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 set ( y = 0 ) in the equation of line AS, which can be derived from line AB:

[ 0 = \frac{1}{3}x + 3 ] [ \frac{1}{3}x = -3 ] [ x = -9 ]

Thus, the coordinates of B are B(-9, 0).

We find the coordinates of C by determining the y-intercept of line AC when ( x = 0 ):

The equation of line AC is given by ( y = 2x - 2 ). So,

[ y = 2(0) - 2 = -2 ]

To find another point on line AC, substitute ( x = 3 ):

[ y = 2(3) - 2 = 6 - 2 = 4 ]

Thus, the coordinates of C are C(3, -8).

First, we need to find the gradient of line BC. Using coordinates B(-9, 0) and C(3, -8):

[ m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-8 - 0}{3 - (-9)} = \frac{-8}{12} = -\frac{2}{3} ]

Since the line we are looking for is parallel to BC, it will have the same gradient. Therefore, using the point S(-15, -2), we apply the point-gradient form:

[ y - y_1 = m(x - x_1) ] [ y - (-2) = -\frac{2}{3}(x + 15) ]

Hence, the equation simplifies to [ y = -\frac{2}{3}x - 12 ].

To find ( \angle BAC ), we use the tangent of the angles formed by the lines. The slopes of AB and AC are ( m_{AB} = \frac{1}{3} ) and ( m_{AC} = 2 ). Applying the formula:

[ \tan(\alpha) = \frac{m_{AC} - m_{AB}}{1 + (m_{AB} \cdot m_{AC})} ]

Substituting the values gives: [ \tan(\angle BAC) = \frac{2 - \frac{1}{3}}{1 + (\frac{1}{3} \cdot 2)} = \frac{\frac{6}{3} - \frac{1}{3}}{1 + \frac{2}{3}} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1 ]

Thus, ( \angle BAC = 45^{\circ} ).

We can find the areas of the triangles using the formula for the area of a triangle:

[ Area = \frac{1}{2} \times base \times height ]

The area of ( \triangle AABD ) using height AD and base AB computes:

[ Area_{AABD} = \frac{1}{2} \times |AB| \times h_{AD} ]

For ( \triangle ASC ), the equations yield similar substitutions. The triangle's areas can be represented automatically with the given lengths and height values, leading to conclusion on how these relate to find the total area of both triangles as they are derived from the same points.