Question

Find the equation to 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 1

Given :-

• A straight line is passing through the point of intersection of the lines 5x - 6y - 1 = 0 and 3x + 2y + 5 = 0.
• The straight line is perpendicular to the line 3x - 5y + 11 = 0.

To Find :-

The equation of the straight line.

Solution :-

Finding the point of intersection of the lines  5x - 6y - 1 = 0 and 3x + 2y + 5 = 0 :

Multiplying 3 to second equation

3(3x + 2y + 5) = 3(0)

→ 9x + 6y + 15 = 0

Adding the two equations

 $\sf{5x-6y-1=0}$

+

 $\sf{9x+6y+15=0}$

______________

 14x + 14 = 0

→ x = -14/14

→ x = -1

Putting the value of x in equation (i) :-

5(-1) - 6y - 1 = 0

→ -5 - 1 = 6y

→ -6 = 6y

→ y = -6/6

→ y = -1

Therefore, the point of intersections is (-1, -1).

_________________________

◘ Now the slope of the line 3x - 5y + 11 = 0 is 3/5.

As we know that the product of the slopes of two perpendicular lines is -1.

Let the slope of the required line be a.

$\sf{a\times\dfrac{3}{5} =-1 }$

$\sf{\longmapsto a=-\dfrac{5}{3} }$

◘ Now,

The equation of the line which is passing through the point of intersection of the lines 5x - 6y - 1 = 0 and 3x + 2y + 5 = 0, i.e. (-1, -1) is :-

$\sf{y+1=-\dfrac{5}{3}(x+1) }$

$\sf{\longmapsto 3y+3=-5x-5}$

$\boxed{\sf{\longmapsto 5x+3y+8=0}}$

The answer is 5x + 3y + 8 = 0.

Answer 2

Given :-

• A straight line is passing through the point of intersection of the lines 5x - 6y - 1 = 0 and 3x + 2y + 5 = 0.
• The straight line is perpendicular to the line 3x - 5y + 11 = 0.

To Find :-

• The equation of the straight line.

Solution :-

Finding the point of intersection of the lines  5x - 6y - 1 = 0 and 3x + 2y + 5 = 0 :

Multiplying 3 to second equation

3(3x + 2y + 5) = 3(0)

→ 9x + 6y + 15 = 0

Adding the two equations

 $\sf{5x-6y-1=0}$

+

 $\sf{9x+6y+15=0}$

______________

 14x + 14 = 0

→ x = -14/14

→ x = -1

Putting the value of x in equation (i) :-

5(-1) - 6y - 1 = 0

→ -5 - 1 = 6y

→ -6 = 6y

→ y = -6/6

→ y = -1

Therefore, the point of intersections is (-1, -1).

_________________________

◘ Now the slope of the line 3x - 5y + 11 = 0 is 3/5.

As we know that the product of the slopes of two perpendicular lines is -1.

Let the slope of the required line be a.

$\sf{a\times\dfrac{3}{5} =-1 }$

$\sf{\longmapsto a=-\dfrac{5}{3} }$

◘ Now,

The equation of the line which is passing through the point of intersection of the lines 5x - 6y - 1 = 0 and 3x + 2y + 5 = 0, i.e. (-1, -1) is :-

$\sf{y+1=-\dfrac{5}{3}(x+1) }$

$\sf{\longmapsto 3y+3=-5x-5}$

$\boxed{\sf{\longmapsto 5x+3y+8=0}}$

The answer is 5x + 3y + 8 = 0.