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 y
is 
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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