Hi, I would be really grateful to anyone who can help me with this problem...

**(1)** An aeroplane flies at a constant speed of 800km per hour in a straight line from (-100, -200) to (300, 1000).

(positions are given with ref to Cartesian coordinate system whose x- and y- axes point due East and due North respectively)

I think the equation of the line of flight of the aeroplane is:

rise= 1000 -(-200) = 1200

run= 300-(-100) = 400

slope = 1200/400 = 3

so y - (-200) = 3 (x -(-100))

y = 3x -500

Am I right so far? ........**No**

**(2)** Next I need to find the direction of travel of the aeroplane as a bearing with the angle correct to one decimal place...

**(3)** The final part of this question is as follows. A radar station at (0, 200) has a range of 100km.

What is the equation of the circle at the limit of the radar stations range?

I also need to find the points where the line of flight of the aeroplane intersects this circle and the time during which the aeroplane is within range of the radar station in minutes to one decimal place.

to (1): The line passing through the points \(\displaystyle A(x_A, y_A)\) and \(\displaystyle B(x_B, y_B)\) has the equation:

\(\displaystyle \dfrac{y-y_A}{x-x_A}=\dfrac{y_B-y_A}{x_B-x_A}\)

Plug in the coordinates of the points and solve for y:

\(\displaystyle y = 3x+100\)

To (2): The bearing is an angle between the North (= 0° or 360°) direction and direction of flight, measured clockwise.

The slope of the line is the tangens of the angle between the line and the x-axis, with your problem between the flight direction and the East.

\(\displaystyle m= 3 ~\implies~\alpha = \arctan(3)\approx 71.565^\circ\)

Consequently the bearing is: \(\displaystyle 90^\circ-71.565^\circ = \boxed{18.4^\circ}\)

to (3): A circle around the point C(h, k) has the equation

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

Plug in the coordinates of the radar station (that's the center of the circle):

\(\displaystyle \boxed{x^2+(y-200)^2=100^2}\)

To calculate the coordinates of the points of intersection solve the system of equations:

\(\displaystyle \left|\begin{array}{l}y = 3x+100 \\ x^2+(y-200)^2=100^2\end{array}\right.\)

\(\displaystyle x^2+(3x+100-200)^2=100^2~\implies~10x^2-600x=0~\implies~x=0\vee x=60\)

Plug in the x-coordinates into the equation of the straight line to get the y-coordinates: A(0, 100), B(60, 280)