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Straight Lines and Pair of Straight Lines question

2021 · 27 Jul · Shift 2 · Q32
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Straight Lines and Pair of Straight Lines question

2021 · 27 Jul · Shift 2 · Q32

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals of the parallelogram is 11x + 7y = 9, then other diagonal passes through the point :
  1. A
    (1, 2)
  2. B
    (2, 2)
  3. C
    (2, 1)
  4. D
    (1, 3)
View written solutionFree

Correct answer: B

  1. Given sides of the parallelogram

The two sides lie along 4x+5y=04x+5y=04x+5y=0 and 7x+2y=0.7x+2y=0.7x+2y=0.

Since both pass through the origin, they can be taken as the two adjacent sides of the parallelogram meeting at the origin.

So let the adjacent side vectors be along these lines:

  • along 4x+5y=04x+5y=04x+5y=0, a direction vector is (5,−4)(5,-4)(5,−4)
  • along 7x+2y=07x+2y=07x+2y=0, a direction vector is (2,−7)(2,-7)(2,−7)

Hence take the two side vectors as a⃗=λ(5,−4),b⃗=μ(2,−7)\vec a = \lambda(5,-4), \qquad \vec b = \mu(2,-7)a=λ(5,−4),b=μ(2,−7) for some scalars λ,μ\lambda,\muλ,μ.

Then the four vertices of the parallelogram are O=(0,0),A=a⃗,B=b⃗,C=a⃗+b⃗.O=(0,0),\quad A=\vec a,\quad B=\vec b,\quad C=\vec a+\vec b.O=(0,0),A=a,B=b,C=a+b.

  1. Use the given diagonal

One diagonal is given by 11x+7y=9.11x+7y=9.11x+7y=9.

Since this is not passing through the origin, it must be the diagonal joining AAA and BBB.

Therefore both points AAA and BBB lie on the line 11x+7y=911x+7y=911x+7y=9.

  1. Find AAA on both lines

Point AAA lies on 4x+5y=04x+5y=04x+5y=0 and also on 11x+7y=9.11x+7y=9.11x+7y=9.

Solve:

From 4x+5y=04x+5y=04x+5y=0, we get x=−5y4.x=-\frac{5y}{4}.x=−45y​.

Substitute into 11x+7y=911x+7y=911x+7y=9: 11(−5y4)+7y=911\left(-\frac{5y}{4}\right)+7y=911(−45y​)+7y=9 −55y4+28y4=9-\frac{55y}{4}+\frac{28y}{4}=9−455y​+428y​=9 −27y4=9-\frac{27y}{4}=9−427y​=9 y=−43.y=-\frac{4}{3}.y=−34​.

Then x=−5(−4/3)4=53.x=-\frac{5(-4/3)}{4}=\frac{5}{3}.x=−45(−4/3)​=35​.

So A=(53,−43).A=\left(\frac{5}{3},-\frac{4}{3}\right).A=(35​,−34​).

  1. Find BBB on both lines

Point BBB lies on 7x+2y=07x+2y=07x+2y=0 and also on 11x+7y=9.11x+7y=9.11x+7y=9.

From 7x+2y=07x+2y=07x+2y=0, we get y=−7x2.y=-\frac{7x}{2}.y=−27x​.

Substitute into 11x+7y=911x+7y=911x+7y=9: 11x+7(−7x2)=911x+7\left(-\frac{7x}{2}\right)=911x+7(−27x​)=9 11x−49x2=911x-\frac{49x}{2}=911x−249x​=9 22x−49x2=9\frac{22x-49x}{2}=9222x−49x​=9 −27x2=9-\frac{27x}{2}=9−227x​=9 x=−23.x=-\frac{2}{3}.x=−32​.

Then y=−7(−2/3)2=73.y=-\frac{7(-2/3)}{2}=\frac{7}{3}.y=−27(−2/3)​=37​.

So B=(−23,73).B=\left(-\frac{2}{3},\frac{7}{3}\right).B=(−32​,37​).

  1. Find the midpoint of diagonals

Diagonals of a parallelogram bisect each other.

So midpoint of diagonal ABABAB is also the midpoint of the other diagonal.

Midpoint of ABABAB is

=\left(\frac{1}{2},\frac{1}{2}\right).$$ 6. **Equation of the other diagonal** The other diagonal joins $O=(0,0)$ and $C=A+B$, so it must pass through the origin and the midpoint $M\left(\frac12,\frac12\right)$. Hence its equation is $$y=x.$$ 7. **Check which option lies on $y=x$** - A: $(1,2)$ gives $2\ne1$ ❌ - B: $(2,2)$ gives $2=2$ ✅ - C: $(2,1)$ gives $1\ne2$ ❌ - D: $(1,3)$ gives $3\ne1$ ❌ Therefore, the other diagonal passes through $$\boxed{(2,2)}.$$
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