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

Question

Accepted Answer

>

> [!TLDR]
> Since parallel lines have equal slopes, writing the point-slope equation through $(x_1,y_1)$ with the slope of $Ax+By+C=0$ directly gives $A(x-x_1)+B(y-y_1)=0$.
>
> The given line is $Ax+By+C=0$. Writing it in slope-intercept form: $By=-Ax-C \Rightarrow y=-\dfrac{A}{B}x-\dfrac{C}{B}$ (assuming $B
eq0$), so its slope is:
> $$m = -\frac{A}{B}$$
>
> Any line parallel to this one has the same slope, $-\dfrac{A}{B}$.
>
> Using the point-slope form of a line through $(x_1,y_1)$ with slope $m$:
> $$y-y_1 = m(x-x_1)$$
>
> $$y-y_1 = -\frac{A}{B}(x-x_1)$$
> …

Quadrant	x-sign	y-sign	Example
I	$+$	$+$	$(2, 3)$
II	$-$	$+$	$(-1, 4)$
III	$-$	$-$	$(-3, -2)$
IV	$+$	$-$	$(5, -1)$

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 understanding — Coordinate Geometry

Coordinate Geometry: Where Algebra Meets Geometry

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The Coordinate System

A Concrete Example

The Four Quadrants

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The Precise Statement

Q.Prove that the line through the point (x1,y1)(x_1, y_1)(x1​,y1​) and parallel to the line Ax+By+C=0Ax + By + C = 0Ax+By+C=0 is A(x−x1)+B(y−y1)=0A(x - x_1) + B(y - y_1) = 0A(x−x1​)+B(y−y1​)=0.

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

Unlock everything free for 14 days