Q. 174.8( 5 Votes )

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.

Answer :

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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