Determine the ratio in which the line 2x + y - 4 = 0 divide the line segment joining the points A(2, -2) and B(3, 7). Also find the coordinates of the point of the division.

Let the line 2x + y - 4 divide the line segment AB in the ratio m : n at P. So, by section-formula coordinates of P are,

x-coordinate = x = $\dfrac{m \times x$2 + n \times x1}{m + n}

$= \dfrac{m \times 3 + n \times 2}{m + n} \\[1em] = \dfrac{3m + 2n}{m + n} \qquad \text{….[Eq 1]} \\[1em]$

Similarly,

y-coordinate = y = $\dfrac{m \times y$2 + n \times y1}{m + n}

$= \dfrac{m \times 7 + n \times (-2)}{m + n} \\[1em] = \dfrac{7m - 2n}{m + n} \qquad \text{….[Eq 2]} \\[1em]$

Since, P lies on the line 2x + y - 4 = 0.

$\therefore 2\dfrac{(3m + 2n)}{m + n} + \dfrac{7m - 2n}{m + n} - 4 = 0 \\[1em] \Rightarrow \dfrac{6m + 4n + 7m - 2n -4(m + n)}{m + n} = 0 \\[1em] \Rightarrow 6m + 4n + 7m - 2n - 4m - 4n = 0 \\[1em] \Rightarrow 9m - 2n = 0 \\[1em] \Rightarrow 9m = 2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{9} \\[1em]$

Putting values in Eq 1 for x-coordinate we get,

$\therefore x = \dfrac{3 \times 2 + 2 \times 9}{2 + 9} \\[1em] = \dfrac{6 + 18}{11} \\[1em] = \dfrac{24}{11}. \\[1em]$

Putting values in Eq 2 for y-coordinate we get,

$\text{y = } \dfrac{7 \times 2 - 2 \times 9}{2 + 9} \\[1em] = \dfrac{14 - 18}{11} \\[1em] = -\dfrac{4}{11}.$

Hence, coordinates of P will be $\Big(\dfrac{24}{11}, -\dfrac{4}{11}\Big)$ and 2 : 9 is the ratio in which the line 2x + y - 4 divides AB.