Find the equations of the two straight lines through (1, 2) forming two sides of a square of which 4x+ 7y = 12 is one diagonal.

#### Solution

Let A (1, 2) be the vertex of square ABCD and BD be one diagonal, whose equation is 4x + 7y = 12

Here, we have to find the equations of sides AB and AD, each of which makes an angle of \[{45}^\circ\] with line 4x + 7y = 12

We know the equations of two lines passing through a point \[\left( x_1 , y_1 \right)\] and making an angle \[\alpha\] with the line whose slope is m.

Equation of the given line is

\[4x + 7y = 9\]

\[ \Rightarrow y = - \frac{4}{7}x + \frac{9}{4}\]

\[\therefore x_1 = 1, y_1 = 2, \alpha = {45}^\circ , m = - \frac{4}{7}\]

So, the equations of the required sides are

\[y - 2 = \frac{- \frac{4}{7} + \tan {45}^\circ}{1 + \frac{4}{7}\tan {45}^\circ}\left( x - 1 \right) \text { and } y - 2 = \frac{- \frac{4}{7} - \tan {45}^\circ}{1 - \frac{4}{7}\tan {45}^\circ}\left( x - 1 \right)\]

\[ \Rightarrow y - 2 = \frac{- \frac{4}{7} + 1}{1 + \frac{4}{7}}\left( x - 1 \right) \text { and } y - 2 = \frac{- \frac{4}{7} - 1}{1 - \frac{4}{7}}\left( x - 1 \right)\]

\[ \Rightarrow y - 2 = \frac{3}{11}\left( x - 1 \right) \text { and } y - 2 = \frac{- 11}{3}\left( x - 1 \right)\]

\[ \Rightarrow 3x - 11y + 19 = 0\text { and } 11x + 3y - 17 = 0\]