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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