r over denominator square root of 2 end fraction close parentheses Since space straight P space lies space on space 2 straight x space minus space 3 straight y space plus space 9 space equals space 0 comma space then space 2 open parentheses 2 plus fraction numerator straight r over denominator square root of 2 end fraction close parentheses space minus space 3 open parentheses 3 plus fraction numerator straight r over denominator square root of 2 end fraction close parentheses space plus space 9 space equals space 0 rightwards double arrow 4 plus fraction numerator 2 straight r over denominator square root of 2 end fraction space minus 9 minus fraction numerator 3 straight r over denominator square root of 2 end fraction space plus space 9 space equals space 0 rightwards double arrow fraction numerator negative straight r over denominator square root of 2 end fraction equals negative 4 rightwards double arrow straight r equals 4 square root of 2 Hence comma space the space distance space AP space equals 4 square root of 2. end style
Answered by Expert 7th November 2016, 8:57 AM
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