Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Straight Lines and Pair of Straight Lines question

2025 · 8 Apr · Shift 2 · Q29
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Straight Lines and Pair of Straight Lines
  5. /2025 · 8 Apr · Shift 2 · Q29

Straight Lines and Pair of Straight Lines question

2025 · 8 Apr · Shift 2 · Q29

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle α\alphaα with the positive x-axis and the equations of its diagonals are (3+1)x+(3−1)y=0(\sqrt{3}+1)x+(\sqrt{3}-1)y=0(3​+1)x+(3​−1)y=0 and (3−1)x−(3+1)y+83=0(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0(3​−1)x−(3​+1)y+83​=0. Then aaa 2 is equal to :
  1. A
    48
  2. B
    16
  3. C
    24
  4. D
    32
View written solutionFree

Correct answer: A

  1. Interpret the geometry

Let the square be OABCOABCOABC with O=(0,0)O=(0,0)O=(0,0) and side length aaa.

Since OAOAOA makes an acute angle α\alphaα with the positive xxx-axis, we can write A=(acos⁡α,asin⁡α).A=(a\cos\alpha,a\sin\alpha).A=(acosα,asinα). A side perpendicular to OAOAOA gives OC→=a(−sin⁡α,cos⁡α).\overrightarrow{OC}=a(-\sin\alpha,\cos\alpha).OC=a(−sinα,cosα). Hence the fourth vertex is B=A+C=(a(cos⁡α−sin⁡α), a(sin⁡α+cos⁡α)).B=A+C=(a(\cos\alpha-\sin\alpha),\ a(\sin\alpha+\cos\alpha)).B=A+C=(a(cosα−sinα), a(sinα+cosα)).

The diagonals of the square are:

  • OBOBOB, joining OOO and BBB
  • ACACAC, joining AAA and CCC
  1. Identify which given line is which diagonal

The given diagonal equations are (3+1)x+(3−1)y=0...(1)(\sqrt3+1)x+(\sqrt3-1)y=0 \qquad ...(1)(3​+1)x+(3​−1)y=0...(1) (3−1)x−(3+1)y+83=0...(2)(\sqrt3-1)x-(\sqrt3+1)y+8\sqrt3=0 \qquad ...(2)(3​−1)x−(3​+1)y+83​=0...(2)

Since diagonal OBOBOB passes through the origin, equation (1) must represent OBOBOB.

So its slope is mOB=−3+13−1.m_{OB}=-\frac{\sqrt3+1}{\sqrt3-1}.mOB​=−3​−13​+1​. Rationalizing, mOB=−(3+1)23−1=−4+232=−(2+3).m_{OB}=-\frac{(\sqrt3+1)^2}{3-1}=-\frac{4+2\sqrt3}{2}=-(2+\sqrt3).mOB​=−3−1(3​+1)2​=−24+23​​=−(2+3​). But tan⁡75∘=2+3,\tan 75^\circ=2+\sqrt3,tan75∘=2+3​, so mOB=−(2+3)=tan⁡(−75∘).m_{OB}=-(2+\sqrt3)=\tan(-75^\circ).mOB​=−(2+3​)=tan(−75∘). Thus the diagonal OBOBOB makes angle 105∘105^\circ105∘ with the positive xxx-axis.

Now in a square, diagonal OBOBOB bisects the right angle at OOO, so if side OAOAOA makes angle α\alphaα, then angle of OB=α+45∘.\text{angle of }OB=\alpha+45^\circ.angle of OB=α+45∘. Therefore, α+45∘=105∘  ⟹  α=60∘.\alpha+45^\circ=105^\circ \implies \alpha=60^\circ.α+45∘=105∘⟹α=60∘.

  1. Find side equations for diagonal ACACAC

Now using α=60∘\alpha=60^\circα=60∘, A=(acos⁡60∘,asin⁡60∘)=(a2,3a2),A=\left(a\cos60^\circ,a\sin60^\circ\right)=\left(\frac a2,\frac{\sqrt3 a}{2}\right),A=(acos60∘,asin60∘)=(2a​,23​a​), C=(−asin⁡60∘,acos⁡60∘)=(−3a2,a2).C=\left(-a\sin60^\circ,a\cos60^\circ\right)=\left(-\frac{\sqrt3 a}{2},\frac a2\right).C=(−asin60∘,acos60∘)=(−23​a​,2a​).

The diagonal ACACAC passes through the midpoint of AAA and CCC, or we can use slope.

Slope of ACACAC:

=\frac{1-\sqrt3}{-(\sqrt3+1)} =\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3.$$ Now from equation (2): $$(\sqrt3-1)x-(\sqrt3+1)y+8\sqrt3=0$$ $$\Rightarrow y=\frac{\sqrt3-1}{\sqrt3+1}x+\frac{8\sqrt3}{\sqrt3+1}.$$ So its slope is also $$\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3,$$ confirming that (2) is indeed the diagonal $AC$. 4. **Use the center of square** The diagonals of a square bisect each other. So their intersection point is the center of the square. Solve (1) and (2): $$(\sqrt3+1)x+(\sqrt3-1)y=0 \qquad ...(1)$$ $$(\sqrt3-1)x-(\sqrt3+1)y+8\sqrt3=0 \qquad ...(2)$$ From (1), $$y=-\frac{\sqrt3+1}{\sqrt3-1}x=-(2+\sqrt3)x.$$ Substitute into (2): $$(\sqrt3-1)x-(\sqrt3+1)(-(2+\sqrt3)x)+8\sqrt3=0.$$ Since $$(\sqrt3+1)(2+\sqrt3)=3\sqrt3+5,$$ we get $$\big((\sqrt3-1)+(3\sqrt3+5)\big)x+8\sqrt3=0$$ $$=(4\sqrt3+4)x+8\sqrt3=0$$ $$4(\sqrt3+1)x=-8\sqrt3$$ $$x=-\frac{2\sqrt3}{\sqrt3+1}=3-\sqrt3.$$ Then from (1), $$(\sqrt3+1)(3-\sqrt3)+(\sqrt3-1)y=0.$$ Now $$(\sqrt3+1)(3-\sqrt3)=2\sqrt3,$$ so $$2\sqrt3+(\sqrt3-1)y=0$$ $$y=-\frac{2\sqrt3}{\sqrt3-1}=-(3+\sqrt3).$$ Thus center is $$\left(3-\sqrt3, -(3+\sqrt3)\right).$$ 5. **Relate center to side length** In a square, the center is the midpoint of diagonal $OB$. Hence $$B=2\times \text{center}=(6-2\sqrt3, -6-2\sqrt3).$$ But the diagonal length of a square is $a\sqrt2$, so $$OB=a\sqrt2.$$ Now $$OB^2=(6-2\sqrt3)^2+(-6-2\sqrt3)^2.$$ Compute: $$(6-2\sqrt3)^2=36+12-24\sqrt3=48-24\sqrt3,$$ $$(6+2\sqrt3)^2=36+12+24\sqrt3=48+24\sqrt3.$$ Hence $$OB^2=(48-24\sqrt3)+(48+24\sqrt3)=96.$$ So $$(a\sqrt2)^2=96$$ $$2a^2=96$$ $$a^2=48.$$ 6. **Check options** Thus, $$a^2=48.$$ So the correct option is **A**.
PreviousNext

More from Straight Lines and Pair of Straight Lines

  • A line passing through the point P(a, 0) makes an acute angle α with the positive x-axis. Let this line be rotated about the point P through an angle 2α​ in the clockwise direction. If in the new position, the slope…2025 · MCQ
  • Let the triangle PQR be the image of the triangle with vertices (1,3),(3,1) and (2,4) in the line x+2y=2. If the centroid of △PQR is the point (α,β), then 15(α−β) is equal to :2025 · MCQ
  • Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on…2025 · Numerical
  • A rod of length eight units moves such that its ends A and B always lie on the lines x−y+2=0 and y+2=0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2:1 is 9(x2+αy2+βxy+γx+28y)−76=0…2025 · MCQ
  • Let the lines 3x−4y−α=0,8x−11y−33=0, and 2x−3y+λ=0 be concurrent. If the image of the point (1,2) in the line 2x−3y+λ=0 is (1357​,13−40​), then ∣αλ∣ is equal to2025 · MCQ
  • Let the points (211​,α) lie on or inside the triangle with sides x+y=11,x+2y=16 and 2x+3y=29. Then the product of the smallest and the largest values of α is equal to :2025 · MCQ
  • Two equal sides of an isosceles triangle are along −x+2y=4 and x+y=4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is:2025 · MCQ
  • If A and B are the points of intersection of the circle x2+y2−8x=0 and the hyperbola 9x2​−4y2​=1 and a point P moves on the line 2x−3y+4=0, then the centroid of ΔPAB lies on the line :2025 · MCQ