Exercise 3.3 on page 110, the first question of the last question in group B I have worked out is (4), the title is that the vertex a (5,1) of △ ABC is known, the equation of the line where the central line cm on the edge of AB is 2x-y-5 = 0, and the equation of the line where the high BH on the edge of AC is x-2y-5 = 0 (1) The coordinates of vertex C; (2) The equation of the straight line BC

Exercise 3.3 on page 110, the first question of the last question in group B I have worked out is (4), the title is that the vertex a (5,1) of △ ABC is known, the equation of the line where the central line cm on the edge of AB is 2x-y-5 = 0, and the equation of the line where the high BH on the edge of AC is x-2y-5 = 0 (1) The coordinates of vertex C; (2) The equation of the straight line BC

The linear equation of high BH on AC side is x-2y-5 = 0, y = 1 / 2x-5 / 2
∵ AC ⊥ BH, AC slope is the negative reciprocal of BH, i.e. - 2
∵ a (5,1), so the straight line of AC is y = - 2x + 11
The linear equation of the central line cm on the edge of AB is 2x-y-5 = 0
The coordinate of point C (4,3) is obtained by cm and AC
The linear equation of BH is x-2y-5 = 0
Let B (2W + 5, w)
So the straight line AB is: y = (W-1) / 2W * (X-5) + 1
The method is: (X-5) / (2W + 5-5) = (Y-1) / (W-1)
The linear equation of CM is 2x-y-5 = 0
The abscissa of M obtained simultaneously is (7W + 5) / (3w + 1)
∵ m is the midpoint of AB side, and the abscissa of ∵ m is half of abscissa of a and B, that is (5 + 2W + 5) / 2
The calculation process W is the denominator, or w = - 3
So B (- 1, - 3)
Because C (4,3)
So BC: 6x-5y-9 = 0