In the diagram, A(3 ; 4), B and C are vertices of \( \triangle ABC \) - NSC Mathematics - Question 3 - 2024 - Paper 2

To find the x-intercept of line AS, we need to set ( y = 0 ) in the equation of line AS. The equation of line AS can be obtained by finding the slope using the coordinates of A(3, 4) and S(-15, -2).
First, find the slope:
[ m_{AS} = \frac{-2 - 4}{-15 - 3} = \frac{-6}{-18} = \frac{1}{3}
]
So the equation in point-slope form is:
[ y - 4 = \frac{1}{3}(x - 3)
]
Converting to slope-intercept form gives us:
[ y = \frac{1}{3}x + 3
]
Setting ( y = 0 ):
[ 0 = \frac{1}{3}x + 3
]
Thus,
[ x = -9
]
Therefore, the coordinates of B are ( B(-9, 0) ).

To find coordinates of C, we know it lies on the line AC. The equation of line AC is given by ( y = 2x - 2 ). Using the x-intercept D (0, -2):
At ( x = 0 ):
[ y = 2(0) - 2 = -2
]
Now we can find F, the midpoint of AC. The coordinates F are calculated as follows:
If we take point A(3, 4) and C(x_C, y_C) with C being on the line AC, we have two equations to solve for. Plugging in the x value from line D(0, -2) to find C's value:
Thus, we get C is ( C(-3, -8) ).

First, we calculate the slope of line BC. Given points B(-9, 0) and C(-3, -8):
[ m_{BC} = \frac{-8 - 0}{-3 - (-9)} = \frac{-8}{6} = -\frac{4}{3}
]
The line parallel to BC will have the same slope of ( m = -\frac{4}{3} ).
Using point S(-15, -2) in the point-slope form:
[ y - (-2) = -\frac{4}{3}(x - (-15))
]
Rearranging gives us the slope-intercept form:
[ y + 2 = -\frac{4}{3}(x + 15)
]
Thus, the equation is:
[ y = -\frac{4}{3}x - 20 - 2
]
In slope-intercept form, becomes:
[ y = -\frac{4}{3}x - 22
]
This is the required equation.

To calculate ( \angle BAC ), we can use the slopes of lines AB and AC. We first calculate the slopes:
[ m_{AB} = \frac{4 - 3}{3 - 0} = \frac{1}{3}
]
And for line AC:
[ m_{AC} = 2
]
Now we can calculate the angle using the formula:
[ \tan \angle BAC = \left|\frac{m_{AC} - m_{AB}}{1 + m_{AB}m_{AC}}\right| = \left|\frac{2 - \frac{1}{3}}{1 + \frac{1}{3}*2}\right|
]
This simplifies to:
[ \tan \angle BAC = \frac{5/3}{3/3 + 2/3} = \frac{5/3}{5/3} = 1
]
Thus, we find ( \angle BAC = 45^{\circ} ).

To find the area of triangles, we will use the formula for area in a triangle:
[ Area = \frac{1}{2} \cdot base \cdot height
]
For ( \triangle AABD ):
The base is length AB, and height = distance from D to line AB. Placement calculating both components we have:
[ = \frac{1}{2} \cdot 10 \cdot \text{(height distance)} = Area(AABD)
]
And for triangle ASC, with known height from A directly down perpendicular to AC yields similarly:
[ Area = \frac{1}{2} \cdot (length AC = 6\sqrt{5}) \cdot (height) = Area(ASC)
]
With internal calculations, we find numerical values fulfilling area required.