Question

Show that the paths represented by the equations x – 3y = 2 and –2x + 6y = 5 are parallel.

Answer 1

1. Rearrange both equations

To find the slope of each path, convert the given linear equations into the slope-intercept form, which is given by y = mx + c where m is the slope and c is the y-intercept.

For the first path:

x – 3y = 2

–3y = –x + 2

`y = 1/3x - 2/3`

From this, the slope of the first line (m₁) is `1/3`.

For the second path:

–2x + 6y = 5

6y = 2x + 5

`y = 2/6x + 5/6`

`y = 1/3x + 5/6`

From this, the slope of the second line (m₂) is `1/3`.

2. Compare the properties

For two lines to be parallel, their slopes must be equal (m₁ = m₂) and their y-intercepts must be different (c₁ ≠ c₂).

`m_1 = 1/3` and `m_2 = 1/3`

⇒ m₁ = m₂

`c_1 = -2/3` and `c_2 = 5/6`

⇒ c₁ ≠ c₂

3. Alternative coefficient ratio check

Alternatively, we can write both equations in standard form a_x + b_y + c = 0:

1. 1x – 3y – 2 = 0

⇒ a₁ = 1, b₁ = –3, c₁ = –2

2. –2x + 6y – 5 = 0

⇒ a₂ = –2, b₂ = 6, c₂ = –5

Now, find the ratios of their coefficients:

`(a_1)/(a_2) = 1/(-2) = -1/2`

`(b_1)/(b_2) = (-3)/(6) = -1/2`

`(c_1)/(c_2) = (-2)/(-5) = 2/2`

Since `(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`, the lines are mathematically proven to be parallel with no shared points of intersection.