The Distance Between the Two Points A and A’ Which Lie on y = 2 Such That Both the Line Segments AB and A’B, Where B Is the Point (2, 3), Subtend an Angle π/4 at the Origin Is Equal To

JEE Main 2022 — 29 June, Morning Shift. Previous Year Question.

Problem

The distance between the two points $A$ and $A’$ which lie on $y=2$ such that both the line segments $AB$ and $A’B$ (where $B$ is the point $(2,3)$) subtend an angle $\dfrac{\pi}4$ at the origin, is equal to:

Key insight. “Subtends angle π/4 at the origin” means the angle between the lines $OA$ and $OB$ is π/4 — a condition captured cleanly by the tangent-of-angle-between-two-lines formula once both points are described by their slopes from the origin.

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Approach

Let $A=(x,2)$ be a general point on the line $y=2$. The condition that $AB$ subtends angle $\dfrac{\pi}4$ at the origin means the angle between line $OA$ (slope $\dfrac2x$) and line $OB$ (slope $\dfrac32$, fixed since $B=(2,3)$) is exactly $\dfrac{\pi}4$. Applying the tangent-of-angle-between-two-lines formula gives an equation in $x$ with two solutions — the $x$-coordinates of $A$ and $A’$ — and the distance between them (since both share $y=2$) is just the difference of these two $x$-values.

Solution

Step 1 — Set up the two slopes

Let $A=(x,2)$. The slope of $OA$ is $m_1=\dfrac2x$, and the slope of $OB$ (fixed, since $B=(2,3)$) is $m_2=\dfrac32$.

Step 2 — Apply the angle-between-two-lines formula

$$\tan\left(\frac{\pi}4\right) = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

Since $\tan\left(\dfrac{\pi}4\right)=1$:

$$\left|\frac{\frac2x-\frac32}{1+\frac2x\cdot\frac32}\right| = 1 \implies \left|\frac{\frac2x-\frac32}{1+\frac3x}\right|=1$$

Step 3 — Clear denominators

Multiplying numerator and denominator by $2x$:

$$\left|\frac{4-3x}{2x+6}\right| = 1$$

This gives two cases, since the absolute value can resolve either way.

Step 4 — Solve both cases

Case 1: $4-3x = 2x+6$

$$-5x = 2 \implies x = -\frac25$$

Case 2: $4-3x = -(2x+6) = -2x-6$

$$4-3x+2x+6 = 0 \implies -x+10=0 \implies x=10$$

Step 5 — Identify A and A’

So the two points are $A=(10,2)$ and $A’=\left(-\dfrac25,2\right)$.

Step 6 — Compute the distance AA’

Since both points share the same $y$-coordinate, the distance is simply the difference of their $x$-coordinates:

$$AA’ = \left|10-\left(-\frac25\right)\right| = 10+\frac25 = \frac{50}5+\frac25 = \frac{52}5$$

Answer

Common mistakes

• Dropping the absolute value and solving only one case. The tangent formula’s absolute value genuinely produces two valid solutions here (one on each side of $B$’s direction from the origin) — solving only one case misses one of the two points needed to compute the distance.
• Sign errors when clearing the compound fraction in Step 3. Multiplying both the numerator $\left(\dfrac2x-\dfrac32\right)$ and denominator $\left(1+\dfrac3x\right)$ by $2x$ has to be done term by term — a dropped sign here changes both roots.
• Forgetting that both points share y=2, so the distance is a simple subtraction. Since $A$ and $A’$ both lie on the horizontal line $y=2$, there’s no need for the full distance formula — the vertical coordinates cancel and only the $x$-coordinates matter.

Practise next

• Find the distance between points $A$ and $A’$ on $y=2$ such that both $AB$ and $A’B$ (where $B=(3,4)$ this time) subtend angle $\dfrac{\pi}4$ at the origin.