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kap26 [50]
3 years ago
15

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

our claim. If it is not an isosceles triangle, what changes can be made to make it isosceles? Be specific.
Mathematics
1 answer:
kirill115 [55]3 years ago
8 0

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

substitute in the formula

d=\sqrt{(5-2)^{2}+(2-0)^{2}}

d=\sqrt{(3)^{2}+(2)^{2}}

dAB=\sqrt{13}\ units

step 2

Find the distance BC

substitute in the formula

d=\sqrt{(7-5)^{2}+(-1-2)^{2}}

d=\sqrt{(2)^{2}+(-3)^{2}}

dBC=\sqrt{13}\ units

step 3

Find the distance AC

substitute in the formula

d=\sqrt{(7-2)^{2}+(-1-0)^{2}}

d=\sqrt{(5)^{2}+(-1)^{2}}

dAC=\sqrt{26}\ units

step 4

Compare the length sides

dAB=\sqrt{13}\ units

dBC=\sqrt{13}\ units

dAC=\sqrt{26}\ units

dAB=dBC

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

(AC)^{2} =(AB)^{2}+(BC)^{2}

substitute

(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}

26=13+13

26=26 -----> is true

therefore

Is an isosceles right triangle

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