Given the vertex coordinates, the equation of the straight line where the hypotenuse is, and the slope of the straight line where the hypotenuse center line is, calculate the other two coordinates. See below for details Given the vertex C (- 2,3) of a right triangle, the linear equation of the hypotenuse AB is 4x-3y-7 = 0, and the slope of the straight line of the middle line on the hypotenuse is - 4 / 3, the coordinates of point a and B are obtained

Given the vertex coordinates, the equation of the straight line where the hypotenuse is, and the slope of the straight line where the hypotenuse center line is, calculate the other two coordinates. See below for details Given the vertex C (- 2,3) of a right triangle, the linear equation of the hypotenuse AB is 4x-3y-7 = 0, and the slope of the straight line of the middle line on the hypotenuse is - 4 / 3, the coordinates of point a and B are obtained

∵ let the D coordinate of the middle point of the hypotenuse be (x, y)
∴ (y-3)/(x+2)=(-4)/3
The D coordinate of the middle point of the hypotenuse can be (3m-2,3-4m)
Substitute 4x-3y-7 = 0 to get 4 (3m-2) - 3 (3-4m) - 7 = 0
The solution is m = 1
Ψ point d (1, - 1)
So CD = 5
∴ AD＝BD＝5
The slope of line AB is 4 / 3
The coordinates of a and B are (- 2, - 5), (4,3)