Triangle ABC: y=x Bisects ∠B, AC is 2x−y=2, 2AB=BC, A=(4,6) — Find α+2β

JEE Main 2024 — 29 January, Morning Shift. Previous Year Question.

Problem

In a triangle $ABC$, suppose $y=x$ is the equation of the bisector of angle $B$, and the equation of side $AC$ is $2x-y=2$. If $2AB=BC$, and the points $A$ and $B$ are respectively $(4,6)$ and $(\alpha,\beta)$, then $\alpha+2\beta$ is equal to:

Key insight. Two separate facts pin down $B$’s coordinates. First, $B$ lies on its own angle bisector $y=x$, so $\alpha=\beta$ immediately. Second, the Angle Bisector Theorem says the bisector from $B$ divides $AC$ in the same ratio as the two adjacent sides $AB:BC$ — turning $2AB=BC$ into a concrete section-formula ratio that locates the third vertex $C$, after which the given length condition solves for the one remaining unknown.

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Approach

Find where the bisector $y=x$ meets side $AC$ — call this point $D$. By the Angle Bisector Theorem, $D$ divides $AC$ in the ratio $AD:DC = AB:BC = 1:2$, which locates $C$ using the section formula (since $A$ and $D$ are both known). Since $B$ lies on $y=x$, write $B=(\alpha,\alpha)$, then use $4AB^2=BC^2$ (from $2AB=BC$) to solve for $\alpha$.

Solution

Step 1 — Find D, where the bisector meets AC

Solving $y=x$ and $2x-y=2$ simultaneously: substituting $y=x$ gives $2x-x=2 \implies x=2$, so $D=(2,2)$.

Step 2 — Apply the Angle Bisector Theorem

The bisector from $B$ divides $AC$ in the ratio $AB:BC$. Since $2AB=BC$, this ratio is $AB:BC=1:2$, so $AD:DC=1:2$.

Step 3 — Use the section formula to find C

Since $D$ divides $AC$ with $AD:DC=1:2$:

$$D = \frac{1\cdot C+2\cdot A}{1+2}$$

$$(2,2) = \frac{C+2(4,6)}{3} \implies (6,6) = C+(8,12) \implies C = (-2,-6)$$

Step 4 — Use B’s position on the bisector

Since $B=(\alpha,\beta)$ lies on $y=x$, we get $\beta=\alpha$ directly — so $B=(\alpha,\alpha)$.

Step 5 — Apply the length condition 4AB² = BC²

$$AB^2 = (4-\alpha)^2+(6-\alpha)^2, \qquad BC^2 = (\alpha+2)^2+(\alpha+6)^2$$

$$4\left[(4-\alpha)^2+(6-\alpha)^2\right] = (\alpha+2)^2+(\alpha+6)^2$$

Expanding the left side:

$$4\left[(16-8\alpha+\alpha^2)+(36-12\alpha+\alpha^2)\right] = 4(2\alpha^2-20\alpha+52) = 8\alpha^2-80\alpha+208$$

Expanding the right side:

$$(\alpha^2+4\alpha+4)+(\alpha^2+12\alpha+36) = 2\alpha^2+16\alpha+40$$

Step 6 — Solve the resulting quadratic

$$8\alpha^2-80\alpha+208 = 2\alpha^2+16\alpha+40$$

$$6\alpha^2-96\alpha+168=0 \implies \alpha^2-16\alpha+28=0$$

$$(\alpha-2)(\alpha-14)=0 \implies \alpha=2 \text{ or } \alpha=14$$

Step 7 — Reject the invalid root

At $\alpha=2$, $B$ would coincide with $D=(2,2)$ — a point on side $AC$ itself, which can’t be a triangle’s vertex. So $\alpha=14$, and since $\beta=\alpha$, $\beta=14$ too.

Step 8 — Compute α+2β

$$\alpha+2\beta = 14+2(14) = 14+28 = 42$$

Answer

Common mistakes

• Forgetting that B lies on its own bisector. Since the angle bisector of $\angle B$ passes through $B$ itself, $\beta=\alpha$ is a free fact — missing it leaves two unknowns instead of one.
• Not rejecting $\alpha=2$. The quadratic naturally has two roots, but one of them places $B$ exactly on side $AC$, which is geometrically impossible for a genuine triangle vertex.

Practise next

• In a triangle $PQR$, suppose $y=2x$ bisects $\angle Q$, and side $PR$ has equation $x-y=1$. If $3PQ=QR$ and $P=(5,4)$, find $\alpha+\beta$ for $Q=(\alpha,\beta)$, using the same bisector-and-section-formula approach.