Question

prove that the line through the point (x1, y1)and parallel to the line ax+by+c=0, is A(x-x1)+B(y-y1)=0​

Answer 1

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Solution

Answer 2

Answer with step-by-step explanation:

Let m be the slop of line which is passing through the point ($x_1,y_1$)

Point-slope form:$y-y_1=m(x-x_1)$

Using the formula

The equation of line which is passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

$m=\frac{y-y_1}{x-x_1}$

The equation of other line

$ax+by+c=0$

$by=-ax-c$

$y=\frac{-a}{b}x-\frac{c}{b}$

Compare with

$y=mx+C$

Where m =Slope of line

We get

Slope of line=$m'=-\frac{a}{b}$

When two lines are parallel then their slopes are equal.

Therefore,

$m=m'$

$\frac{y-y_1}{x-x_1}=-\frac{a}{b}$

$b(y-y_1)=-a(x-x_1)$

$a(x-x_1)+b(y-y_1)=0$

Hence, proved.

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