# Find the equation of the straight line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.

Given: the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.

To find:

The equation of the required line.

Concept Used:

Point of intersection of two lines.

Explanation:

5x – 6y – 1 = 0 …… (i)

3x + 2y + 5 = 0 …… (ii)

By solving equation (i) and (ii) ,By cross multiplication,

x , y

Point of intersection ( - 1, - 1)

Now, the slope of the line 3x – 5y + 11 = 0 or yis

Now, we know that rhe product of the slope of two perpendicular lines is - 1.

Assuming: the slope of required line is m

Now, the equation of the required line passing through ( - 1, - 1) and having slope is given by,

Y + 1

3y + 3 = - 5x – 5

5x + 3y + 8 = 0

Hence, equation of required line is 5x + 3y + 8 = 0.

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