Question

name the missing coordinate of the triangle.
a(0, a)
b(a, 0)
c(?, ?)

Explanation:

Step1: Analyze triangle symmetry

The triangle has two equal sides (marked with ticks), so it's isosceles with respect to the y - axis? Wait, no, looking at points A(0, a) and B(a, 0), and the symmetry of the triangle. The point C should be symmetric to point A with respect to the x - axis (since the triangle is symmetric about the line that is the angle bisector, and also, the distance from C to B should be equal to the distance from A to B, and the x - coordinate of C should be 0 (since it's on the y - axis, as A is (0, a) and the vertical line is the y - axis). The y - coordinate of C should be the negative of the y - coordinate of A because of the symmetry about the x - axis (since B is (a, 0) on the x - axis).

Step2: Determine coordinates of C

Since A is (0, a) and the triangle is isosceles with AB = BC and AC is vertical (on the y - axis), the x - coordinate of C is 0 (same as A's x - coordinate, on the y - axis). The y - coordinate of C should be - a because it's below the x - axis (opposite of A's y - coordinate which is above the x - axis) to maintain the isosceles property (distance from B(a, 0) to C(0, - a) should be equal to distance from B(a, 0) to A(0, a)). Let's verify the distance: Distance between A(0, a) and B(a, 0) is $\sqrt{(a - 0)^2+(0 - a)^2}=\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}}=\vert a\vert\sqrt{2}$. Distance between B(a, 0) and C(0, - a) is $\sqrt{(0 - a)^2+(-a - 0)^2}=\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}}=\vert a\vert\sqrt{2}$, which is equal. So the coordinates of C are (0, - a).

Answer:

The missing coordinate of point C is \((0, - a)\)