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 45⁰.

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:

Here,
x₁ = 3, y₁ = - 5, α = 45∘, m

So, the equations of the required sides are

and

and

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.