Respuesta :
Use the distance formula.
[tex] \sf\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
(1, 3), (9, 8)
x1 y1 x2 y2
[tex] \sf\sqrt{(9 - 1)^2 + (8 - 3)^2} [/tex]
[tex] \sf\sqrt{(8)^2 + (5)^2} [/tex]
[tex] \sf\sqrt{64 + 25} [/tex]
[tex] \sf\sqrt{89} [/tex]
[tex] \sf\approx 9.4 [/tex]
[tex] \sf\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
(1, 3), (9, 8)
x1 y1 x2 y2
[tex] \sf\sqrt{(9 - 1)^2 + (8 - 3)^2} [/tex]
[tex] \sf\sqrt{(8)^2 + (5)^2} [/tex]
[tex] \sf\sqrt{64 + 25} [/tex]
[tex] \sf\sqrt{89} [/tex]
[tex] \sf\approx 9.4 [/tex]
Answer
Find out the distance between points D and H .
To prove
Now apply the distance formula
[tex]Distance\ formula = \sqrt{(x_{2} -x_{1})^{2} +(y_{2} -y_{1})^{2}}[/tex]
The points be D (1,3)and H (9,8)
Put in the formula
[tex]DH = \sqrt{(9 -1)^{2} +(8 -3)^{2}}[/tex]
[tex]DH = \sqrt{(8)^{2} +(5)^{2}}[/tex]
[tex]DH = \sqrt{64+25}[/tex]
[tex]DH = \sqrt{89}[/tex]
As
[tex]\sqrt{89} = 9.4(approx)[/tex]
DH = 9.4 units (approx)
Option (D) is correct .