Is the line through (-2, 3) and (4, 1) perpendicular to the line 3x = y + 1? Does the line 3x = y + 1 bisect the join of (-2, 3) and (4, 1)?

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

$y - y$1 = \dfrac{y2 - y1}{x2 - x1}(x - x1)

Putting values in above formula we get,

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

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,

$= m$1 \times m2 \\[1em] = -\dfrac{1}{3} \times 3 \\[1em] = -1.

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

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

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

Putting (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 is perpendicular to the line joining (-2, 3) and (4, 1) and also bisects it.