Question

prove that A(8,-10) ,B(7,-3) and C(0,-4) are the vertices of the right angled triangle​

Answer 1

Answer:

Let, the points are A (8, -10), B (7, -3) and C (0, -4)

• AB = \sqrt{(8-7)^{2}+(-10+3)^{2}}

(8−7)

2

+(−10+3)

2

units

=\sqrt{1^{2}+7^{2}}=

1

2

+7

2

units

=\sqrt{1+49}=

1+49

units

=\sqrt{50}=

50

units

• BC = \sqrt{(7-0)^{2}+(-3+4)^{2}}

(7−0)

2

+(−3+4)

2

units

=\sqrt{7^{2}+1^{2}}=

7

2

+1

2

units

=\sqrt{49+1}=

49+1

units

=\sqrt{50}=

50

units

• CA = \sqrt{(0-8)^{2}+(-4+10)^{2}}

(0−8)

2

+(−4+10)

2

units

=\sqrt{8^{2}+6^{2}}=

8

2

+6

2

units

=\sqrt{64+36}=

64+36

units

=\sqrt{100}=

100

units

We see that, AB^{2}+BC^{2}=CA^{2}AB

2

+BC

2

=CA

2

where (\sqrt{50})^{2}+(\sqrt{50})^{2}=(\sqrt{100})^{2}(

50

)

2

+(

50

)

2

=(

100

)

2

By Pythagorian theorem, we can say that the triangle ABC is a right angled triangle.

Hence, proved.

Step-by-step explanation:

Hope it helps.

Answer 2

Step-by-step explanation:

let A(8 , -10) , B(7,-3 ) ,c (0 , -4)

( AB) distance =

$\sqrt{50}$

(BC) distance =

$\sqrt{50}$

and (AC ) =

$\sqrt{100}$

so, hence proved

as. , AB +BC =AC