Question

What is the midpoint of the longest side of the triangle with vertices (1,4) ,(3,4), and (3,6)?

Answer 1

Answer:

The answer to your question is: C

Step-by-step explanation:

Data

A (1 , 4)

B (3, 4)

C(3, 6)

Formula

d = $\sqrt{(x2-x1)^{2} + (y2 - y1)^{2} }$

Process

dAB = $\sqrt{(3 - 1)^{2} + (4 - 4)^{2} }$

dAB = $\sqrt{(2)^{2} + (0)^{2} }$

dAB = 2

dBC = $\sqrt{(3-3)^{2} + (6 - 4)^{2} }$

dBC = $\sqrt{(0)^{2} + (2)^{2} }$

dBC = 2

dAC = $\sqrt{(3-1)^{2} + (6 - 4)^{2} }$

dAC = $\sqrt{(2)^{2} + (2)^{2} }$

dAC = √8

Longest side = AC

midpoint  x = $\frac{1 + 3}{2} = \frac{4}{2} = 2$

midpoint y = $\frac{4 + 6}{2} = \frac{10}{2} = 5$

Midpoint = ( 2, 5)