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Straight Lines and Pair of Straight Lines question

2020 · 7 Jan · Shift 2 · Q25
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Straight Lines and Pair of Straight Lines question

2020 · 7 Jan · Shift 2 · Q25

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is :
  1. A
    3x - 2y = 0
  2. B
    7x - 5y = 0
  3. C
    2x - 3y = 0
  4. D
    5x - 7y = 0
View written solutionFree

Correct answer: D

  1. Given lines
  • Points lie on the line x=2yx=2yx=2y.
  • From each such point, a perpendicular is drawn to the line x=yx=yx=y.
  • We need the locus of the midpoint of that perpendicular segment.
  1. Take a general point on x=2yx=2yx=2y

Let the point be P(2t,t)P(2t,t)P(2t,t), since x=2yx=2yx=2y.

The line x=yx=yx=y can be written as x−y=0.x-y=0.x−y=0.

  1. Find the foot of perpendicular from P(2t,t)P(2t,t)P(2t,t) to x−y=0x-y=0x−y=0

For a point (x1,y1)(x_1,y_1)(x1​,y1​) and line ax+by+c=0ax+by+c=0ax+by+c=0, the foot of perpendicular is

y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2}\right).$$ Here, $$a=1,\quad b=-1,\quad c=0,\quad (x_1,y_1)=(2t,t).$$ Compute: $$ax_1+by_1+c=2t-t=t,$$ $$a^2+b^2=1+1=2.$$ So foot of perpendicular $Q$ is $$Q\left(2t-\frac{t}{2},\ t-\frac{-t}{2}\right) =\left(\frac{3t}{2},\frac{3t}{2}\right).$$ --- 4. **Find midpoint of** $PQ$ Let midpoint be $M(X,Y)$. Then $$X=\frac{2t+\frac{3t}{2}}{2}= rac{\frac{7t}{2}}{2}=\frac{7t}{4},$$ $$Y=\frac{t+\frac{3t}{2}}{2}= rac{\frac{5t}{2}}{2}=\frac{5t}{4}.$$ So $$M\left(\frac{7t}{4},\frac{5t}{4}\right).$$ --- 5. **Eliminate parameter** $t$ From $$X=\frac{7t}{4},\qquad Y=\frac{5t}{4},$$ we get $$\frac{X}{7}=\frac{Y}{5}.$$ Hence, $$5X=7Y$$ or $$5x-7y=0.$$ --- 6. **Match with options** The required locus is $$\boxed{5x-7y=0}.$$ So the correct option is **D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.
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