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

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

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=√((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=√((-1+2)^2+(1-4)^2)`

`AB=√(10)`

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

`(x_2,y_2)=(2,2)`

Substitute the vales in the formula

`BC=√((2+1)^2+(2-1)^2)`

`BC=√(10)`

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

`(x_2,y_2)=(2,2)`

Substitute the vales in the formula

`AC=√((2+2)^2+(2-4)^2)`

`AC=√(20)`

So,
`AB=√(10)` ,
`BC=√(10)` and
`AC=√(20)`

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

We will use Pythagoras theorem

`Hypotenuse^2=Perpendicular^2+Base^2`

`(√(20))^2=(√(10))^2+(√(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=√((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=√((-1+2)^2+(1-4)^2)`

`AB=√(10)`

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

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

Substitute the vales in the formula

`BC=√((0+1)^2+(4-1)^2)`

`BC=√(10)`

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

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

Substitute the vales in the formula

`AC=√((0+2)^2+(4-4)^2)`

`AC=4`

So,
`AB=√(10)` ,
`BC=√(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=(√(10))^2+(√(10))^2`

`16=10+10`

`16 \\eq20`

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=√((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=√((-1+2)^2+(1-4)^2)`

`AB=√(10)`

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

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

Substitute the vales in the formula

`BC=√((-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=√((-2+2)^2+(1-4)^2)`

`AC=3`

So,
`AB=√(10)` ,
`BC=1` and
`AC=3`

We will use Pythagoras theorem

`Hypotenuse^2=Perpendicular^2+Base^2`

`(√(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.

Answer 2

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

Slope of AB is :
`m=(1-4)/(-1-\left(-2\right)) =-3.`

With coordinate C(2,2).

Slope of BC is:
`m=(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=(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=(1-1)/(-2-\left(-1\right))=0`

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