The hypotenuse of a right triangle has endpoints a(4, 1) and b(–1, –2).

at which points could the vertex of the right angle in the triangle be located? check all that apply.
(–1, 1)
(4, –2)
(1, 1)
(2, –2)
(4, –1)
(–1, 4)

(–1, 1)

(4, –2)

Step-by-step explanation:

Given that the hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2), then two triangles can be formed (see figure attached).

The vertex of the right angle in each triangle are on points (-1, 1) and (4, -2), as can be seen in the figure.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12