Known: point a (- 2,0), B (3,0), point C on the Y-axis and the area of △ ABC is 15, find the coordinates of point C

Let the origin of coordinate be o, then | OC | is the height of AB in △ ABC
So s △ ABC = 1 / 2 * | OC | * | ab|
Let C (0, c), then | OC | = | C |, and | ab | = | - 2-3 | = 5
So there is 1 / 2 * | C | * 5 = 15, and the solution is C = ± 6
So C (0,6) or (0, - 6)

Given points a (- 3,0), B (- 1, - 2) and point C on the x-axis, if the area of △ ABC is 15, the coordinates of point C can be obtained

Let the coordinates of point C be (a, 0)
1. a>-3
Area = 15 = 1 / 2 * 2 * (a + 3)
a+3=15
a=12
2.a

As shown in the figure, point a (1, - 2), B (3,4), point C is on the Y axis and AC + BC is the shortest

The symmetric point of B (3,4) about y axis is B '(- 3,4),
A straight line ab ': y = - 3x / 2-1 / 2, intersecting Y-axis at C (0, - 1 / 2),
Area of triangle ABC = trapezoidal area - area of two triangles
=(1/2)(1+3)*5-(1/2)*3*(9/2)-(1/2)*1*(1/2)
=3

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