Answer :
Answer:
The triangles are congruent
Step-by-step explanation:
Given
[tex]P(5,-4), Q(-3, 7), R(0, 2), X(-2,-1), Y(9, 7), Z(3, 2)[/tex]
Required
Determine if PQR = XYZ
To do this, we make use of distance formula.
[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
For PQ:
[tex](x_1,y_1) = (5,-4)[/tex] and [tex](x_2,y_2) = (-3,7)[/tex]
So:
[tex]PQ = \sqrt{(-3-5)^2+(7-(-4))^2}[/tex]
[tex]PQ = \sqrt{(-8)^2+(7+4)^2}[/tex]
[tex]PQ = \sqrt{(-8)^2+(11)^2}[/tex]
[tex]PQ = \sqrt{64+121}[/tex]
[tex]PQ = \sqrt{185}[/tex]
For XY:
[tex](x_1,y_1) = (-2,-1)[/tex]
[tex](x_2,y_2) = (9,7)[/tex]
So:
[tex]XY = \sqrt{(9-(-2))^2+(7-(-1))^2}[/tex]
[tex]XY = \sqrt{(9+2)^2+(7+1)^2}[/tex]
[tex]XY = \sqrt{(11)^2+(8)^2}[/tex]
[tex]XY = \sqrt{121+64}[/tex]
[tex]XY = \sqrt{185}[/tex]
For QR:
[tex](x_1,y_1) = (-3,7)[/tex] and [tex](x_2,y_2) = (0,2)[/tex]
So:
[tex]QR = \sqrt{(0-(-3))^2 + (2-7)^2}[/tex]
[tex]QR = \sqrt{(0+3)^2 + (-5)^2}[/tex]
[tex]QR = \sqrt{(3)^2 + 25}[/tex]
[tex]QR = \sqrt{9 + 25}[/tex]
[tex]QR = \sqrt{34}[/tex]
For YZ:
[tex](x_1,y_1) = (9,7)[/tex] and [tex](x_2,y_2) = (3,2)[/tex]
So:
[tex]YZ = \sqrt{(3-9)^2+(2-7)^2}[/tex]
[tex]YZ = \sqrt{(-6)^2+(-5)^2}[/tex]
[tex]YZ = \sqrt{36+25}[/tex]
[tex]YZ = \sqrt{61}[/tex]
For PR:
[tex](x_1,y_1) = (5,-4)[/tex] and [tex](x_2,y_2) = (0,2)[/tex]
So:
[tex]PR = \sqrt{(0-5)^2+(2-(-4))^2}[/tex]
[tex]PR = \sqrt{(-5)^2+(2+4)^2}[/tex]
[tex]PR = \sqrt{(-5)^2+(6)^2}[/tex]
[tex]PR = \sqrt{25+36}[/tex]
[tex]PR = \sqrt{61}[/tex]
For XZ:
[tex](x_1,y_1) = (-2,-1)[/tex] and [tex](x_2,y_2) = (3,2)[/tex]
So:
[tex]XZ = \sqrt{(3-(-2))^2+(2-(-1))^2}[/tex]
[tex]XZ = \sqrt{(3+2)^2+(2+1)^2}[/tex]
[tex]XZ = \sqrt{5^2+3^2}[/tex]
[tex]XZ = \sqrt{34}[/tex]
From the calculations above:
[tex]PQ = XY = \sqrt{185[/tex]
[tex]QR = XZ = \sqrt{34[/tex]
[tex]PR = YZ = \sqrt{61[/tex]
Hence, the triangles are congruent base on SSS (Sides- Sides-Sides)