Math  /  Geometry

Question3. DEF\triangle D E F has vertices at D(8,2),E(1,3)D(8,-2), E(1,-3), and F(9,9)F(9,-9). Use special segments to determine if EFE F is the base of an isosceles triangle. ρ(8,2)\rho(8,-2)

Studdy Solution

STEP 1

1. We are given a triangle DEF\triangle DEF with vertices D(8,2)D(8, -2), E(1,3)E(1, -3), and F(9,9)F(9, -9).
2. We need to determine if EFEF can be the base of an isosceles triangle.
3. An isosceles triangle has at least two equal sides.

STEP 2

1. Calculate the length of segment EFEF.
2. Calculate the lengths of segments DEDE and DFDF.
3. Compare the lengths to determine if EFEF is the base of an isosceles triangle.

STEP 3

Calculate the length of segment EFEF using the distance formula:
EF=(x2x1)2+(y2y1)2 EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Substitute the coordinates of points E(1,3)E(1, -3) and F(9,9)F(9, -9):
EF=(91)2+(9+3)2 EF = \sqrt{(9 - 1)^2 + (-9 + 3)^2} =82+(6)2 = \sqrt{8^2 + (-6)^2} =64+36 = \sqrt{64 + 36} =100 = \sqrt{100} =10 = 10

STEP 4

Calculate the length of segment DEDE using the distance formula:
DE=(x2x1)2+(y2y1)2 DE = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Substitute the coordinates of points D(8,2)D(8, -2) and E(1,3)E(1, -3):
DE=(18)2+(3+2)2 DE = \sqrt{(1 - 8)^2 + (-3 + 2)^2} =(7)2+(1)2 = \sqrt{(-7)^2 + (-1)^2} =49+1 = \sqrt{49 + 1} =50 = \sqrt{50} =52 = 5\sqrt{2}

STEP 5

Calculate the length of segment DFDF using the distance formula:
DF=(x2x1)2+(y2y1)2 DF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Substitute the coordinates of points D(8,2)D(8, -2) and F(9,9)F(9, -9):
DF=(98)2+(9+2)2 DF = \sqrt{(9 - 8)^2 + (-9 + 2)^2} =12+(7)2 = \sqrt{1^2 + (-7)^2} =1+49 = \sqrt{1 + 49} =50 = \sqrt{50} =52 = 5\sqrt{2}

STEP 6

Compare the lengths of the sides:
- EF=10EF = 10 - DE=52DE = 5\sqrt{2} - DF=52DF = 5\sqrt{2}
Since DE=DFDE = DF, DEF\triangle DEF is isosceles with EFEF as the base.
The segment EFEF is the base of an isosceles triangle.

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