Q. 115.0( 3 Votes )

# 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 :

__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^{0}.

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,

(ax_{1} + by_{1} + c_{1})(ax_{2} + by_{2} + c_{1}) > 0

Here, x_{1} = 0, y_{1} = -4, x_{2} = 3, y_{2} = -5, a = 2, b = 3, c1 = -7

Now,

(ax_{1} + by_{1} + c_{1})(ax_{2} + by_{2} + c_{1}) = (2 × 0 – 3 × 4 – 7) (2 × 3 – 3 × 5 – 7)

(ax_{1} + by_{1} + c_{1})(ax_{2} + by_{2} + c_{1}) = -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,

x_{1} = 3, y_{1} = - 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.

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