In the figure, L(-4 ; 1), S(4 ; 5) and N(-2 ; -3) are the vertices of a triangle having $ ext{LN} = 90^\circ$ - NSC Mathematics - Question 3 - 2023 - Paper 2

The gradient of SN is found using the formula:

$m_{SN} = \frac{y_n - y_s}{x_n - x_s}$

Substituting the coordinates of S(4, 5) and N(-2, -3):

$m_{SN} = \frac{-3 - 5}{-2 - 4} = \frac{-8}{-6} = \frac{4}{3}.$

Thus, the gradient of SN is rac{4}{3}.

The angle of inclination $\theta$ can be found using:

$\tan(\theta) = m_{SN}$

Here, we already calculated $m_{SN} = \frac{4}{3}$:

$\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ.$

Since $\text{LN} \perp \text{SN}$, we can find the angle $\angle LNS$ as:

$\angle LNS = 90^\circ - \theta \approx 90^\circ - 53.13^\circ = 36.87^\circ.$

The equation of a line in slope-intercept form is $y = mx + c$. Since the line through L is parallel to SN, it has the same gradient:

$m = \frac{4}{3}$

Using the point L(-4, 1) to find c:

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

Solving for c gives:

$c = 1 + \frac{16}{3} = \frac{19}{3}$

Thus, the line equation is:

$y = \frac{4}{3}x + \frac{19}{3}.$

The area of a triangle given vertices can be calculated using:

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

Substituting the coordinates of S(4, 5), N(-2, -3), L(-4, 1):

$\text{Area} = \frac{1}{2} \left| 4(-3-1) + (-2)(1-5) + (-4)(5+3) \right|$ $= \frac{1}{2} \left| 4(-4) + (-2)(-4) + (-4)(8) \right|$ $= \frac{1}{2} \left| -16 + 8 - 32 \right| = \frac{1}{2} \left| -40 \right| = 20.$

Therefore, the area of $\Delta SNL$ is 20 square units.

Let the coordinates of point P be (a, b). The equations based on the distances are:

1. Distance from P to L: $\sqrt{(a + 4)^2 + (b - 1)^2} = \sqrt{(a - 4)^2 + (b - 5)^2}$

2. Distance from P to N: $\sqrt{(a + 2)^2 + (b + 3)^2} = \sqrt{(a - 4)^2 + (b - 5)^2}$

Solving these simultaneously gives the coordinates for P.