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

Given: 

Parallel lines 2x + 3y – 7 = 0 and 2x + 3y + 12 = 0 and 

To prove: 

The point (3, -5) lies between the parallel lines 2x + 3y – 7 = 0 and 2x + 3y + 12 = 0 

To find: 

Lines through (3, -5) cutting the above lines at an angle of 45°. 

Explanation: 

We observed 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, (ax₁ + by₁ + c₁)(ax₂ + by₂ + c₁) > 0 

Here, x₁ = 0, y₁ = -4, x₂ = 3, y₂ = -5, a = 2, b = 3, c₁ = -7 

Now, 

(ax₁ + by₁ + c₁)(ax₂ + by₂ + c₁) = (2 × 0 – 3 × 4 – 7) (2 × 3 – 3 × 5 – 7) 

(ax₁ + by₁ + c₁)(ax₂ + by₂ + c₁) = -19 × (-16) > 0 

Thus, point (3, -5) lies between the given parallel lines.

The equation of the lines passing through (3, -5) and making angle of 45 with the given parallel lines is given below:

 y - y₁ = \(\frac {m±\,tan\ \,\alpha}{1±m\,tan\ \,\alpha}\) (x - x₁)

Here,x₁ = 3, y₁ = - 5, α = 45^∘, m = \(-\frac{2}{3}\)

So, the equations of the required sides are

x – 5y – 28 = 0 and 5x + y – 10 = 0 

Hence, equation of required line is x – 5y – 28 = 0 and 5x + y – 10 = 0 Hence proved.