Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.

Does the line 3x = y + 1 bisect the line segment joining the two given points?

Let A = (-2, 3) and B = (4, 1)

$\text{Slope of AB }(\text{m}$1) = \dfrac{y2 - y1}{x2 - x_1} \\[1em] = \dfrac{1 - 3}{4 - (-2)} \\[1em] = \dfrac{1 - 3}{4 + 2} \\[1em] = \dfrac{-2}{6} \\[1em] = -\dfrac{1}{3}

By point-slope form, the equation of line AB is

⇒ y – y₁ = m(x – x₁)

⇒ y – 3 = $-\dfrac{1}{3}$[x - (-2)]

⇒ 3(y – 3) = -1(x + 2)

⇒ 3y - 9 = -x - 2

⇒ 3y + x = -2 + 9

⇒ x + 3y = 7 ………(1)

Given,

⇒ 3x = y + 1

⇒ y = 3x - 1

Comparing above equation with y = mx + c we get,

⇒ Slope (m₂) = 3

Since, m₁ × m₂ = $-\dfrac{1}{3} \times 3$ = -1.

Hence, the line through points A and B is perpendicular to the 3x = y + 1.

Given line is 3x = y + 1 ……..(2)

Let P be the mid-point of AB,

The co-ordinates of the mid-point of AB (i.e. P) are

$\Big(\dfrac{-2 + 4}{2}, \dfrac{3 + 1}{2}\Big) = \Big(\dfrac{2}{2}, \dfrac{4}{2}\Big)$ = (1, 2).

Now, Let’s check if point P satisfies the line equation (2)

⇒ 3(1) = 2 + 1

⇒ 3 = 3

Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.