In the figure, L(−4 ; 1), S(4 ; 5) and N(−2 ; −3) are the vertices of a triangle having ∠LNS=90˚ - NSC Mathematics - Question 3 - 2023 - Paper 2

The gradient (m) of the line passing through points S(4,5) and N(−2,−3) is given by:

$m_{SN} = \\frac{y_S - y_N}{x_S - x_N} = \\frac{5 - (-3)}{4 - (-2)} = \\frac{5 + 3}{4 + 2} = \\frac{8}{6} = \\frac{4}{3}$

Thus, the gradient of SN is $\\frac{4}{3}$.

The angle of inclination (θ) of SN can be calculated using the tangent function:

$\tan(θ) = m_{SN} = \frac{4}{3}$

So, to find θ, we apply the arctangent function:

$θ = \tan^{-1}(\\frac{4}{3}) ≈ 53.13^ ext{°}$

Hence, the size of θ is approximately $53.13^ ext{°}$.

Since angle ∠LNS is a right angle and SN is a diameter of the circle that subtends this angle,

we have:

$∠LNS = 90^ ext{°}$

Since the line parallel to SN will have the same gradient, we can use the point-slope form:

$y - y_L = m_{SN}(x - x_L)$

Using L(−4, 1):

$y - 1 = \frac{4}{3}(x + 4)$
Expanding gives:

$y - 1 = \frac{4}{3}x + \frac{16}{3}$

Thus:

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

The area of triangle ΔLSN can be calculated using the formula:

$Area_{ΔLSN} = \\frac{1}{2} \\times base \\times height$

We can take the base as SN and height as the perpendicular distance from L to line SN.
Since the length of SN = 10 units, we need to find the height:

distance = 4 units, we get:

$Area_{ΔLSN} = \\frac{1}{2} \\times 10 \\times 4 = 20 ext{ units}^2$

Let P(a,b) be a point equidistant from L, S, and N.
We can set up the following equations based on the distance:

1. Distance PL:
   $PL = \\sqrt{(a - (-4))^2 + (b - 1)^2}$
2. Distance PS:
   $PS = \\sqrt{(a - 4)^2 + (b - 5)^2}$
3. Distance PN:
   $PN = \\sqrt{(a - (-2))^2 + (b - (-3))^2}$

Setting PL = PS:

$(a + 4)^2 + (b - 1)^2 = (a - 4)^2 + (b - 5)^2$

Also, setting PL = PN will provide another equation:

Solving these simultaneously yields the coordinates of point P.