Question

Show that the point (3, −5) lies between the parallel lines 2x + 3y − 7 = 0 and 2x + 3y + 12 = 0 and find the equation of lines through (3, −5) cutting the above lines at an angle of 45°.

Answer 1

We observe that (0,−4) lies on the line 2x + 3y + 12 = 0
If (3, −5) lies between the lines 2x + 3y − 7 = 0 and 2x + 3y + 12 = 0, then we have,

\[\left( a x_1 + b y_1 + c_1 \right)\left( a x_2 + b y_2 + c_1 \right) > 0\]

Here,

\[x_1 = 0, y_1 = - 4, x_2 = 3, y_2 = - 5, a = 2, b = 3 \text { and } c_1 = - 7\]

Now,

\[\left( a x_1 + b y_1 + c_1 \right)\left( a x_2 + b y_2 + c_1 \right) = \left( 2 \times 0 - 3 \times 4 - 7 \right)\left( 2 \times 3 - 3 \times 5 - 7 \right)\]

\[\left( a x_1 + b y_1 + c_1 \right)\left( a x_2 + b y_2 + c_2 \right) = - 19 \times \left( - 16 \right) > 0\]

Thus, point (3,−5) lies between the given parallel lines.
The equation of the lines passing through (3,−5) and making an angle of 45° with the given parallel lines is given below:

\[y - y_1 = \frac{m \pm tan\alpha}{1 \mp mtan\alpha}\left( x - x_1 \right)\]

Here,

\[x_1 = 3, y_1 = - 5, \alpha = {45}^\circ \text { and } m = - \frac{2}{3}\]

\[\therefore y + 5 = \frac{- \frac{2}{3} \pm \tan {45}^\circ}{1 \mp \left( - \frac{2}{3} \right)\tan {45}^\circ}\left( x - 3 \right)\]

\[ \Rightarrow y + 5 = \frac{- \frac{2}{3} + 1}{1 + \frac{2}{3}}\left( x - 3 \right)\text {  and }y + 5 = \frac{- \frac{2}{3} - 1}{1 - \frac{2}{3}}\left( x - 3 \right)\]

\[ \Rightarrow y + 5 = \frac{1}{5}\left( x - 3 \right) \text { and } y + 5 = - 5\left( x - 3 \right)\]

\[ \Rightarrow x - 5y - 28 = 0 \text { and } 5x + y - 10 = 0\]