Given coordinates of vertices A and B are A(−2,4) and B(−1,1).
Slope of AB is : [tex]m=\frac{1-4}{-1-\left(-2\right)} =-3.[/tex]
With coordinate C(2,2).
Slope of BC is: [tex]m=\frac{1-4}{-1-\left(-2\right)}=-3[/tex]
With coordinate C(0,4)
Slope of BC is [tex]m=\frac{4-1}{0-\left(-1\right)}=3[/tex].
Slopes of AB and BC are not negative reciprocals.
With coordinate (-2,1)
Slope of BC is [tex]m=\frac{1-1}{-2-\left(-1\right)}=0[/tex]
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: [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(-2,4)[/tex]
[tex](x_2,y_2)=(-1,1)[/tex]
Substitute the vales in the formula
[tex]AB=\sqrt{(-1+2)^2+(1-4)^2}[/tex]
[tex]AB=\sqrt{10}[/tex]
[tex](x_1,y_1)=(-1,1)[/tex]
[tex](x_2,y_2)=(2,2)[/tex]
Substitute the vales in the formula
[tex]BC=\sqrt{(2+1)^2+(2-1)^2}[/tex]
[tex]BC=\sqrt{10}[/tex]
[tex](x_1,y_1)=(-2,4)[/tex]
[tex](x_2,y_2)=(2,2)[/tex]
Substitute the vales in the formula
[tex]AC=\sqrt{(2+2)^2+(2-4)^2}[/tex]
[tex]AC=\sqrt{20}[/tex]
So, [tex]AB=\sqrt{10}[/tex] , [tex]BC=\sqrt{10}[/tex] and [tex]AC=\sqrt{20}[/tex]
Now to check whether it is a right angled triangle or not
We will use Pythagoras theorem
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex](\sqrt{20})^2=(\sqrt{10})^2+(\sqrt{10})^2[/tex]
[tex]20=10+10[/tex]
[tex]20=20[/tex]
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: [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(-2,4)[/tex]
[tex](x_2,y_2)=(-1,1)[/tex]
Substitute the vales in the formula
[tex]AB=\sqrt{(-1+2)^2+(1-4)^2}[/tex]
[tex]AB=\sqrt{10}[/tex]
[tex](x_1,y_1)=(-1,1)[/tex]
[tex](x_2,y_2)=(0,4)[/tex]
Substitute the vales in the formula
[tex]BC=\sqrt{(0+1)^2+(4-1)^2}[/tex]
[tex]BC=\sqrt{10}[/tex]
[tex](x_1,y_1)=(-2,4)[/tex]
[tex](x_2,y_2)=(0,4)[/tex]
Substitute the vales in the formula
[tex]AC=\sqrt{(0+2)^2+(4-4)^2}[/tex]
[tex]AC=4[/tex]
So, [tex]AB=\sqrt{10}[/tex] , [tex]BC=\sqrt{10}[/tex] and [tex]AC=4[/tex]
Now to check whether it is a right angled triangle or not
We will use Pythagoras theorem
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex](4)^2=(\sqrt{10})^2+(\sqrt{10})^2[/tex]
[tex]16=10+10[/tex]
[tex]16 \neq20[/tex]
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: [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(-2,4)[/tex]
[tex](x_2,y_2)=(-1,1)[/tex]
Substitute the vales in the formula
[tex]AB=\sqrt{(-1+2)^2+(1-4)^2}[/tex]
[tex]AB=\sqrt{10}[/tex]
[tex](x_1,y_1)=(-1,1)[/tex]
[tex](x_2,y_2)=(-2,1)[/tex]
Substitute the vales in the formula
[tex]BC=\sqrt{(-2+1)^2+(1-1)^2}[/tex]
[tex]BC=1[/tex]
[tex](x_1,y_1)=(-2,4)[/tex]
[tex](x_2,y_2)=(-2,1)[/tex]
Substitute the vales in the formula
[tex]AC=\sqrt{(-2+2)^2+(1-4)^2}[/tex]
[tex]AC=3[/tex]
So, [tex]AB=\sqrt{10}[/tex] , [tex]BC=1[/tex] and [tex]AC=3[/tex]
We will use Pythagoras theorem
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex](\sqrt{10})^2=(3)^2+(1)^2[/tex]
[tex]10=9+1[/tex]
[tex]10=10[/tex]
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.
