Login/Sign Up
Home » Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.
Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.
CBSE | Class 11 | Exercise 23.12 | Maths | RD Sharma | Straight Lines
Find the equation of a line which is perpendicular to the line√3x – y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.

According to ques,:

The equation is perpendicular to:

    \[~\surd 3x\text{ }\text{ }y\text{ }+\text{ }5\text{ }=\text{ }0~\]

equation and cuts off an intercept of 4 units with the negative direction of y-axis.

The line perpendicular to:

    \[~\surd 3x\text{ }\text{ }y\text{ }+\text{ }5\text{ }=\text{ }0~is\text{ }x\text{ }+\text{ }\surd 3y\text{ }+\text{ }\lambda \text{ }=\text{ }0\]

It is given that the line,

    \[~x\text{ }+\text{ }\surd 3y\text{ }+\text{ }\lambda \text{ }=\text{ }0~\]

cuts off an intercept of 4 units with the negative direction of the y-axis.

This means that the line passes through (0,-4).

So,

Let us substitute the values in the equation

    \[~x\text{ }+\text{ }\surd 3y\text{ }+\text{ }\lambda \text{ }=\text{ }0\]

, we have

    \[0\text{ }\text{ }\surd 3\text{ }\left( 4 \right)\text{ }+\text{ }\lambda \text{ }=\text{ }0~\]

    \[\lambda \text{ }=\text{ }4\surd 3\]

Now, substitute the value of λ back, we get

    \[x\text{ }+\text{ }\surd 3y\text{ }+\text{ }4\surd 3\text{ }=\text{ }0\]

∴ The required equation of line is

    \[x\text{ }+\text{ }\surd 3y\text{ }+\text{ }4\surd 3\text{ }=\text{ }0.\]

 

 

i

More Material