Point-Line distance

In the figure above, the parabola $y=x^2+6x+9$ is the blue curve. The straight line $3x + 4y = 6$ or $4y = 6-3x$ is drawn in orange. The loci of points distance 1 unit away from $4y=6-3x$ are two lines parallel to and equal distance from $4y=6-3x$ . Let the two straight lines be $4y = 6-3x+k$ (grey) and \ $4y=6-3x-k$ (yellow), where $k$ is a constant (in this case it is 1.2). Then point $P$ satisfies the following two systems of equations.

$\begin{cases} y = x^2+6x+9 \\ 4y = 6-3x+k \end{cases}$ and $\begin{cases} y = x^2+6x+9 \\ 4y = 6-3x-k \end{cases}$

They are the intersection points of the parabola and the two parallel lines, $P_1(x_1, y_1)$ , $P_2(x_2, y_2)$ , $P_3(x_3, y_3)$ , and $P_4(x_4, y_4)$ , Solving the two systems of equations, we have:

$\begin{cases} 4x^2+24x+36 = 6-3x+k & \implies 4x^2 + 27x + 30 - k = 0 \\ 4x^2+24x+36 = 6-3x+k & \implies 4x^2 + 27x + 30 + k = 0 \end{cases}$

By Vieta's formula , $x_1+x_2 = - \frac {27}4$ and $x_3+x_4 = - \frac {27}4$ . Therefore, $x_1+x_2 + x_3+x_4 = - \frac {27}2$ and $6(x_1+x_2 + x_3+x_4) = \boxed{-81}$ .

The formula for the distance between a point $P=(m,n)$ and the line $ax+by+c=0$ is given as $D=|am+bn+c|/√(a^2+b^2)$ where $D$ is the distance. Here, the line is $3x+4y-6=0$ so $a=3$ , $b=4$ , and $c=-6$ . The $x$ -coordinate of the point $P$ is $m$ , but since $m$ is on the parabola $y=x^2+6x+9$ , $n=m^2+6m+9=(m+3)^2$ . Also, $D=1$ .

As a result, $|3m+4(m+3)^2-6|=√(3^2+4^2)=5$ . This can be split into $3m+4(m^2+6m+9)-6=5$ and $3m+4(m^2+6m+9)-6=-5$ . By simplification, $4m^2+27m+25=0$ and $4m^2+27m+35=0$ . Then, by the quadratic formula, the sum of the roots from both equations is $-13.5$ , so $-13.5*6=-81$ which is the final answer.