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JEE Main & Advanced Mathematics Straight Line Question Bank Mock Test - Straight Lines

  • question_answer
    If each of the points, \[({{x}_{1}},4)\], \[(-2,{{y}_{1}})\]lies on the line joining the points (2, -1) and (5, -3), then the point \[p({{x}_{1}},{{y}_{1}})\]lies on the line

    A) \[6(x+y)-25=0\]

    B) \[2x+6y+1=0\]

    C) \[2x+3y-6=0\]

    D) \[6(x+y)+25=0\]

    Correct Answer: B

    Solution :

    [b] The equation of the line joining the points (2,-1) and (5,-3) is given by \[y+1=\frac{-1+3}{2-5}(x-2)\] Or \[2x+3y-1=0\] Since \[({{x}_{1}},4)\]and \[(-2,{{y}_{1}})\]lie on \[2x+3y-1=0\], we have \[2{{x}_{1}}+12-1=0\]or \[{{x}_{1}}=-\frac{11}{2}\] And \[-4+3{{y}_{1}}-1=0\]or \[{{y}_{1}}=\frac{5}{3}\] Thus, \[({{x}_{1}},{{y}_{1}})\]satisfies \[2x+6y+1=0\].


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