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Math  /  Geometry

QuestionSlope of line DE=−1/3∨D E=-1 / 3 \vee
Slope of line EF= 3 □\square
Slope of line DF=−1/3∼D F=-1 / 3 \sim □\square
Length of the line DE == □\square
Length of the line EF= □\square
Length of the line DF= □\square Determine the type of the triangle □\square

Studdy Solution

STEP 1

1. We are given a triangle with vertices D,E,F D, E, F .
2. The coordinates of the points are approximately D(−40,40) D(-40, 40) , E(20,40) E(20, 40) , and F(10,0) F(10, 0) .
3. We need to determine the type of triangle formed by these points.
4. The type of triangle can be determined by examining the slopes of the sides and their lengths.

STEP 2

1. Calculate the slopes of the lines DE DE , EF EF , and DF DF .
2. Calculate the lengths of the sides DE DE , EF EF , and DF DF .
3. Determine the type of triangle based on the slopes and lengths.

STEP 3

Calculate the slope of line DE DE using the formula for slope:
slope of DE=y2−y1x2−x1 \text{slope of } DE = \frac{y_2 - y_1}{x_2 - x_1}
Substitute the coordinates of D(−40,40) D(-40, 40) and E(20,40) E(20, 40) :
slope of DE=40−4020−(−40)=060=0 \text{slope of } DE = \frac{40 - 40}{20 - (-40)} = \frac{0}{60} = 0

STEP 4

Calculate the slope of line EF EF using the formula for slope:
slope of EF=y2−y1x2−x1 \text{slope of } EF = \frac{y_2 - y_1}{x_2 - x_1}
Substitute the coordinates of E(20,40) E(20, 40) and F(10,0) F(10, 0) :
slope of EF=0−4010−20=−40−10=4 \text{slope of } EF = \frac{0 - 40}{10 - 20} = \frac{-40}{-10} = 4

STEP 5

Calculate the slope of line DF DF using the formula for slope:
slope of DF=y2−y1x2−x1 \text{slope of } DF = \frac{y_2 - y_1}{x_2 - x_1}
Substitute the coordinates of D(−40,40) D(-40, 40) and F(10,0) F(10, 0) :
slope of DF=0−4010−(−40)=−4050=−45 \text{slope of } DF = \frac{0 - 40}{10 - (-40)} = \frac{-40}{50} = -\frac{4}{5}

STEP 6

Calculate the length of line DE DE using the distance formula:
Length of DE=(x2−x1)2+(y2−y1)2 \text{Length of } DE = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Substitute the coordinates of D(−40,40) D(-40, 40) and E(20,40) E(20, 40) :
Length of DE=(20−(−40))2+(40−40)2=602+02=60 \text{Length of } DE = \sqrt{(20 - (-40))^2 + (40 - 40)^2} = \sqrt{60^2 + 0^2} = 60

STEP 7

Calculate the length of line EF EF using the distance formula:
Length of EF=(x2−x1)2+(y2−y1)2 \text{Length of } EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Substitute the coordinates of E(20,40) E(20, 40) and F(10,0) F(10, 0) :
Length of EF=(10−20)2+(0−40)2=(−10)2+(−40)2=100+1600=1700 \text{Length of } EF = \sqrt{(10 - 20)^2 + (0 - 40)^2} = \sqrt{(-10)^2 + (-40)^2} = \sqrt{100 + 1600} = \sqrt{1700}

STEP 8

Calculate the length of line DF DF using the distance formula:
Length of DF=(x2−x1)2+(y2−y1)2 \text{Length of } DF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Substitute the coordinates of D(−40,40) D(-40, 40) and F(10,0) F(10, 0) :
Length of DF=(10−(−40))2+(0−40)2=502+(−40)2=2500+1600=4100 \text{Length of } DF = \sqrt{(10 - (-40))^2 + (0 - 40)^2} = \sqrt{50^2 + (-40)^2} = \sqrt{2500 + 1600} = \sqrt{4100}

STEP 9

Determine the type of triangle by examining the slopes and lengths:
- The slope of DE DE is 0 0 , indicating it is a horizontal line. - The slopes of EF EF and DF DF are not negative reciprocals, so they are not perpendicular. - The lengths of the sides are DE=60 DE = 60 , EF=1700 EF = \sqrt{1700} , DF=4100 DF = \sqrt{4100} .
Since none of the slopes are perpendicular, the triangle is not a right triangle. The sides are of different lengths, so it is a scalene triangle.
The triangle DEF DEF is a scalene triangle.

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