JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area of an equilateral triangle with symmetric base vertices

Q: Why must the altitude pass through the origin?

Because B and C are symmetric about the origin, the origin is the midpoint of BC. In any equilateral triangle, the altitude from the opposite vertex bisects the base, so the altitude line from A to BC must pass through the midpoint of BC, which is the origin.

Let B and C be the two points on the line y + x = 0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y - 2x = 2 such that Δ ABC is an equilateral triangle. Then, the area of the Δ ABC is

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A10/(√(3))B3√(3)C2√(3)D8/(√(3))

Step-by-step solutionView

Correct answer

The area of $Δ A B C$ is $\frac{8}{\sqrt{3}}$ , found by locating vertex $A$ on the perpendicular bisector of $B C$ (the line $y = x$ ) and using the height to compute the area.

Option analysis

Why each option works or fails

A · 10/(√(3))

This arises from miscalculating the coordinates of vertex $A$ or using an incorrect formula relating the altitude $h$ to the area, such as $\frac{h^{2}}{2}$ instead of $\frac{h^{2}}{\sqrt{3}}$ . Ensure the altitude of an equilateral triangle is related to its area by $Area = \frac{h^{2}}{\sqrt{3}}$ , where $h$ is the distance from the opposite vertex to the midpoint of the base.

B · 3√(3)

This comes from an algebraic error when solving the system $y - 2 x = 2$ and $y = x$ , leading to $x = - 3$ instead of $x = - 2$ . Substitute $y = x$ directly into $y - 2 x = 2$ to obtain $x - 2 x = 2 ⟹ - x = 2 ⟹ x = - 2$ .

C · 2√(3)

This results from taking the distance from $A$ to the origin as the side length $s$ rather than the altitude $h$ , giving $\frac{\sqrt{3}}{4} (2 \sqrt{2})^{2} = 2 \sqrt{3}$ . Recognize that the origin is the midpoint of $B C$ , so the distance from $A$ to the origin is the altitude $h$ , not the side length $s$ .

D · 8/(√(3))

This is the correct calculation. The midpoint of $B C$ is the origin $(0, 0)$ . The altitude through $A$ is the perpendicular bisector of $B C$ , which is the line $y = x$ . Intersecting $y = x$ with $y - 2 x = 2$ gives $A (- 2, - 2)$ . The altitude is $h = O A = \sqrt{(- 2)^{2} + (- 2)^{2}} = 2 \sqrt{2}$ , giving $Area = \frac{h^{2}}{\sqrt{3}} = \frac{8}{\sqrt{3}}$ .

Reviewed route

Solution

StepWorking

01Given

Line containing $B$ and $C$ is $y + x = 0$ . $B$ and $C$ are symmetric with respect to the origin $O (0, 0)$ , which makes $O$ the midpoint of segment $B C$ . Point $A$ lies on the line $y - 2 x = 2$ . Triangle $A B C$ is equilateral.

02Goal

Find the area of the equilateral triangle $Δ A B C$ .

03Approach

In an equilateral triangle, the median to any side is also its perpendicular bisector and altitude. Since $O$ is the midpoint of $B C$ , the altitude from $A$ passes through $O$ . The line $B C$ has slope $- 1$ , so the perpendicular line passing through $O (0, 0)$ has slope $1$ , giving the line $y = x$ . Thus, point $A$ is the intersection of $y = x$ and $y - 2 x = 2$ . Once $A$ is known, the altitude length $p = O A$ gives the area via $Area = \frac{p^{2}}{\sqrt{3}}$ .

04Execute

Find coordinates of $A$ by solving the system $y = x$ and $y - 2 x = 2$ : $x - 2 x = 2 ⟹ - x = 2 ⟹ x = - 2$ $y = - 2$ Thus, $A = (- 2, - 2)$ .

05Execute

Calculate the altitude $p = O A$ from origin $O (0, 0)$ to $A (- 2, - 2)$ : $p = \sqrt{(- 2 - 0)^{2} + (- 2 - 0)^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$

06Execute

For an equilateral triangle of altitude $p$ , the side length is $s = \frac{2 p}{\sqrt{3}}$ , so the area is: $Area = \frac{\sqrt{3}}{4} s^{2} = \frac{\sqrt{3}}{4} {(\frac{2 p}{\sqrt{3}})}^{2} = \frac{p^{2}}{\sqrt{3}}$ Substitute $p = \sqrt{8}$ : $Area = \frac{(\sqrt{8})^{2}}{\sqrt{3}} = \frac{8}{\sqrt{3}}$

✓Verify

Side length $s = \frac{2 \sqrt{8}}{\sqrt{3}} = \frac{4 \sqrt{2}}{\sqrt{3}}$ . Length $O C = s / 2 = \frac{2 \sqrt{2}}{\sqrt{3}}$ . Distance from $O$ along line $y = - x$ gives $C = (\frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}})$ and $B = (- \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}})$ . Distance $A B^{2} = (- 2 + 2 / \sqrt{3})^{2} + (- 2 - 2 / \sqrt{3})^{2} = 2 (4 + 4 / 3) = 2 (16 / 3) = 32 / 3 = s^{2}$ . All side lengths match.

Hints that build this answer step by step

Since $B$ and $C$ lie on $y + x = 0$ and are symmetric with respect to the origin, which line must contain vertex $A$ ?
The line $y = x$ , which is perpendicular to $y + x = 0$ at the midpoint $(0, 0)$ .
What are the coordinates of vertex $A$ obtained by intersecting $y = x$ with $y - 2 x = 2$ ?
$(- 2, - 2)$
Given the altitude $h$ is the distance from $A (- 2, - 2)$ to the midpoint $(0, 0)$ , what is the area of $Δ A B C$ ?
$\frac{8}{\sqrt{3}}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why must the altitude pass through the origin?

Because $B$ and $C$ are symmetric about the origin, the origin is the midpoint of $B C$ . In any equilateral triangle, the altitude from the opposite vertex bisects the base, so the altitude line from $A$ to $B C$ must pass through the midpoint of $B C$ , which is the origin.

Answer

The area of $Δ A B C$ is $\frac{8}{\sqrt{3}}$ , found by locating vertex $A$ on the perpendicular bisector of $B C$ (the line $y = x$ ) and using the height to compute the area.

Why each option works or fails

A: 10/(√(3)) - This arises from miscalculating the coordinates of vertex $A$ or using an incorrect formula relating the altitude $h$ to the area, such as $\frac{h^{2}}{2}$ instead of $\frac{h^{2}}{\sqrt{3}}$ . Ensure the altitude of an equilateral triangle is related to its area by $Area = \frac{h^{2}}{\sqrt{3}}$ , where $h$ is the distance from the opposite vertex to the midpoint of the base.
B: 3√(3) - This comes from an algebraic error when solving the system $y - 2 x = 2$ and $y = x$ , leading to $x = - 3$ instead of $x = - 2$ . Substitute $y = x$ directly into $y - 2 x = 2$ to obtain $x - 2 x = 2 ⟹ - x = 2 ⟹ x = - 2$ .
C: 2√(3) - This results from taking the distance from $A$ to the origin as the side length $s$ rather than the altitude $h$ , giving $\frac{\sqrt{3}}{4} (2 \sqrt{2})^{2} = 2 \sqrt{3}$ . Recognize that the origin is the midpoint of $B C$ , so the distance from $A$ to the origin is the altitude $h$ , not the side length $s$ .
D · correct: 8/(√(3)) - This is the correct calculation. The midpoint of $B C$ is the origin $(0, 0)$ . The altitude through $A$ is the perpendicular bisector of $B C$ , which is the line $y = x$ . Intersecting $y = x$ with $y - 2 x = 2$ gives $A (- 2, - 2)$ . The altitude is $h = O A = \sqrt{(- 2)^{2} + (- 2)^{2}} = 2 \sqrt{2}$ , giving $Area = \frac{h^{2}}{\sqrt{3}} = \frac{8}{\sqrt{3}}$ .

Step-by-step solution

Given: Line containing $B$ and $C$ is $y + x = 0$ . $B$ and $C$ are symmetric with respect to the origin $O (0, 0)$ , which makes $O$ the midpoint of segment $B C$ . Point $A$ lies on the line $y - 2 x = 2$ . Triangle $A B C$ is equilateral.
Goal: Find the area of the equilateral triangle $Δ A B C$ .
Approach: In an equilateral triangle, the median to any side is also its perpendicular bisector and altitude. Since $O$ is the midpoint of $B C$ , the altitude from $A$ passes through $O$ . The line $B C$ has slope $- 1$ , so the perpendicular line passing through $O (0, 0)$ has slope $1$ , giving the line $y = x$ . Thus, point $A$ is the intersection of $y = x$ and $y - 2 x = 2$ . Once $A$ is known, the altitude length $p = O A$ gives the area via $Area = \frac{p^{2}}{\sqrt{3}}$ .
Execute: Find coordinates of $A$ by solving the system $y = x$ and $y - 2 x = 2$ : $x - 2 x = 2 ⟹ - x = 2 ⟹ x = - 2$ $y = - 2$ Thus, $A = (- 2, - 2)$ .
Execute: Calculate the altitude $p = O A$ from origin $O (0, 0)$ to $A (- 2, - 2)$ : $p = \sqrt{(- 2 - 0)^{2} + (- 2 - 0)^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$
Execute: For an equilateral triangle of altitude $p$ , the side length is $s = \frac{2 p}{\sqrt{3}}$ , so the area is: $Area = \frac{\sqrt{3}}{4} s^{2} = \frac{\sqrt{3}}{4} {(\frac{2 p}{\sqrt{3}})}^{2} = \frac{p^{2}}{\sqrt{3}}$ Substitute $p = \sqrt{8}$ : $Area = \frac{(\sqrt{8})^{2}}{\sqrt{3}} = \frac{8}{\sqrt{3}}$
Verify: Side length $s = \frac{2 \sqrt{8}}{\sqrt{3}} = \frac{4 \sqrt{2}}{\sqrt{3}}$ . Length $O C = s / 2 = \frac{2 \sqrt{2}}{\sqrt{3}}$ . Distance from $O$ along line $y = - x$ gives $C = (\frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}})$ and $B = (- \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}})$ . Distance $A B^{2} = (- 2 + 2 / \sqrt{3})^{2} + (- 2 - 2 / \sqrt{3})^{2} = 2 (4 + 4 / 3) = 2 (16 / 3) = 32 / 3 = s^{2}$ . All side lengths match.

Area of an equilateral triangle with symmetric base vertices

What feels right?

The area of $Δ A B C$ is $\frac{8}{\sqrt{3}}$ , found by locating vertex $A$ on the perpendicular bisector of $B C$ (the line $y = x$ ) and using the height to compute the area.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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