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Find the Equation of the Straight Line Perpendicular to 2x − 3y = 5 and Cutting off an Intercept 1 on the Positive Direction of the X-axis. - Mathematics

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प्रश्न

Find the equation of the straight line perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.

संक्षेप में उत्तर

उत्तर

The line perpendicular to 2x − 3y = 5 is \[3x + 2y + \lambda = 0\]

It is given that the line \[3x + 2y + \lambda = 0\] cuts off an intercept of 1 on the positive direction of the x-axis.
This means that the line \[3x + 2y + \lambda = 0\]  passes through the point (1, 0).

\[\therefore 3 + 0 + \lambda = 0\]

\[ \Rightarrow \lambda = - 3\]

Substituting the value of \[\lambda\],we get

\[3x + 2y - 3 = 0\] which is equation of the required line.

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अध्याय 23: The straight lines - Exercise 23.12 [पृष्ठ ९३]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.12 | Q 8 | पृष्ठ ९३

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