Animated Solution for Mathematics - Straight Lines: Let the equation $x(x+2)(12-k)=2$ have equal roots. Then the distance of the point $(k,\frac{k}{2})$ from the line $3x+4y+5=0$ is

Imagine you are standing on the coordinate plane, looking at a line defined by $3x+4y+5=0$. You have a point $P$ hovering somewhere, defined by the coordinates $(k,\frac{k}{2})$.

To find the distance between this point and the line, we must first determine the value of $k$. We are given the algebraic equation $x(x+2)(12-k)=2$, which is stated to have equal roots.

When you see an expression like $12-k$ embedded in a quadratic, the secret to advanced problem-solving is to reduce cognitive load. Let us define a new variable, $\lambda =12-k$.

The equation $x(x+2)(12-k)=2$ transforms into $\lambda x(x+2)=2$. Expanding this, we obtain:

This is a standard quadratic equation $a{x}^2+bx+c=0$, where $a=\lambda$, $b=2\lambda$, and $c=-2$.

The condition for equal roots is the heartbeat of this problem. For any quadratic equation to have equal roots, its discriminant $D$ must be zero. We invoke the formula:

Substituting our coefficients, we get:

This simplifies to $4{\lambda }^2+8\lambda =0$. Factoring this, we find $4\lambda (\lambda +2)=0$.

This gives us two candidates: $\lambda =0$ or $\lambda =-2$. Since $\lambda =0$ would destroy the quadratic nature of our equation, we are left with the elegant result: $\lambda =-2$.

Now, we reverse our substitution. Since $\lambda =12-k$ and $\lambda =-2$, we have $-2=12-k$.

Solving for $k$, we find $k=14$. With $k$ in hand, the coordinates of point $P$ are revealed:

We have moved from the abstract world of algebra into the concrete world of coordinate geometry. We now know exactly where point $P$ sits.

We are now at the finish line. We need the perpendicular distance from $P(14,7)$ to the line $3x+4y+5=0$.

We use the classic distance formula:

Plugging in our values, we get:

The numerator becomes $|42+28+5|=|75|=75$. The denominator is $\sqrt{9+16}=\sqrt{25}=5$.

Finally, we calculate the distance:

The distance is 15 units. You have successfully navigated the intersection of quadratic theory and coordinate geometry.