Animated Solution for Mathematics - Straight Lines: If the line $l_1:3y-2x=3$ is the angular bisector of the lines $l_2:x-y+1=0$ and $l_3:\alpha x+\beta y+17=0$, then ${\alpha }^2+{\beta }^2-\alpha -\beta$ is equal to

Welcome, future engineer. Today, we are not merely solving for coefficients $\alpha$ and $\beta$. We are embarking on a journey to understand the hidden symmetry of lines.

When you look at a problem involving an angular bisector, do not see it as a dry algebraic exercise. See it as a dance of geometry, where lines converge and distances balance in perfect harmony. Let us break this down, step by step, and uncover the elegance hidden within the equations.

Imagine three lines on a plane. We have $l_1:3y-2x=3$, which acts as our anchor, the angular bisector. Then we have $l_2:x-y+1=0$ and $l_3:\alpha x+\beta y+17=0$. The problem states that $l_1$ bisects the angle between $l_2$ and $l_3$.

Before we dive into heavy calculations, pause and visualize the geometry. Where does an angular bisector begin? It originates from the very point where the two lines it bisects intersect.

This is a crucial realization: $l_1$, $l_2$, and $l_3$ must all be concurrent. They must meet at a single, shared point. If you miss this, you are fighting the problem instead of flowing with it.

Let us find this point of intersection by solving the system of $l_1$ and $l_2$:

From the second equation, we can express $x$ as $x=y-1$. Substituting this into the first equation, we get $3y-2(y-1)=3$, which simplifies to $y=1$. Consequently, $x=0$.

Our intersection point is $(0,1)$. This point is the key that unlocks the door to our unknown coefficients.

Since $l_3$ must also pass through this intersection point $(0,1)$, we can substitute these coordinates directly into the equation of $l_3$:

Just like that, the $\alpha$ term vanishes, and we are left with $\beta =-17$. We have already conquered half the problem! It is moments like these—where a complex-looking variable simply disappears—that remind us why we love mathematics.

Now, we must find $\alpha$. This is where we invoke the soul of the angular bisector: the distance property. Any point on an angular bisector is equidistant from the two lines forming the angle.

Let us pick a point $P$ on $l_1$. If we set $y=3$, then $3(3)-2x=3$, which gives $9-2x=3$, so $2x=6$, and $x=3$. Our point $P$ is $(3,3)$.

Now, we calculate the perpendicular distance $d_1$ from $P(3,3)$ to $l_2:x-y+1=0$:

Next, we calculate the distance $d_2$ from $P(3,3)$ to $l_3:\alpha x-17y+17=0$:

Because $P$ lies on the bisector, $d_1$ must equal $d_2$. We set them equal:

To eliminate the square roots and the absolute value, we square both sides:

Cross-multiplying gives us ${\alpha }^2+289=2(9{\alpha }^2-204\alpha +1156)$. Expanding this, we get ${\alpha }^2+289=18{\alpha }^2-408\alpha +2312$. Rearranging everything to one side, we arrive at the quadratic equation:

Dividing the entire equation by $17$, we get:

This factors elegantly into $(\alpha -7)(\alpha -17)=0$. We have two candidates: $\alpha =7$ and $\alpha =17$.

We must be critical thinkers. If $\alpha =17$, the line $l_3$ becomes $17x-17y+17=0$, which is $x-y+1=0$. This is identical to $l_2$. An angular bisector cannot exist between two identical lines.

Thus, we reject $17$. Our true value is $\alpha =7$.

Finally, we calculate the target expression: ${\alpha }^2+{\beta }^2-\alpha -\beta$. Substituting $\alpha =7$ and $\beta =-17$:

We have arrived at the destination. The answer is 348. Remember, the math is not just about the final number; it is about the logical path you carved to get there.