(i) Is the line passing through the points A(–2, 3) and B(4, 1) perpendicular to the line 3x – y = 1?

(ii) Does the line 3x – y = 1 bisect the join of A(–2, 3) and B(4, 1)?

(i) Equation of line through (-2, 3) and (4, 1) can be given by two-point form i.e.,

y - y₁ = $\dfrac{y$2 - y1}{x2 - x1} (x - x₁)

Substituting values we get,

$\Rightarrow y - 3 = \dfrac{1 - 3}{4 - (-2)} [x - (-2)] \\[1em] \Rightarrow y - 3 = \dfrac{-2}{6} [x + 2] \\[1em] \Rightarrow y - 3 = \dfrac{-1}{3} [x + 2] \\[1em] \Rightarrow y = \dfrac{-x}{3} - \dfrac{2}{3} + 3 \\[1em] \Rightarrow y = \dfrac{-x}{3} - \dfrac{2 + 9}{3} \\[1em] \Rightarrow y = \dfrac{-x}{3} - \dfrac{11}{3}.$

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

Slope = m₁ = $-\dfrac{1}{3}$

The other equation is 3x = y + 1 or y = 3x - 1, comparing this with y = mx + c we get,

slope = m₂ = 3

Product of slopes,

= m₁ × m₂

= $-\dfrac{1}{3}$ × 3

= -1.

Since, the product of slopes is -1 hence, the lines are perpendicular to each other.

Hence, the line joining A(–2, 3) and B(4, 1) is perpendicular to the line 3x – y = 1.

(ii) Mid-point of (-2, 3) and (4, 1) can be given by mid-point formula i.e.,

$(x, y) = \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) = \Big(\dfrac{-2 + 4}{2}, \dfrac{3 + 1}{2}\Big) = (1, 2).

Line 3x = y + 1 bisects the line joining (-2, 3) and (4, 1) if the mid-point i.e., (1, 2) satisfies the equation.

Substituting (1, 2) in 3x = y + 1.

L.H.S. = 3x = 3(1) = 3.

R.H.S. = y + 1 = 2 + 1 = 3.

Since, L.H.S. = R.H.S. hence, (1, 2) satisfies 3x = y + 1.

Hence, the line 3x – y = 1 bisects the join of A(–2, 3) and B(4, 1).