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Straight Lines and Pair of Straight Lines question

2019 · 9 Apr · Shift 1 · Q44
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Straight Lines and Pair of Straight Lines question

2019 · 9 Apr · Shift 1 · Q44

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is :
  1. A
    7−17+1{{\sqrt 7 - 1} \over {\sqrt 7 + 1}}7​+17​−1​
  2. B
    5−15+1{{\sqrt 5 - 1} \over {\sqrt 5 + 1}}5​+15​−1​
  3. C
    1−51+5{{1 - \sqrt 5 } \over {1 + \sqrt 5 }}1+5​1−5​​
  4. D
    1−71+7{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}1+7​1−7​​
View written solutionFree

Correct answer: D

  1. Let the required line through P(2,3)P(2,3)P(2,3) have slope mmm.

    Its equation is y−3=m(x−2).y-3=m(x-2).y−3=m(x−2).

  2. This line intersects the line x+y=7x+y=7x+y=7 at some point QQQ.

    Since QQQ lies on both lines, substitute y=3+m(x−2)y=3+m(x-2)y=3+m(x−2) into x+y=7x+y=7x+y=7: x+3+m(x−2)=7x+3+m(x-2)=7x+3+m(x−2)=7 x+mx+3−2m=7x+mx+3-2m=7x+mx+3−2m=7 x(1+m)=4+2mx(1+m)=4+2mx(1+m)=4+2m xQ=4+2m1+m=2(m+2)m+1.x_Q=\frac{4+2m}{1+m}=\frac{2(m+2)}{m+1}.xQ​=1+m4+2m​=m+12(m+2)​.

    Then

    =\frac{7(m+1)-2(m+2)}{m+1} =\frac{5m+3}{m+1}.$$

  3. Given that the intersection point QQQ is at distance 444 from P(2,3)P(2,3)P(2,3): PQ=4.PQ=4.PQ=4.

    So, (xQ−2)2+(yQ−3)2=16.\left(x_Q-2\right)^2+\left(y_Q-3\right)^2=16.(xQ​−2)2+(yQ​−3)2=16.

    Compute the differences:

    =\frac{2m+4-2m-2}{m+1} =\frac{2}{m+1},$$ $$y_Q-3=\frac{5m+3}{m+1}-3 =\frac{5m+3-3m-3}{m+1} =\frac{2m}{m+1}.$$ Hence, $$\left(\frac{2}{m+1}\right)^2+\left(\frac{2m}{m+1}\right)^2=16.$$ $$\frac{4+4m^2}{(m+1)^2}=16.$$ Divide by $4$: $$\frac{1+m^2}{(m+1)^2}=4.$$ Therefore, $$1+m^2=4(m+1)^2=4(m^2+2m+1).$$ $$1+m^2=4m^2+8m+4$$ $$0=3m^2+8m+3.$$

  4. Solve the quadratic: 3m2+8m+3=0.3m^2+8m+3=0.3m2+8m+3=0.

    =\frac{-8\pm\sqrt{28}}{6} =\frac{-8\pm 2\sqrt7}{6} =\frac{-4\pm\sqrt7}{3}.$$ So the two possible slopes are $$m=\frac{-4+\sqrt7}{3},\quad m=\frac{-4-\sqrt7}{3}.$$

  5. Match with options.

    Check option D: 1−71+7\frac{1-\sqrt7}{1+\sqrt7}1+7​1−7​​ Rationalizing,

    =\frac{(1-\sqrt7)^2}{1-7} =\frac{1-2\sqrt7+7}{-6} =\frac{8-2\sqrt7}{-6} =\frac{-4+\sqrt7}{3}.$$ This matches one of the slopes. Check option A: $$\frac{\sqrt7-1}{\sqrt7+1} =\frac{(\sqrt7-1)^2}{7-1} =\frac{8-2\sqrt7}{6} =\frac{4-\sqrt7}{3},$$ which is not equal to either root. Thus among the given options, the correct one is **D**.

  6. Note: geometrically there are two possible lines through PPP meeting x+y=7x+y=7x+y=7 at distance 444, with slopes −4+73and−4−73.\frac{-4+\sqrt7}{3} \quad \text{and} \quad \frac{-4-\sqrt7}{3}.3−4+7​​and3−4−7​​. But only the first one appears in the options, namely option D.

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