Question

Prove that the lines given by the equations 3x + 2y + 5 = 0 and 3x + 2y − 1 = 0 are parallel. At what points do they intersect the x-axis? And the y-axis?

Answer 1

We know that two lines are parallel if they have equal slope.

Given: Equation of lines 3x + 2y + 5 = 0 and 3x + 2y − 1 = 0

Taking the coordinates of the two points on the first line as (x₁, y₁) and (x₂, y₂):

3x₁ + 2y₁ + 5 = 0

3x₂ + 2y₂ + 5 = 0

(3x₁ + 2y₁ + 5) − (3x₂ + 2y₂ + 5) = 0

⇒ 3(x₁ − x₂) + 2(y₁ − y₂) = 0

⇒

Thus, the slope of the first line is

Taking the coordinates of the two points on the second line as (x₁, y₁) and (x₂, y₂):

3x₁ + 2y₁ − 1 = 0

3x₂ + 2y₂ − 1 = 0

(3x₁ + 2y₁ − 1) − (3x₂ + 2y₂ − 1) = 0

⇒ 3(x₁ − x₂) + 2(y₁ − y₂) = 0

⇒

Thus, the slope of the second line is

Thus, the given lines are parallel to each other.

If a line cuts the x-axis, then the y-coordinate of that point must be zero.

For the equation, 3x + 2y + 5 = 0, putting y = 0, we get:

3x + 2(0) + 5 = 0

⇒ 3x + 5= 0

⇒ x =

Therefore, the point at which the line 3x + 2y + 5 = 0 cuts the x-axis is

Similarly, for the equation 3x + 2y − 1= 0, putting y = 0, we get:

3x + 2(0) − 1 = 0

⇒ 3x − 1 = 0

⇒ x =

Therefore, the point at which the line 3x + 2y − 1 = 0 cuts the x-axis is

Similarly, if a line cuts the y-axis then the x-coordinate of that point must be zero.

For the equation, 3x + 2y + 5 = 0, putting x = 0, we get:

3(0) + 2y + 5 = 0

⇒ 2y + 5 = 0

⇒ y =

Therefore, the point at which the line 3x + 2y + 5 = 0 cuts the y-axis is

For the equation 3x + 2y − 1 = 0, putting x = 0, we get:

3(0) + 2y − 1 = 0

⇒ 2y − 1= 0

⇒ y =

Therefore, the point at which the line 3x + 2y − 1 = 0 cuts the y-axis is