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 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.


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