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In △ABC, the coordinates of vertices A and B are A(1,−1) and B(3,2).

For each of the given coordinates of vertex C, is △ABC a right triangle?

Select Right Triangle or Not a Right Triangle for each set of coordinates.
C(0,2)
C(3,−1)
C(0,4)

User Matt Ryall
by
8.1k points

1 Answer

3 votes

we know that

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



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


In this problem we have


A(1,-1)\ B(3,2)\ C1(0,2)\ C2(3,-1)\ C3(0,4)

Step 1

Find the distance AB


A(1,-1)\ B(3,2)

Substitute the values in the formula


d=\sqrt{(2+1)^(2)+(3-1)^(2)}



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



dAB=√(13)\ units


Step 2

Find the distance AC1


A(1,-1)\ C1(0,2)

Substitute the values in the formula


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



d=\sqrt{(3)^(2)+(-1)^(2)}



dAC1=√(10)\ units


Step 3

Find the distance BC1


B(3,2)\ C1(0,2)

Substitute the values in the formula


d=\sqrt{(2-2)^(2)+(0-3)^(2)}



d=\sqrt{(0)^(2)+(-3)^(2)}



dBC1=3\ units


Step 4

Find the distance AC2


A(1,-1)\ C2(3,-1)

Substitute the values in the formula


d=\sqrt{(-1+1)^(2)+(3-1)^(2)}



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



dAC2=2\ units


Step 5

Find the distance BC2


B(3,2)\ C2(3,-1)

Substitute the values in the formula


d=\sqrt{(-1-2)^(2)+(3-3)^(2)}



d=\sqrt{(-3)^(2)+(0)^(2)}



dBC2=3\ units

Step 6

Find the distance AC3


A(1,-1)\ C3(0,4)

Substitute the values in the formula


d=\sqrt{(4+1)^(2)+(0-1)^(2)}



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



dAC3=√(26)\ units


Step 7

Find the distance BC3


B(3,2)\ C3(0,4)

Substitute the values in the formula


d=\sqrt{(4-2)^(2)+(0-3)^(2)}



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



dBC3=√(13)\ units


we know that

If the length sides of the triangle satisfy the Pythagoras Theorem. then the triangle is a right triangle

The formula of the Pythagoras Theorem is equal to


c^(2) =a^(2)+b^(2)

where

c is the hypotenuse (the greater side)

a and b are the legs of the triangle

Step 8

Verify if the triangle ABC1 is a right triangle

we have


dAB=√(13)\ units



dAC1=√(10)\ units



dBC1=3\ units


Applying Pythagoras theorem


AB^(2)=AC1^(2)+BC1^(2)


√(13)^(2) =√(10)^(2)+3^(2)


13 =10+9


13 =19 --------> is not true

therefore

the triangle ABC1 is not a right triangle

Step 9

Verify if the triangle ABC2 is a right triangle

we have


dAB=√(13)\ units



dAC2=2\ units



dBC2=3\ units


Applying Pythagoras theorem


AB^(2)=AC2^(2)+BC2^(2)


√(13)^(2) =2^(2)+3^(2)


13 =4+9


13 =13 --------> is true

therefore

the triangle ABC2 is a right triangle

Step 10

Verify if the triangle ABC3 is a right triangle

we have


dAB=√(13)\ units



dAC3=√(26)\ units



dBC3=√(13)\ units


Applying Pythagoras theorem


AC3^(2)=AB^(2)+BC3^(2)


√(26)^(2) =√(13)^(2)+√(13)^(2)


26 =13+13


26=26 --------> is true

therefore

the triangle ABC3 is a right triangle

therefore

the answer is


C(0,2)\ Not\ a\ right\ triangle\\C(3,-1)\ A\ right\ triangle\\C(0,4)\ A\ right\ triangle

User AXM
by
8.4k points

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