Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 450 to the lines 3x + y – 5 = 0.
Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 450 to the lines 3x + y – 5 = 0.

The equation passes through (2, 3) and make an angle of 450with the line 3x + y – 5 = 0.

Since, the equations of two lines passing through a point x1,y1 and making an angle α with the given line y = mx + c are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 102

Equation of the given line is,

    \[\begin{array}{*{35}{l}} 3x\text{ }+\text{ }y\text{ }-\text{ }5\text{ }=\text{ }0  \\ y\text{ }=\text{ }-\text{ }3x\text{ }+\text{ }5  \\ \end{array}\]

Comparing this equation with y = mx + c we get, m = – 3

x1 = 2, y1 = 3, α = 45∘, m = – 3.

So, the equations of the required lines are

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 103

x + 2y – 8 = 0 and 2x – y – 1 = 0

∴ The equation of given line is x + 2y – 8 = 0 and 2x – y – 1 = 0