Q.Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax + By + C = 0$ is $A(x - x_1) + B(y - y_1) = 0$.

Concept: Parallel lines share the same slope; point-slope form.

A line parallel to $Ax + By + C = 0$ must have identical coefficients for $x$ and $y$ (same slope). The general form of any line parallel to it is $Ax + By + k = 0$ for some constant $k$.

To find $k$, substitute the given point $(x_1, y_1)$ through which our line passes:

Substituting back into $Ax + By + k = 0$:

$Ax + By - Ax_1 - By_1 = 0$ …

Why this works: the geometry of parallel lines

Two lines are parallel if and only if they have identical slopes. When a line is written in the form $Ax + By + C = 0$, the coefficients $A$ and $B$ encode its direction: the slope is $m = -\frac{A}{B}$ (when $B \neq 0$), and more generally the ratio $A : B$ determines the line's orientation regardless of whether it is vertical.

If we want a line parallel to $Ax + By + C = 0$, we must preserve the same $A$ and $B$ but allow the constant term to change. The new line will have the form $Ax + By + C' = 0$ for some different constant $C'$. The task is to find $C'$ such that the line passes through the given point $(x_1, y_1)$.

Step-by-step derivation

1. Start with the general form of a parallel line.

Any line parallel to $Ax + By + C = 0$ can be written as

for some constant $k$. The coefficients $A$ and $B$ are unchanged because they control the direction.

2. Impose the condition that the line passes through $(x_1, y_1)$.

Substitute $x = x_1$ and $y = y_1$ into the equation:

3. Solve for the constant $k$.

Rearranging gives

4. Substitute $k$ back into the equation.

The equation of the parallel line becomes

5. Factor to obtain the desired form.

Group the terms:

$A(x - x_1) + B(y - y_1) = 0.$ …