The diagram shows the line PQ and the line QR - Leaving Cert Mathematics - Question 2 - 2019

Using the point-slope form of the equation:

$y - y_1 = m(x - x_1)$

Substituting $m = \frac{3}{4}$ and point P(4, 2):

$y - 2 = \frac{3}{4}(x - 4)$

Expanding this gives:

$y - 2 = \frac{3}{4}x - 3$

Rearranging to standard form: $\frac{3}{4}x - y + 1 = 0\text{ (Multiplying through by 4 to eliminate the fraction:)}$ $3x - 4y + 4 = 0$

So the equation of the line PQ is:

$3x - 4y + 4 = 0$

The slope of a line perpendicular to another is the negative reciprocal of the original slope.

Given the slope of PQ is $\frac{3}{4}$, the slope of a line perpendicular to PQ is:

$-\frac{1}{(\frac{3}{4})} = -\frac{4}{3}$

To find the area of triangle PQR, we can use the formula:

$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substituting the points P(4, 2), Q(8, 5), R(2, 11):

$= \frac{1}{2} |4(5 - 11) + 8(11 - 2) + 2(2 - 5)|$

Calculating:
$= \frac{1}{2} |4(-6) + 8(9) + 2(-3)|$

$= \frac{1}{2} |-24 + 72 - 6|$
$= \frac{1}{2} |42| = 21$

Thus, the area of triangle PQR is $21$ square units.