Mathematics problems in senior one, straight line and equation In the triangle ABC, the point a (1,1), B (4,2) and point C are on the line X-Y 5 = 0, and the linear equation of the height on the BC side is 5x-2y-3 = 0 (1) Find point C (2) Is triangle ABC a right triangle? It's a process. Thank you x-y+5

Mathematics problems in senior one, straight line and equation In the triangle ABC, the point a (1,1), B (4,2) and point C are on the line X-Y 5 = 0, and the linear equation of the height on the BC side is 5x-2y-3 = 0 (1) Find point C (2) Is triangle ABC a right triangle? It's a process. Thank you x-y+5

(1)
The linear equation of height is 5x-2y-3 = 0, and its slope is k = 5 / 2
So the slope of the line BC perpendicular to the height is k = - 1 / k = - (2 / 5)
Since BC passes through point B, let the equation be: (Y-2) = - (2 / 5) * (x-4)
The simplified formula is: 2x + 5y-18 = 0
Then find the intersection point C of 2x + 5y-18 = 0 and X-Y + 5 = 0
To solve this system of equations:
2x+5y-18=0
x-y+5=0
The solution is as follows
x=-1,y=4
Point C: (- 1,4)
(2)
AC^2=(-1-1)^2+(4-1)^2=4+9=13
BC^2=(-1-4)^2+(4-2)^2=25+4=29
AB^2=(4-1)^2+(2-1)^2=9+1=10
Therefore, triangle is not right triangle

If three straight lines X + y = 0, X-Y = 0, x + ay = 3 form a triangle, then the range of a is?
If three lines x-2y + 1 = 0, x + 3y-1 = 0 and ax + 2y-3 = 0 have two different intersections, then a =?

1. If a = 1, then x + ay = 3 and X + y = 2 are parallel. If a = - 1, then x + ay = 3 and X-Y = 0 are parallel. The first two intersections are (1,1) substituted, 1 + a = 3, a = 2, so a ≠ - 1 and a ≠ 1 and a ≠ 2. Because there are only two different intersections of three lines, we can infer that two of them are parallel lines, because the first

The line L passing through point (2, 3) is cut by two parallel lines l1:2x-5y + 9 = 0 and l2:2x-5y-7 = 0. The midpoint of line AB is just on line x-4y-1 = 0. The equation of line L is obtained

Let the coordinates (a, b) of the midpoint P of the line AB be equal from P to L1 and L2, then | 2A − 5B + 9 | 22 + 52 = | 2A − 5B − 7 | 22 + 52. After sorting out, 2a-5b + 1 = 0, and point P is on the line x-4y-1 = 0, so a-4b-1 = 0. Solve the equations 2A − 5B + 1 = 0A − 4B − 1 = 0, then a = − 3B = − 1, that is, the coordinates of point P (- 3, - 1), and the line L passes through the point (2, 3), so the equation of line L is y − (− 1) 3 − (− 1) ）The equation of line L is: 4x-5y + 7 = 0

(1) Given that the center of circle C is on the straight line L: x-2y-1 = 0, and passing through the origin and a (2,1), the standard equation of circle C is obtained
(2) There are four points a (0,1) B (2,1) C (3,4) d (- 1,2) in the plane rectangular coordinate system. Can these four points be on the same circle? Why?
(3) Given that the ratio of distance between point m and two fixed points o (0,0) a (3,0) is 1 / 2, the trajectory equation of point m is obtained

Let (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2 pass through the origin and a (2,1), so a ^ 2 + B ^ 2 = R ^ 2. ① (2-A) ^ 2 + (1-B) ^ 2 = R ^ 2. ② because the center of the circle is on the straight line L: x-2y-1 = 0, so a-2b-1 = 0