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Straight Lines and Pair of Straight Lines question

2018 · 15 Apr · Shift 2 · Q36
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Straight Lines and Pair of Straight Lines question

2018 · 15 Apr · Shift 2 · Q36

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
The sides of a rhombus ABCD are parallel to the lines, x −-− y + 2 = 0 and 7x −-− y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :
  1. A
    52{5 \over 2}25​
  2. B
    74{7 \over 4}47​
  3. C
    2
  4. D
    72{7 \over 2}27​
View written solutionFree

Correct answer: A

  1. Interpret the question

Since vertex AAA is on the yyy-axis, let A=(0,a),a≠0.A=(0,a), \quad a\neq 0.A=(0,a),a=0.

The sides of the rhombus are parallel to the lines x−y+2=0and7x−y+3=0.x-y+2=0 \quad \text{and} \quad 7x-y+3=0.x−y+2=0and7x−y+3=0. So the two side directions have slopes:

  • from x−y+2=0⇒y=x+2x-y+2=0 \Rightarrow y=x+2x−y+2=0⇒y=x+2, slope 111
  • from 7x−y+3=0⇒y=7x+37x-y+3=0 \Rightarrow y=7x+37x−y+3=0⇒y=7x+3, slope 777

Thus, from AAA, the two adjacent sides of the rhombus go along directions with slopes 111 and 777.


  1. Take side vectors along these directions

Let the two side vectors from AAA be AB⃗=t(1,1),AD⃗=s(1,7).\vec{AB}=t(1,1), \qquad \vec{AD}=s(1,7).AB=t(1,1),AD=s(1,7).

Because ABCDABCDABCD is a rhombus, adjacent sides are equal: ∣AB⃗∣=∣AD⃗∣.|\vec{AB}|=|\vec{AD}|.∣AB∣=∣AD∣.

Now, ∣(1,1)∣=2,∣(1,7)∣=50=52.|(1,1)|=\sqrt{2}, \qquad |(1,7)|=\sqrt{50}=5\sqrt{2}.∣(1,1)∣=2​,∣(1,7)∣=50​=52​. So, ∣t∣2=∣s∣⋅52⇒∣t∣=5∣s∣.|t|\sqrt{2}=|s|\cdot 5\sqrt{2} \Rightarrow |t|=5|s|.∣t∣2​=∣s∣⋅52​⇒∣t∣=5∣s∣.

Hence we may take t=5k,s=kt=5k, \qquad s=kt=5k,s=k for some real kkk (sign can be absorbed into kkk).

Therefore, AB⃗=5k(1,1),AD⃗=k(1,7).\vec{AB}=5k(1,1), \qquad \vec{AD}=k(1,7).AB=5k(1,1),AD=k(1,7).


  1. Use the midpoint of diagonals

In a rhombus (parallelogram), diagonals bisect each other. Since they intersect at P=(1,2),P=(1,2),P=(1,2), we have P=A+C2.P=\frac{A+C}{2}.P=2A+C​.

Also, for a parallelogram, C=A+AB⃗+AD⃗.C=A+\vec{AB}+\vec{AD}.C=A+AB+AD. So, P=A+AB⃗+AD⃗2.P=A+\frac{\vec{AB}+\vec{AD}}{2}.P=A+2AB+AD​.

Now, AB⃗+AD⃗=5k(1,1)+k(1,7)=(6k,12k).\vec{AB}+\vec{AD}=5k(1,1)+k(1,7)=(6k,12k).AB+AD=5k(1,1)+k(1,7)=(6k,12k). Thus, P=(0,a)+(6k,12k)2=(0,a)+(3k,6k)=(3k,a+6k).P=(0,a)+\frac{(6k,12k)}{2}=(0,a)+(3k,6k)=(3k,a+6k).P=(0,a)+2(6k,12k)​=(0,a)+(3k,6k)=(3k,a+6k).

Given P=(1,2)P=(1,2)P=(1,2), compare coordinates: 3k=1⇒k=13.3k=1 \Rightarrow k=\frac13.3k=1⇒k=31​. Then, a+6(13)=2⇒a+2=2⇒a=0.a+6\left(\frac13\right)=2 \Rightarrow a+2=2 \Rightarrow a=0.a+6(31​)=2⇒a+2=2⇒a=0. This gives A=(0,0)A=(0,0)A=(0,0), but the question says AAA is different from the origin.

So we must also allow the second side direction to be taken in the opposite sense: AD⃗=−k(1,7),AB⃗=5k(1,1).\vec{AD}=-k(1,7), \qquad \vec{AB}=5k(1,1).AD=−k(1,7),AB=5k(1,1). Then AB⃗+AD⃗=5k(1,1)−k(1,7)=(4k,−2k).\vec{AB}+\vec{AD}=5k(1,1)-k(1,7)=(4k,-2k).AB+AD=5k(1,1)−k(1,7)=(4k,−2k). Hence P=(0,a)+(4k,−2k)2=(0,a)+(2k,−k)=(2k,a−k).P=(0,a)+\frac{(4k,-2k)}{2}=(0,a)+(2k,-k)=(2k,a-k).P=(0,a)+2(4k,−2k)​=(0,a)+(2k,−k)=(2k,a−k).

Now use P=(1,2)P=(1,2)P=(1,2): 2k=1⇒k=12,2k=1 \Rightarrow k=\frac12,2k=1⇒k=21​, a−k=2⇒a=2+12=52.a-k=2 \Rightarrow a=2+\frac12=\frac52.a−k=2⇒a=2+21​=25​.

Therefore, A=(0,52).A=(0,\tfrac52).A=(0,25​).

So the coordinate asked is 52.\boxed{\frac52}.25​​.


  1. Check options

The yyy-coordinate of AAA is 52\dfrac5225​, which matches Option A.


  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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