Question

The area of the traingle formed by the midpoints of line segments PQ, QR, and RP where the coordinates of P, Q, and R (0,0),(3,0), and (3, 4) respectively, is?

Answer 1

Step-by-step explanation:

your answer is above....................

Answer 2

Given:

P ( 0 , 0 ) Q ( 3 , 0 ) R ( 3 , 4 )

To find :

The area of the triangle formed by the midpoints of line segments PQ, QR, and RP .

Formula to be used:

Midpoint of a line = $(\frac{x_{1} +x_{2} }{2} , \frac{y_{1} +y_{2} }{2} )$

Area of the triangle ABC =  $\frac{1}{2} [x_{1} (y_{2} - y_{3} ) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{2} - y_{1} )]$

Solution:

Step 1 of 2:

Let A , B , C are the midpoint of line segments  PQ , QR , RS of a triangle respectively.

To find the coordinates of A , B and C using midpoint formula,

A is the midpoint of PQ. Therefore ,

Coordinate of A = $(\frac{0 +3 }{2} , \frac{0+0 }{2} )$

Coordinate of A = $({\frac{3}{2} ,0)$

B is the midpoint of QR. Therefore ,

Coordinate of B = $(\frac{3 +3 }{2} , \frac{0+4 }{2} )$

Coordinate of B = $(\frac{6 }{2} , \frac{4 }{2} )$  

C is the midpoint of RS. Therefore ,

Coordinate of C = $(\frac{0 +3 }{2} , \frac{0+4 }{2} )$  

Coordinate of C =  $(\frac{3 }{2} , \frac{4 }{2} )$

Step 2 of 2:

Area of the triangle ABC =  $\frac{1}{2} [x_{1} (y_{2} - y_{3} ) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{2} - y_{1} )]$

$x_{1} = \frac{3}{2} , y_{1} = 0 \\x_{2} = \frac{6}{2} , y_{2} = \frac{4}{2} \\\x_{3} = \frac{3}{2} , y_{3} =\frac{4}{2}$

Area of a triangle ABC = $\frac{1}{2} [\frac{3}{2}(\frac{4}{2}-\frac{4}{2}) +\frac{6}{2}(\frac{4}{2}-0) + \frac{3}{2}(\frac{4}{2}-0)]$

Area of a triangle ABC = $\frac{1}{2} [\frac{3}{2}(0) +\frac{6}{2}(\frac{4}{2}) + \frac{3}{2}(\frac{4}{2})]$

Area of a triangle ABC = [ 0 + 6 + 3 ]

Area of a triangle ABC = [9]

Area of a triangle ABC = 4.5 square units.

Final answer:

The area of the triangle formed by the midpoints of line segments PQ, QR, and RP is 4.5 square units.