![\bold{\huge{\underline{ Solution }}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/jdi2w7914cic76zpb2xuxp7e51pz44d9g8.png)
Given :-
- We have given the coordinates of the triangle PQR that is P(-4,6) , Q(6,1) and R(2,9)
To Find :-
- We have to calculate the length of the sides of given triangle and also we have to determine whether it is right angled triangle or not
Let's Begin :-
Here, we have
- Coordinates of P =( x1 = -4 , y1 = 6)
- Coordinates of Q = ( x2 = 6 , y2 = 1 )
- Coordinates of R = ( x3 = 2 , y3 = 9 )
By using distance formula
![\pink{\bigstar}\boxed{\sf{Distance=√((x_1-x_2)^2+(y_1-y_2)^2\;)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8q00mdnfnkvma8trl2lupzom63vaz29fxv.png)
Subsitute the required values in the above formula :-
Length of side PQ
![\sf{ = }{\sf\sqrt{ (6 - (-4))^(2) + (1 - 6)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/r16makhqfygdifzo2zuv32njfv6rlnmj0t.png)
![\sf{ = }{\sf\sqrt{ (6 + 4 )^(2) + (- 5)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/p4x2mto5vwcqrqlgj8eygzgyub5dbhs5xd.png)
![\sf{ = }{\sf\sqrt{ (10)^(2) + (- 5)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/6s69wql93f3moabnhak7efxzl4qal3tqj7.png)
![\sf{ = }{\sf√( 100 + 25 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/gm4isxu7rlv90i1d09aeb348ed5h1xe465.png)
![\sf{ = }{\sf√( 125 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vfmn9nhqwozlk4zmt7w164kxwxle3u72im.png)
![\sf{ = 5 }{\sf√( 5 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/847gnwlulboql7bambl4a2kn73q7n5wgt3.png)
Length of QR
![\sf{ = }{\sf\sqrt{(2 - 6)^(2) + (9 - 1)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/z6euzippnz4yqeekriclylvgeb2djilfg9.png)
![\sf{ = }{\sf\sqrt{(- 4 )^(2) + (8)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/o46ahbt2tfoh8fbkba2fdq4w17njq6n892.png)
![\sf{ = }{\sf√(16 + 64 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ffjb0kcefm2nagjpwhtkm2k92jh83ggw8u.png)
![\sf{ = }{\sf√(80 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2g4nhxu6t17b58ruw8qcwi6n5rndmq29r1.png)
![\sf{ = 4 }{\sf√(5 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/7ueghlmfs0wy20tj4tie40vyn9pcnytvf5.png)
Length of RP
![\sf{ = }{\sf\sqrt{ (-4 - 2 )^(2) + (6 - 9)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/grhpfvebj1899z6y7jmy13seihu4zjhz0w.png)
![\sf{ = }{\sf\sqrt{ (-6 )^(2) + (-3)^(2)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ndaelie9xvfjdnwy5juf41jmlef2zarbhq.png)
![\sf{ = }{\sf√( 36 + 9 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/85fsob245t2a1ylyph3c463cy1ee5r36c0.png)
![\sf{ = }{\sf√( 45 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/a3aqghlxdr19pfvojg88h5tn66r047o4y8.png)
![\sf{ = 3}{\sf√( 5 )}](https://img.qammunity.org/2023/formulas/mathematics/high-school/wv00koearrscpnqhqu0lspitf0yiezm12u.png)
Now,
We have to determine whether the triangle PQR is right angled triangle
Therefore,
By using Pythagoras theorem :-
- Pythagoras theorem states that the sum of squares of two sides that is sum of squares of 2 smaller sides of triangle is equal to the square of hypotenuse that is square of longest side of triangle
That is,
![\bold{ PQ^(2) + QR^(2) = PR^(2)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/oheb6nbtq0lzbfauyd7lm42ho6jdwavp4t.png)
Subsitute the required values,
![\bold{ 125 + 80 = 45 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/ttf24472ugbc7x8qs8k7vyw9asrqe0yl9l.png)
![\bold{ 205 = 45 }](https://img.qammunity.org/2023/formulas/mathematics/high-school/25mibnvu55jjwny3aoe3y3rsi1d6yzeqmy.png)
From above we can conclude that,
- The triangle PQR is not a right angled triangle because 205 ≠ 45 .