Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y

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.

Class 11 | Exercise 10.3 | Maths | NCERT | Straight Lines

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.

Let the slant of line \[\mathbf{Ax}\text{ }+\text{ }\mathbf{By}\text{ }+\text{ }\mathbf{C}\text{ }=\text{ }\mathbf{0}\] be m

\[\mathbf{Ax}\text{ }+\text{ }\mathbf{By}\text{ }+\text{ }\mathbf{C}\text{ }=\text{ }\mathbf{0}\]

In this way, \[\mathbf{y}\text{ }=\text{ }-\text{ }\mathbf{A}/\mathbf{Bx}\text{ }\text{ }\mathbf{C}/\mathbf{B}\]

\[\mathbf{m}\text{ }=\text{ }-\text{ }\mathbf{A}/\mathbf{B}\]

By utilizing the equation,

Condition of the line going through point (x1, y1) and having slant \[\mathbf{m}\text{ }=\text{ }-\text{ }\mathbf{A}/\mathbf{B}\] is

\[\mathbf{y}\text{ }\text{ }\mathbf{y1}\text{ }=\text{ }\mathbf{m}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{x1} \right)\]

\[\mathbf{y}\text{ }\text{ }\mathbf{y1}=\text{ }-\text{ }\mathbf{A}/\mathbf{B}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{x1} \right)\]

\[\mathbf{B}\text{ }\left( \mathbf{y}\text{ }\text{ }\mathbf{y1} \right)\text{ }=\text{ }-\text{ }\mathbf{A}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{x1} \right)\]

\[\therefore \mathbf{A}\left( \mathbf{x}\text{ }\text{ }\mathbf{x1} \right)\text{ }+\text{ }\mathbf{B}\left( \mathbf{y}\text{ }\text{ }\mathbf{y1} \right)\text{ }=\text{ }\mathbf{0}\]

In this way, the line through point (x1, y1) and corresponding to the line \[\mathbf{Ax}\text{ }+\text{ }\mathbf{By}\text{ }+\text{ }\mathbf{C}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{is}\text{ }\mathbf{A}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{x1} \right)\text{ }+\text{ }\mathbf{B}\text{ }\left( \mathbf{y}\text{ }\text{ }\mathbf{y1} \right)\text{ }=\text{ }\mathbf{0}\]

Subsequently demonstrated.

i

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