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Straight Lines and Pair of Straight Lines question

2019 · 8 Apr · Shift 2 · Q44
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Straight Lines and Pair of Straight Lines question

2019 · 8 Apr · Shift 2 · Q44

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Suppose that the points (h,k), (1,2) and (–3,4) lie on the line L1 . If a line L2 passing through the points (h,k) and (4,3) is perpendicular to L1 , then khk \over hhk​ equals :
  1. A
    13{1 \over 3}31​
  2. B
    3
  3. C
    0
  4. D
    -17{1 \over 7}71​
View written solutionFree

Correct answer: A

  1. Since the points (1,2)(1,2)(1,2) and (−3,4)(-3,4)(−3,4) lie on line L1L_1L1​, its slope is m1=4−2−3−1=2−4=−12.m_1=\frac{4-2}{-3-1}=\frac{2}{-4}=-\frac12.m1​=−3−14−2​=−42​=−21​.

  2. The point (h,k)(h,k)(h,k) also lies on L1L_1L1​, so using point-slope form through (1,2)(1,2)(1,2): y−2=−12(x−1).y-2=-\frac12(x-1).y−2=−21​(x−1). Thus (h,k)(h,k)(h,k) satisfies k−2=−12(h−1).k-2=-\frac12(h-1).k−2=−21​(h−1). So, 2k−4=−(h−1)=−h+12k-4=-(h-1)=-h+12k−4=−(h−1)=−h+1 2k=−h+52k=-h+52k=−h+5 h+2k=5.(1)h+2k=5. \qquad (1)h+2k=5.(1)

  3. Line L2L_2L2​ passes through (h,k)(h,k)(h,k) and (4,3)(4,3)(4,3) and is perpendicular to L1L_1L1​. Since m1=−12m_1=-\frac12m1​=−21​, the perpendicular slope is m2=2.m_2=2.m2​=2.

So slope of the line through (h,k)(h,k)(h,k) and (4,3)(4,3)(4,3) is 3−k4−h=2.\frac{3-k}{4-h}=2.4−h3−k​=2. Hence, 3−k=2(4−h)3-k=2(4-h)3−k=2(4−h) 3−k=8−2h3-k=8-2h3−k=8−2h 2h−k=5.(2)2h-k=5. \qquad (2)2h−k=5.(2)

  1. Solve equations (1)(1)(1) and (2)(2)(2): (1): h+2k=5(1):\ h+2k=5(1): h+2k=5 (2): 2h−k=5(2):\ 2h-k=5(2): 2h−k=5

From (2)(2)(2), k=2h−5.k=2h-5.k=2h−5. Substitute into (1)(1)(1): h+2(2h−5)=5h+2(2h-5)=5h+2(2h−5)=5 h+4h−10=5h+4h-10=5h+4h−10=5 5h=155h=155h=15 h=3.h=3.h=3. Then, k=2(3)−5=1.k=2(3)-5=1.k=2(3)−5=1.

  1. Therefore, kh=13.\frac{k}{h}=\frac{1}{3}.hk​=31​.

So the correct option is A.

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