Answer :

Here we have the straight line which passes through the point of intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and is perpendicular to the straight line 3x – 5y + 11 = 0.


As the straight line passes through the point of intersection of the lines 5x – 6y = 1 and 3x + 2y + 5 = 0, we should find the intersection point by solving these equations:


5x –6y = 1


3x + 2y + 5 = 0


5x – 6y– 1 = 0 –––(1)


and 3x + 2y + 5 = 0–––(2)


Now multiply equation (2) by 3 on both the sides.


Thus we have 9x + 6y + 15 = 0–––(3)


Now, we have



14x = –14


x = –1


Now, substituting x = –1 in the equation 5x–6y = 1, we have,


5(–1)–6y = 1


–5–6y = 1


–6y = 1 + 5 = 6



y = –1


Therefore the point of intersection is (–1,–1).


Slope of the line 3x –5y + 11 = 0 is :



As the line which passes through (–1,–1) is perpendicular to 3x –5y + 11 = 0 ,the product of their slopes will be –1.


Therefore,


(Here we have multiplied on both the sides)



Hence the equation of the line passing through (–1,–1) and slope as is:


(y–y1) = m(x–x1)




3y + 3 = –5x–5


3y + 3 + 5x + 5 = 0


5x + 3y + 8 = 0


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