Question

The locus of the mid-points of the perpendiculars drawn from points on the line $x + y + 5 = 0$ to the line $2x - y + 3 = 0$ is

Answer 1

7x + 4y + 28 = 0

Answer 2

Let $P(x_1, y_1)$ be a point on the line $L_1: x + y + 5 = 0$. Let $Q(h, k)$ be the foot of the perpendicular from $P$ to the line $L_2: 2x - y + 3 = 0$. The mid-point of $PQ$ is $M(x, y)$, where $x = \frac{x_1+h}{2}$ and $y = \frac{y_1+k}{2}$. This implies $x_1 = 2x - h$ and $y_1 = 2y - k$.

Since $P(x_1, y_1)$ lies on $L_1$, we have $(2x - h) + (2y - k) + 5 = 0$, which simplifies to $h + k = 2x + 2y + 5$.

The line $PQ$ is perpendicular to $L_2$. The slope of $L_2$ is $m_2 = 2$. The slope of $PQ$ is $m_{PQ} = -\frac{1}{2}$. The equation of the line $PQ$ passing through $P(x_1, y_1)$ is $y - y_1 = -\frac{1}{2}(x - x_1)$, or $x + 2y = x_1 + 2y_1$. Substituting $x_1$ and $y_1$ in terms of $x, y, h, k$: $h + 2k = (2x - h) + 2(2y - k)$, which simplifies to $2h + 4k = 2x + 4y$, or $h + 2k = x + 2y$.

The foot of the perpendicular $Q(h, k)$ lies on $L_2$, so $2h - k + 3 = 0$, or $k = 2h + 3$.

We have the following system of equations:

1. $h + k = 2x + 2y + 5$
2. $h + 2k = x + 2y$
3. $k = 2h + 3$

Substitute (3) into (1): $h + (2h + 3) = 2x + 2y + 5 \implies 3h + 3 = 2x + 2y + 5 \implies 3h = 2x + 2y + 2$. Substitute (3) into (2): $h + 2(2h + 3) = x + 2y \implies h + 4h + 6 = x + 2y \implies 5h = x + 2y - 6$.

Now we have two expressions for $h$: $h = \frac{2x + 2y + 2}{3}$ and $h = \frac{x + 2y - 6}{5}$.

Equating these: $\frac{2x + 2y + 2}{3} = \frac{x + 2y - 6}{5}$ $5(2x + 2y + 2) = 3(x + 2y - 6)$ $10x + 10y + 10 = 3x + 6y - 18$ $7x + 4y + 28 = 0$.