Show that the triangle formed by the lines y = 2x − 7, x + 2y = 16, and 3x + y = 13 is isosceles. Show also that the lengths of the sides of this triangle fit the Pythagorean equation.

Answer 1

y = 2x-7 vs. x + 2y = 16

x + 2(2x-7) = 16

x + 4x - 14 = 16

5x - 14 = 16

5x = 30

x = 6

they meet at (6,5)

y = 2x-7 vs 3x+y = 13

3x + (2x-7) = 13

5x - 7 = 13

5x = 20

x = 4

they meet at (4,1)

x+2y = 16 vs 3x+y = 13

x = 16-2y ---> 3(16 - 2y) + y = 13

48 - 6y + y = 13

48 - 5y = 13

-5y = -35

y =7

they meet at ( 2,7)

(6,5), (4,1) (2,7)

(6,5) vs ( 4,1)

distance formula says sqrt ( (6-4)^2 + (5-1)^2) = sqrt( 2^2 + 4^2) = sqrt(4+16) = sqrt(20)

(6,5) vs (2,7)

distance formula says sqrt( (6-2)^2 + (7-5)^2) = sqrt( 4^2 + 2^2) = sqrt( 16 + 4 ) = sqrt(20)

(4,1) vs (2,7)

distance formula says sqrt( (4-2)^2 + (7-1)^2) = sqrt( 2^2 + 6^2) = sqrt( 4 + 36) = sqrt(40)= 2 * sqrt(10)

isosceles proven with equal sides sqsrt(20) = 2*sqrt(5)

pythagorean says sqrt(20)^2 + sqrt(20)^2 = 20 + 20 = 40 = a^2 + b^2

c^2 = (2 * sqrt(10))^2 = 4 * 10 = 40

pythagorean proven, so it is an isosceles right triangle, 45-45-90