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निर्देशांक ज्यामिति

Question
CBSEHHIMAH10010060

बिंदुओं A(2, -2) और B(3, 7) को जोड़ने वाले रेखाखंड को रेखा 2x + y - 4 = 0 जिस अनुपात में विभाजित करती है उसे ज्ञात कीजिए।

Solution

मान लीजिए A(2, -2) और B(3, 7) को जोड़ने वाले रेखाखंड को रेखा 2x + y - 4 = 0 को m:1 में विभाजित करती हैं।

विभाजन सूत्र के प्रयोग से: 
    therefore space open parentheses fraction numerator straight m subscript 1 straight x subscript 2 plus straight m subscript 2 straight x subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction comma space fraction numerator straight m subscript 1 straight y subscript 2 plus straight m subscript 2 straight y subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses
straight x with bar on top space equals space fraction numerator 3 straight m plus 1 cross times 2 over denominator straight m plus 1 end fraction space space तथ ा space space straight y with bar on top space equals space fraction numerator 7 straight m plus 1 left parenthesis negative 2 right parenthesis over denominator straight m plus 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space straight x with bar on top space equals fraction numerator 3 straight m plus 2 over denominator straight m plus 1 end fraction space space तथ ा space space straight y with bar on top space space equals space fraction numerator 7 straight m minus 2 over denominator straight m plus 1 end fraction
चूँकि left parenthesis straight x with bar on top comma space straight y with bar on top right parenthesis रेखा 2x + y - 4 पर स्थित हैं, 
अत: 2 open parentheses fraction numerator 3 straight m plus 2 over denominator straight m plus 1 end fraction close parentheses plus space open parentheses fraction numerator 7 straight m minus 2 over denominator straight m plus 1 end fraction close parentheses minus 4 space equals space 0
rightwards double arrow space space fraction numerator 6 straight m plus 7 plus 7 straight m minus 2 minus 2 straight m minus 4 over denominator left parenthesis straight m plus 1 right parenthesis end fraction space equals 0
rightwards double arrow space space space space space space 9 straight m minus 2 space equals space 0
rightwards double arrow space space space space space space space space space space straight m space equals space 2 over 9 space equals space 2 colon 9
therefore space space space space space space space space space अभ ी ष ् ट space अन ु प ा त space space equals space 2 space colon space 9

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