In the diagram below, 𝐴𝐵 intersects 𝐷𝐶 at 𝐸(− 3; 2). 𝐵 is a point on the 𝑥 −axis and 𝐹

Question

is a 𝑥-intercept. 𝐷𝐶 intersects the 𝑦-axis at − 1. 𝑋𝐹𝐸 = 135°. The equation of 𝐴𝐷 is
given by 𝑦 − 4 = 0. 𝐷𝐶 ⊥ 𝐴𝐵.
Figure 4: Diagram of the Cartesian plane.
Determine:
4.1 The gradient of 𝐷𝐶. (2)
4.2 The equation of 𝐷𝐶. (2)
4.3 The coordinates of 𝐷. (4)
4.4 The size of the angle 𝐷𝐴𝐵 if it is given that 𝐷𝐴||𝐵𝐹. (3)
[11

Answer 1

To solve the problem, let's analyze each part step-by-step.

Given Data:

1. Point \( E(-3, 2) \) is the intersection of lines \( AB \) and \( DC \).
2. Point \( B \) is on the \( x \)-axis, so its coordinates are \( (b, 0) \).
3. Line \( DC \) intersects the \( y \)-axis at \( -1 \), thus the \( y \)-intercept of line \( DC \) is \( (0, -1) \).
4. \( \angle XFE = 135^\circ \), where \( F \) is the \( x \)-intercept of \( AB \).
5. The equation of line \( AD \) is given as \( y - 4 = 0 \), i.e., \( y = 4 \).
6. Lines \( DC \) and \( AB \) are perpendicular.

Step 4.1: The Gradient of Line \( DC \)

To find the gradient of line \( DC \), we can use the coordinates of the point where it intersects the \( y \)-axis:

• The coordinates are \( (0, -1) \) (y-intercept) and point \( E(-3, 2) \).

The gradient (\( m \)) of a line through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting the points into the formula:

\[ m_{DC} = \frac{2 - (-1)}{-3 - 0} = \frac{2 + 1}{-3} = \frac{3}{-3} = -1 \]

So, the gradient of \( DC \) is \( -1 \) (Answer: -1).

Step 4.2: The Equation of Line \( DC \)

We can use the gradient-intercept form of the equation of a line, which is \( y = mx + c \), where \( m \) is the gradient and \( c \) is the \( y \)-intercept.

Here, since \( m = -1 \) and \( c = -1 \):

\[ y = -1x - 1 \quad \text{or} \quad y = -x - 1 \]

So, the equation of line \( DC \) is \( y = -x - 1 \) (Answer: y = -x - 1).

Step 4.3: The Coordinates of Point \( D \)

To find the coordinates of point \( D \), we know:

1. Line \( AD \) is horizontal (since \( y = 4 \)).
2. Point \( D \) lies on \( y = 4 \) and also lies on line \( DC \).

To find the intersection of line \( DC: y = -x - 1 \) and line \( AD: y = 4 \):

Set \( -x - 1 = 4 \):

\[ -x - 1 = 4 \implies -x = 5 \implies x = -5 \]

Thus, the coordinates of point \( D \) are \( (-5, 4) \) (Answer: (-5, 4)).

Step 4.4: Size of Angle \( DAB \)

Since \( DA \parallel BF \), and \( DC \perp AB \), we can find the angle \( DAB \):

1. The gradient of line \( AB \) is \( m_{AB} = 1 \) (since \( DC \) has a gradient of \( -1 \))
2. We can use the tangent of angles to find angle \( DAB \):

We can find the angles using the gradient relationship:

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

Using our gradients:

1. Gradient \( m_{AB} = 1 \)
2. Gradient \( m_{DC} = -1 \)

To find \( \angle DAB \):

The angle \( DAB \) is complementary to \( XFE \), given \( XFE = 135^\circ \).

Thus:

\[ \theta_{DAB} = 90^\circ - 135^\circ = 45^\circ \]

So, the size of angle \( DAB \) is \( 45^\circ \) (Answer: 45°).

Summary of Answers:

4.1: -1
4.2: y = -x - 1
4.3: (-5, 4)
4.4: 45°