The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

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

Given:

The opposite from the beginning meets the given line at \[\left( \text{ }\mathbf{1},\text{ }\mathbf{2} \right).\]

The condition of line is \[\mathbf{y}\text{ }=\text{ }\mathbf{mx}\text{ }+\text{ }\mathbf{c}\]

The line joining the focuses (0, 0) and (– 1, 2) is opposite to the given line.

Along these lines, the incline of the line joining (0, 0) and (– 1, 2) \[=\text{ }\mathbf{2}/\left( -\text{ }\mathbf{1} \right)\text{ }=\text{ }-\text{ }\mathbf{2}\]

Incline of the given line is m.

\[\mathbf{m}\text{ }\times \text{ }\left( -\text{ }\mathbf{2} \right)\text{ }=\text{ }-\text{ }\mathbf{1}\]

\[\mathbf{m}\text{ }=\text{ }\mathbf{1}/\mathbf{2}\]

Since, point (- 1, 2) lies on the given line,

\[\mathbf{y}\text{ }=\text{ }\mathbf{mx}\text{ }+\text{ }\mathbf{c}\]

\[\mathbf{2}\text{ }=\text{ }\mathbf{1}/\mathbf{2}\text{ }\times \text{ }\left( -\text{ }\mathbf{1} \right)\text{ }+\text{ }\mathbf{c}\]

\[\mathbf{c}\text{ }=\text{ }\mathbf{2}\text{ }+\text{ }\mathbf{1}/\mathbf{2}\text{ }=\text{ }\mathbf{5}/\mathbf{2}\]

∴ The upsides of m and c are 1/2 and 5/2 individually.

i

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