The points (1,2),(7,4) and (3,6) are vertices of a right triangle. Determine the are of the triangle to the nearest hundredth

Question

It's a triangle, I need to find the midpoint of the segment that connects the triangles???? I'm nt exactly sure

Answer 1

First, plot the triangle.  Label the points A(1,2). B(7,4) and C(3,6).

 

Then, be sure that it is a right triangle as stated (note: have you never been "tricked?").  The line segments A(1,2)C(3,6) [with slope 2] and C(3,6)B(7,4) [with slope (-1/2) are perpendicular (inverse reciprocal slope).  O.K., it is a right triangle.

 

The area of the triangle is (1/2)bh.

  h = distance CB is 2√5     [distance formula:  d = ( (y₂-y₁)² + (x₂-x₁)² )  ]

  b = distance AC is 2√5     [distance formula again]

  A = (1/2)bh

  A = (1/2)(2√5)(2√5)

  A = (1/2)(4)(5)

  A = 10 square units           ("to the nearest hundredth", this is 10.0]