Question 3 (a) Show that, for all $k \in \mathbb{R}$, the point $P(4k - 2, 3k + 1)$ lies on the line $l_1: 3x - 4y + 10 = 0.$ (b) The line $l_2$ passes through $P$ and is perpendicular to $l_1$ - Leaving Cert Mathematics - Question 3 - 2014

First, we compute the slope of the line $l_1$. Rearranging its equation gives us:

$y = \frac{3}{4}x + \frac{5}{4}.$

Thus, the slope of $l_1$ is $\frac{3}{4}$, making the slope of $l_2$ (being perpendicular to $l_1$) equal to:

$m_{l_2} = -\frac{4}{3}.$

The equation of line $l_2$, passing through point $P(4k - 2, 3k + 1)$, can be expressed as:

$y - (3k + 1) = -\frac{4}{3}(x - (4k - 2)).$

This simplifies to:

$3y - 3(3k + 1) = -4(x - (4k - 2)).$

After manipulations, we find:

$4x + 3y - 25k - 5 = 0.$

To find the appropriate value of $k$, we substitute the coordinates $(3, 11)$ into the equation derived for $l_2$:

$4(3) + 3(11) - 25k - 5 = 0.$

Solving gives:

$12 + 33 - 25k - 5 = 0 \rightarrow 50 - 25k = 0 \rightarrow k = 2.$

The equation of line $l_2$ using $k = 2$ is:

$4x + 3y - 45 = 0.$

To find the foot of the perpendicular, we solve the system formed by:

$3x - 4y + 10 = 0$

and

$4x + 3y - 45 = 0.$

This leads us to:

12x + 9y - 135 &= 0. \end{align*}$$ Solving gives the coordinates of the foot as $(6, 7)$, after deduction of values for $x$ and $y$.