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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax+By+C=0$ is \[\] $\qquad \qquad A(x-x_1)+B(y-y_1)=0$.

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  • Equation of a line whose slope is $m$ and passing through $(x_1, y_1)$ is $ y-y_1=m(x-x_1)$.
  • Slopes of the parallel lines are equal.
Equation of the line is $Ax+By+C=0$
This can be written as
$y= \bigg( -\large\frac{A}{B} \bigg) x + \bigg( -\large\frac{C}{B} \bigg)$
This is of the form $y=mx+C$
Hence slope of the form $m = -\large\frac{A}{B}$
We know that parallel lines have same slope. Hence the equation of the line passing through the point $(x_1,y_1)$ and slope $ -\large\frac{A}{B}$ is
$ y-y_1= \bigg( -\large\frac{A}{B} \bigg) (x-x_1)$
$ \Rightarrow B(y-y_1)=-A(x-x_1)$
$ \Rightarrow A(x-x_1)+B(y-y_1)=0$
Hence the line through the point $(x_1,y_1)$ and parallel to the line $Ax+By+C=0$ is $A(x-x_1)+B(y-y_1)=0$.
Hence proved.
answered May 13, 2014 by thanvigandhi_1

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