1.△ area of ABC = length of AB line segment * distance from point C to straight line AB /2, and let the coordinates of point C be (x, y).

Because the length of AB line segment is equal to root number 40, the equation of straight line AB is 3x-y+6=0.

The distance from point C to straight line AB is equal to |3x-y+6|/ (root sign 10).

So (root number 40)*|3x-y+6|/ (root number 10)=32,

Simplify and get |3x-y+6|= 16.

So 3x-y+6= 16 or 3x-y+6=- 16, that is, the trajectory equation of point C is 3x-y- 10=0 or 3x-y+22=0.

2. The slope of Ma * the slope of MB =- 1, and the straight line of the surface MA⊥ straight line MB. Let the coordinates of point M be (x, y).

Then vector MA=(-a-x, -y) and vector MB=(a-x, -y), because vector MA is multiplied by vector MB=0.

So (-a-x, -y)*(a-x, -y)=x? -a？ +y？ =0, that is, the trajectory equation of the moving point m is x? +y？ =a？

Write questions 3 and 4 again when you are free.