Given a (- 2,4), B (6,8), find a point C on the x-axis, which is a triangle, ABC is an isosceles triangle

Given a (- 2,4), B (6,8), find a point C on the x-axis, which is a triangle, ABC is an isosceles triangle

(2,0),(5,0)(6,0),(10,0)

As shown in the figure, in the plane rectangular coordinate system, △ ABC is an equilateral triangle, a (0, √ 3). B (1,0). If the line y = KX + 2K intersects the x-axis at D, it is equal to △ ABC

Using the knowledge of triangle, we can get: S &; EC = 1 / 2CD * ce * sin60? S &; EF = 1 / 2aae * af * sina60? {if the area is equal, we only need CD * ce = AE * AF and CD = 1. We can get CE = AE * AF linear equation: y = KX + 2K linear equation: y = KX + 2K linear AC equation: y = √ 3 (x + 1) linear AB equation: y = - 3 (x + 3 (x + 1) AB equation: y = - - 3 (x-3 (x-1) 3 (x-3 (x-1) join to find out e ((3-3-2k) / (k - (k - √ 3) / (K-3) / (K-3) / (K-3) / (K + √ 3) 3, 3 (3-3-3 \\ [(((3 + \\(0, √ 3) C (- 1, 0) using the formula of distance from point to line, we can get: AE = 2 (√ 3-2k) / (k - √ 3) AF = 2 (√ 3-2k) / (K + √ 3) ce = 2K / (k - √ 3) bring in the formula CE = AE * AF to get: 7K ^ 2-9 √ 3K + 6 = 0, the solution is k = 2 √ 3 / 7 or K = √ 3 (rounding off), so there is such a k = 2 √ 3 / 7 that the area of two triangles is equal. All manual calculation, and then all manual marking to the top, tired me

In the plane rectangular coordinate system, point a (0,4), point B (3,0), find a point C on the y-axis, so that △ ABC is an isosceles triangle
In the plane rectangular coordinate system, point a (0,4), point B (3,0), find a point C on the Y axis, so that △ ABC is an isosceles triangle. How many points are there?

As shown in the figure: take a as the center, AB as the radius, intersect Y-axis at C1 (0,9), C2 (0, - 1); take B as the center, Ba as the radius, intersect Y-axis at C3 (0, - 4); make the vertical bisector of AB, intersect y-axis at C4 (0,0.875), that is, there are four such points c, namely C1 (0,9), C2 (0, - 1), C3 (0, - 4), C4 (0,0.875)

In the rectangular coordinate system, a (1,0), B (- 1,0,) △ ABC is a right triangle, AB is the bottom, and the area of △ ABC is 1, then the coordinate of point C is

(0,1) or (0, - 1)