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JEE Main & Advanced Mathematics Straight Line Question Bank Self Evaluation Test - Straight Lines & Pair Striaght Lines

  • question_answer
    A line which makes an acute angle \[\theta \] with the positive direction of x-axis is drawn through the point P(3, 4) to meet the line \[x=6\] at R and \[y=8\] at S, then

    A) \[PR=3\cos \theta \]       

    B) \[PS=-4\cos ec\theta \]

    C) \[PR-PS=\frac{2(3sin\theta +4cos\theta )}{\sin 2\theta }\]

    D) \[\frac{9}{{{(PR)}^{2}}}+\frac{16}{{{(PS)}^{2}}}=1\]

    Correct Answer: D

    Solution :

    [d] The equation of the line in parametric form is \[\frac{x-3}{\cos \theta }=\frac{y-4}{\sin \theta }=r\] Any point on this line is \[(3+r\,cos\theta ,4+r\,sin\theta )\] It lies on \[x=6\] if \[3+r\cos \theta =6\Rightarrow r=3\sec \theta \] \[\therefore \,\,\,\,\operatorname{PR}=3\,sec\,\,\theta \] Again the point lies on \[y=8\] if \[4+r\sin \theta =8\] \[\therefore \,\,\,\,r=4\,\,\cos ec\,\,\theta \] or \[PS=4\cos ec\,\,\theta \] other options can be chocked easily


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