JEE 2023 — Mathematics PYQ

The angular bisector ($l_1$) and the two lines ($l_2$ and $l_3$) must all intersect at the same point.

• $l_1:3y-2x=3$

• $l_2:x-y+1=0⟹x=y-1$

Substitute $x$ in $l_1$:

If $y=1$, then $x=1-1=0$. The point of intersection is $(0,1)$.

Since $(0,1)$ must also lie on $l_3:\alpha x+\beta y+17=0$:

2. Use the Angle Property

The angle between $l_1$ and $l_2$ must be equal to the angle between $l_1$ and $l_3$.

The slopes are:

• $m_1$ (slope of $l_1$) $=\frac{2}{3}$

• $m_2$ (slope of $l_2$) $=1$

• $m_3$ (slope of $l_3$) $=-\frac{\alpha }{\beta }=\frac{\alpha }{17}$

Using the formula $\tan \theta =\left|\frac{m_a-m_b}{1+m_a{m}_b}\right|$:

Case 1:

Case 2:

(If $\alpha =17$, the slope $m_3=1$, which makes $l_3$ parallel to $l_2$, so we take $\alpha =7$).

3. Final Calculation

We have $\alpha =7$ and $\beta =-17$. We need to find ${\alpha }^2+{\beta }^2-\alpha -\beta$:

Final Answer:

The value is 348.