Find the equation of the line perpendicular to the line joining the points A(1, 2) and B(6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.

Let the slope of line joining A(1, 2) and B(6, 7) be s₁. Slope of two points is given by,

$s$1 = \dfrac{y2 - y1}{x2 - x_1} \\[1em] = \dfrac{7 - 2}{6 - 1} \\[1em] = \dfrac{5}{5} \\[1em] = 1.

Let slope of perpendicular line be s₂. So,

$\Rightarrow s$1 \times s2 = -1 \\[1em] \Rightarrow 1 \times s2 = -1 \\[1em] \Rightarrow s2 = -1.

Given, the new line passes through the point which divides the line segment AB in the ratio 3 : 2. By section formula coordinates are,

$= \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] = \Big(\dfrac{3 \times 6 + 2 \times 1}{3 + 2}, \dfrac{3 \times 7 + 2 \times 2}{3 + 2} \Big) \\[1em] = \Big(\dfrac{18 + 2}{5}, \dfrac{21 + 4}{5}\Big) \\[1em] = \Big(\dfrac{20}{5}, \dfrac{25}{5}\Big) \\[1em] = (4, 5).

Equation of the line having slope -1 and passing through (4, 5) can be given by point-slope form i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 5 = -1(x - 4) \\[1em] \Rightarrow y - 5 = -x + 4 \\[1em] \Rightarrow x + y - 9 = 0.

Hence, the equation of the required line is x + y - 9 = 0.