G C S E Gradient - Let's Ramp It Up

[kuiperbelt]

Please see link below:

https://ibb.co/Ybq5rJs

[ZebuSanctuary]

The gradient of the radius OW can be calculated δy/δx thus 8/4 = 2

It is hoped the student is aware of the following points given below;

(1.) Geometrically it can be shown that a tangent to a circle forms a right angle with the circle's radius, at the point of contact with the tangent.

(2.) A property of 2 straight lines crossing at right angles, the gradient of one of the lines is the inverse of the other and vice versa. In otherwords their product = -1

eg. gradient -1/4 would become +4 for the other line hence
-1/4 x 4 = -1

So, since the gradient of (radius) the line OW = 2, invoking the above property (2.) gives a gradient of line AB = -1/2 or -0.5

From the general equation for straight line y = mx + c,
the equation of lineAB in terms of x and y is therefore y = c - 0.5x where c is the y coordinate of point A.

Voila!

[ZebuSanctuary]

Afterthought...

Since the coordinate W(4, 8) sits on the lineAB and lineAB is represented by y = c - 0.5x where c is the y coordinate of point A (intercept on the y-axis), then c can be calculated.

Substituting y = 8 and x =4 into y = c - 0.5x gives;

8 = c - 0.5 x 4

The arithmetic bears out c equal to 10
or the coordinate of point A (0, 10).

This yields lineAB in terms of x and y;

y = 10 - 0.5x

Now we are done.