Question

In △ABC, the coordinates of vertices A and B are A(−2,4) and B(−1,1).



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.

Answer 1

Given coordinates of vertices A and B are A(−2,4) and B(−1,1).

Slope of AB is : $m=\frac{1-4}{-1-\left(-2\right)} =-3.$

With coordinate C(2,2).

Slope of BC is: $m=\frac{1-4}{-1-\left(-2\right)}=-3$

Slopes of AB and BC are same, therefore with C(2,2) is not a right triangle.

With coordinate C(0,4)

Slope of BC is $m=\frac{4-1}{0-\left(-1\right)}=3$.

Slopes of AB and BC are not negative reciprocals.

Therefore, C(0,4) also not a coordinate to make a right triangle.

With coordinate (-2,1)

Slope of BC is $m=\frac{1-1}{-2-\left(-1\right)}=0$

Therefore, C(-2,1) also not a coordinate to make a right triangle.

Answer 2

Answer:

For the given coordinates of vertex C of Option A and Option C ,△ABC is a right triangle.

Step-by-step explanation:

Coordinates of A = (−2,4)

Coordinates of B =(−1,1)

Since we are asked For each of the given coordinates of vertex C, is △ABC a right triangle

So, Option 1)

Coordinates of C = (2,2)

Now to find length of AB, BC and AC

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(-2,4)$

$(x_2,y_2)=(-1,1)$

Substitute the vales in the formula

$AB=\sqrt{(-1+2)^2+(1-4)^2}$

$AB=\sqrt{10}$

$(x_1,y_1)=(-1,1)$

$(x_2,y_2)=(2,2)$

Substitute the vales in the formula

$BC=\sqrt{(2+1)^2+(2-1)^2}$

$BC=\sqrt{10}$

$(x_1,y_1)=(-2,4)$

$(x_2,y_2)=(2,2)$

Substitute the vales in the formula

$AC=\sqrt{(2+2)^2+(2-4)^2}$

$AC=\sqrt{20}$

So, $AB=\sqrt{10}$ , $BC=\sqrt{10}$ and $AC=\sqrt{20}$

Now to check whether it is a right angled triangle or not

We will use Pythagoras theorem

$Hypotenuse^2=Perpendicular^2+Base^2$

$(\sqrt{20})^2=(\sqrt{10})^2+(\sqrt{10})^2$

$20=10+10$

$20=20$

So, For Option 1 , △ABC is a right triangle.

Option 2)

Coordinates of C = (0,4)

Now to find length of AB, BC and AC

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(-2,4)$

$(x_2,y_2)=(-1,1)$

Substitute the vales in the formula

$AB=\sqrt{(-1+2)^2+(1-4)^2}$

$AB=\sqrt{10}$

$(x_1,y_1)=(-1,1)$

$(x_2,y_2)=(0,4)$

Substitute the vales in the formula

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

$BC=\sqrt{10}$

$(x_1,y_1)=(-2,4)$

$(x_2,y_2)=(0,4)$

Substitute the vales in the formula

$AC=\sqrt{(0+2)^2+(4-4)^2}$

$AC=4$

So, $AB=\sqrt{10}$ , $BC=\sqrt{10}$ and $AC=4$

Now to check whether it is a right angled triangle or not

We will use Pythagoras theorem

$Hypotenuse^2=Perpendicular^2+Base^2$

$(4)^2=(\sqrt{10})^2+(\sqrt{10})^2$

$16=10+10$

$16 \neq20$

So, For Option 2, △ABC is not a right triangle.

Option 2)

Coordinates of C = (-2,1)

Now to find length of AB, BC and AC

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(-2,4)$

$(x_2,y_2)=(-1,1)$

Substitute the vales in the formula

$AB=\sqrt{(-1+2)^2+(1-4)^2}$

$AB=\sqrt{10}$

$(x_1,y_1)=(-1,1)$

$(x_2,y_2)=(-2,1)$

Substitute the vales in the formula

$BC=\sqrt{(-2+1)^2+(1-1)^2}$

$BC=1$

$(x_1,y_1)=(-2,4)$

$(x_2,y_2)=(-2,1)$

Substitute the vales in the formula

$AC=\sqrt{(-2+2)^2+(1-4)^2}$

$AC=3$

So, $AB=\sqrt{10}$ , $BC=1$ and $AC=3$

We will use Pythagoras theorem

$Hypotenuse^2=Perpendicular^2+Base^2$

$(\sqrt{10})^2=(3)^2+(1)^2$

$10=9+1$

$10=10$

So, For Option 3, △ABC is  a right triangle.

Hence For the given coordinates of vertex C of Option A and Option C ,△ABC is a right triangle.