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

2023 · 29 Jan · Shift 1 · Q23
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. /2023 · 29 Jan · Shift 1 · Q23

Straight Lines and Pair of Straight Lines question

2023 · 29 Jan · Shift 1 · Q23

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let BBB and CCC be the two points on the line y+x=0y+x=0y+x=0 such that BBB and CCC are symmetric with respect to the origin. Suppose AAA is a point on y−2x=2y-2 x=2y−2x=2 such that △ABC\triangle A B C△ABC is an equilateral triangle. Then, the area of the △ABC\triangle A B C△ABC is :
  1. A
    103\frac{10}{\sqrt{3}}3​10​
  2. B
    232 \sqrt{3}23​
  3. C
    333 \sqrt{3}33​
  4. D
    83\frac{8}{\sqrt{3}}3​8​
View written solutionFree

Correct answer: D

  1. Set coordinates of BBB and CCC

Since BBB and CCC lie on the line x+y=0x+y=0x+y=0, let B=(t,−t).B=(t,-t).B=(t,−t). Because BBB and CCC are symmetric with respect to the origin, we must have C=(−t,t).C=(-t,t).C=(−t,t).

So, BCBCBC is the segment joining (t,−t)(t,-t)(t,−t) and (−t,t)(-t,t)(−t,t).

  1. Find the midpoint and length of BCBCBC

The midpoint of BCBCBC is (t+(−t)2,−t+t2)=(0,0).\left(\frac{t+(-t)}{2},\frac{-t+t}{2}\right)=(0,0).(2t+(−t)​,2−t+t​)=(0,0). So the origin is the midpoint of BCBCBC.

Now,

=\sqrt{(2t)^2+(-2t)^2} =\sqrt{8t^2}=2\sqrt{2}|t|.$$ --- 3. **Use the equilateral triangle condition** If $\triangle ABC$ is equilateral, then $A$ must lie on the perpendicular bisector of $BC$. Now the line $BC$ is $x+y=0$, whose slope is $-1$. Therefore, the perpendicular bisector has slope $1$ and passes through the origin, so its equation is $$y=x.$$ Also, $A$ lies on the line $$y-2x=2.$$ So $A$ is the intersection of $$y=x \quad \text{and} \quad y-2x=2.$$ Substitute $y=x$ into the second equation: $$x-2x=2 \implies -x=2 \implies x=-2.$$ Hence, $$A=(-2,-2).$$ --- 4. **Distance from $A$ to the line $BC$** In an equilateral triangle, the altitude from $A$ to $BC$ passes through the midpoint of $BC$, which is the origin. Thus the altitude is $$AO=\text{distance from }(-2,-2)\text{ to }x+y=0.$$ Using point-to-line distance formula: $$AO=\frac{|(-2)+(-2)|}{\sqrt{1^2+1^2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}.$$ For an equilateral triangle of side $s$, altitude is $$h=\frac{\sqrt{3}}{2}s.$$ So, $$2\sqrt{2}=\frac{\sqrt{3}}{2}s \implies s=\frac{4\sqrt{2}}{\sqrt{3}}.$$ --- 5. **Compute the area** Area of an equilateral triangle is $$\text{Area}=\frac{\sqrt{3}}{4}s^2.$$ Now, $$s^2=\left(\frac{4\sqrt{2}}{\sqrt{3}}\right)^2=\frac{32}{3}.$$ Therefore, $$\text{Area}=\frac{\sqrt{3}}{4}\cdot \frac{32}{3} =\frac{8\sqrt{3}}{3} =\frac{8}{\sqrt{3}}.$$ --- 6. **Check with options** $$\frac{8}{\sqrt{3}}$$ matches **Option D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So the stored answer is correct.
PreviousNext

More from Straight Lines and Pair of Straight Lines

  • A light ray emits from the origin making an angle 30 ∘ with the positive x-axis. After getting reflected by the line x+y=1, if this ray intersects x-axis at Q, then the abscissa of Q is :2023 · MCQ
  • Let A(a​3​,a​),a>0, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D(3cosθ,asinθ) is a point in the fourth quadrant…2022 · Numerical
  • Let the area of the triangle with vertices A(1, α), B(α, 0) and C(0, α) be 4 sq. units. If the points (α, −α), (−α, α) and (α 2, β) are collinear, then β is equal to :2022 · MCQ
  • A line, with the slope greater than one, passes through the point A(4,3) and intersects the line x−y−2=0 at the point B. If the length of the line segment AB is 329​​, then B also lies on the line :2022 · MCQ
  • Let the point P(α,β) be at a unit distance from each of the two lines L1​:3x−4y+12=0, and L2​:8x+6y+11=0. If P lies below L1​ and above L2​, then 100(α+β) is equal to :2022 · MCQ
  • A point P moves so that the sum of squares of its distances from the points (1,2) and (−2,1) is 14. Let f(x,y)=0 be the locus of P, which intersects the x-axis at the points A, B and the y-axis…2022 · MCQ
  • The equations of the sides AB,BC and CA of a triangle ABC are 2x+y=0,x+py=15a and x−y=3 respectively. If its orthocentre is (2,a),−21​<a<2, then…2022 · Numerical
  • Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of Δ PQR is :2022 · MCQ