Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas. He starts by assigning coordinates as given.

A right triangle is graphed on a coordinate plane. The horizontal x-axis and y-axis are solid, and the grid is hidden. The vertices are labeled as M, K, and L. The vertex labeled as M lies on begin ordered pair 0 comma 0 end ordered pair. The vertex labeled as K lies on begin ordered pair 0 comma 2 b end ordered pair. The vertex labeled as L lies on begin ordered pair 2a comma 0 end ordered pair. A bisector is drawn from point M to the line KL. The intersection point on line KL is labeled as N.

Enter the answers to complete the coordinate proof.
N is the midpoint of KL¯¯¯¯¯KL¯ . Therefore, the coordinates of N are (a,
).

To find the area of △KNM△KNM , the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM△KNM is
.

To find the area of △MNL△MNL , the length of the base ML is
, and the length of the height is
. So an expression for the area of △MNL△MNL is ab.

Comparing the expressions for the areas shows that the areas of the triangles are equal.

The coordinates of N is (a,b) using the midpoint formula.
The area for △KNM is (1/2)(a)(2b) = ab
The area of △MNL is ab.
Since the area of △KNM = △MNL and the area of △KML is 2ab, then we have proved that a segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

Answer Link

frika frika · Answer 2 · 2018-08-25T14:01:42+02:00

1. N is a midpoint of the segment KL, then N has coordinates

[tex] \left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b). [/tex]

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

[tex] A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab. [/tex]

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

[tex] A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab. [/tex]

4. Comparing the expressions for the areas you have that the area [tex] A_{KMN} [/tex] is equal to the area [tex] A_{MNL} [/tex]. This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.