Find the Equation of a Line Which is Perpendicular to the Line $\Sqrt{3}X - Y + 5 = 0$ and Which Cuts off an Intercept of 4 Units with the Negative Direction Of Y-axis. - Mathematics

Find the equation of a line which is perpendicular to the line $\sqrt{3}x - y + 5 = 0$ and which cuts off an intercept of 4 units with the negative direction of y-axis.

Solution

The line perpendicular to $\sqrt{3}x - y + 5 = 0$ is $x + \sqrt{3}y + \lambda = 0$.

It is given that the line $x + \sqrt{3}y + \lambda = 0$  cuts off an intercept of 4 units with the negative direction of the y-axis.
This means that the line passes through $\left( 0, - 4 \right)$.

$\therefore 0 - \sqrt{3} \times 4 + \lambda = 0$

$\Rightarrow \lambda = 4\sqrt{3}$

Substituting the value of $\lambda$, we get

$x + \sqrt{3}y + 4\sqrt{3} = 0$, which is the equation of the required line.

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.12 | Q 5 | Page 92