A straight line joins the points [tex]\( A(-2, -3) \)[/tex] and [tex]\( C(1, 9) \)[/tex].

(a) Find the equation of the line [tex]\( AC \)[/tex] in the form [tex]\( y = mx + c \)[/tex].

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

[tex]\[
m = \frac{9 - (-3)}{1 - (-2)} = \frac{12}{3} = 4
\][/tex]

Now, using the point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex] with point [tex]\( A(-2, -3) \)[/tex]:

[tex]\[
y + 3 = 4(x + 2)
\][/tex]

[tex]\[
y + 3 = 4x + 8
\][/tex]

[tex]\[
y = 4x + 5
\][/tex]

(b) Calculate the acute angle between [tex]\( AC \)[/tex] and the [tex]\( x \)[/tex]-axis.

The slope [tex]\( m = 4 \)[/tex] gives us the tangent of the angle [tex]\( \theta \)[/tex]:

[tex]\[
\tan \theta = 4
\][/tex]

[tex]\[
\theta = \tan^{-1}(4)
\][/tex]

(c) [tex]\( ABCD \)[/tex] is a kite, where [tex]\( AC \)[/tex] is the longer diagonal.

[Continue with relevant calculations or details about the kite as necessary.]

[2]



Answer :

Let's tackle the problem step by step to find the solution.

### (a) Find the equation of the line [tex]\( AC \)[/tex] in the form [tex]\( y = mx + c \)[/tex]

Given points [tex]\( A(-2, -3) \)[/tex] and [tex]\( C(1, 9) \)[/tex]:

1. Calculate the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the given points [tex]\( (x_1, y_1) = (-2, -3) \)[/tex] and [tex]\( (x_2, y_2) = (1, 9) \)[/tex]:
[tex]\[ m = \frac{9 - (-3)}{1 - (-2)} = \frac{9 + 3}{1 + 2} = \frac{12}{3} = 4 \][/tex]

2. Find the y-intercept [tex]\( c \)[/tex]:
Using the point-slope form of the line [tex]\( y = mx + c \)[/tex]:
Substitute point [tex]\( A(-2, -3) \)[/tex] and the slope [tex]\( m = 4 \)[/tex]:
[tex]\[ -3 = 4(-2) + c \implies -3 = -8 + c \implies c = -3 + 8 = 5 \][/tex]

3. Equation of the line:
Substitute [tex]\( m = 4 \)[/tex] and [tex]\( c = 5 \)[/tex] into the line equation:
[tex]\[ y = 4x + 5 \][/tex]

So, the equation of the line [tex]\( AC \)[/tex] is:
[tex]\[ y = 4x + 5 \][/tex]

### (b) Calculate the acute angle between [tex]\( AC \)[/tex] and the [tex]\( x \)[/tex]-axis

To find the acute angle between the line [tex]\( AC \)[/tex] and the [tex]\( x \)[/tex]-axis, we use the fact that the tangent of the angle [tex]\( \theta \)[/tex] formed with the [tex]\( x \)[/tex]-axis is equal to the slope [tex]\( m \)[/tex]:
[tex]\[ m = \tan(\theta) \][/tex]

Given the slope [tex]\( m = 4 \)[/tex]:
[tex]\[ \theta = \tan^{-1}(4) \][/tex]

To express [tex]\( \theta \)[/tex] in degrees, we convert it from radians to degrees:
[tex]\[ \theta \approx 75.96^\circ \][/tex]

So, the acute angle between the line [tex]\( AC \)[/tex] and the [tex]\( x \)[/tex]-axis is:
[tex]\[ 75.96^\circ \][/tex]

### (c) [tex]\( ABCD \)[/tex] is a kite, with [tex]\( AC \)[/tex] as the longer diagonal

Without additional specific information about points [tex]\( B \)[/tex] and [tex]\( D \)[/tex] or further details regarding the kite's dimensions and properties, it is not possible to provide a conclusive solution for this part. Typically, a kite has two pairs of adjacent sides that are equal in length, and its diagonals intersect at right angles. With such a geometric figure, knowing the exact coordinates or lengths is essential to proceed with further calculations or conclusions.

However, if the problem or additional context about points [tex]\( B \)[/tex] and [tex]\( D \)[/tex] is provided, the properties and relationships within the kite could be explored further.