Question

Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem L(0,5) M(9,12) N(3,14)

Answer 1

Answer:

Yes the given points form a right angled triangle.

Step-by-step explanation:

Given : Points L(0,5) M(9,12) N(3,14).

To find : Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem?

Solution :

First we find the length of the sides of the triangles using distance formula,

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

The distance between L(0,5) M(9,12)

$LM=\sqrt{(9-0)^2+(12-5)^2}$

$LM=\sqrt{81+49}$

$LM=\sqrt{130}$

The distance between M(9,12) N(3,14)

$MN=\sqrt{(3-9)^2+(14-12)^2}$

$MN=\sqrt{36+4}$

$MN=\sqrt{40}$

The distance between L(0,5) N(3,14)

$LN=\sqrt{(3-0)^2+(14-5)^2}$

$LN=\sqrt{9+81}$

$LN=\sqrt{90}$

Now, If these form a right angle then Pythagoras theorem should satisfy i.e.

$H^2=B^2+P^2$

Where, H is the longest side.

So, Let $H=LM=\sqrt{130}$, $P=MN=\sqrt{40}$, $B=LN=\sqrt{90}$

LHS, $H^2=(\sqrt{130})^2=130$

RHS,  $B^2+P^2=(\sqrt{40})^2+(\sqrt{90})^2=40+90=130$

LHS=RHS.

Therefore, Yes the given points form a right angled triangle.

Answer 2

Step-by-step explanation:

First we find the length of the sides of the triangles using distance formula,

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

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

The distance between L(0,5) M(9,12)

LM=\sqrt{(9-0)^2+(12-5)^2}LM=

(9−0)

2

+(12−5)

2

LM=\sqrt{81+49}LM=

81+49

LM=\sqrt{130}LM=

130

The distance between M(9,12) N(3,14)

MN=\sqrt{(3-9)^2+(14-12)^2}MN=

(3−9)

2

+(14−12)

2

MN=\sqrt{36+4}MN=

36+4

MN=\sqrt{40}MN=

40

The distance between L(0,5) N(3,14)

LN=\sqrt{(3-0)^2+(14-5)^2}LN=

(3−0)

2

+(14−5)

2

LN=\sqrt{9+81}LN=

9+81

LN=\sqrt{90}LN=

90

Now, If these form a right angle then Pythagoras theorem should satisfy i.e.

H^2=B^2+P^2H

2

=B

2

+P

2

Where, H is the longest side.

So, Let H=LM=\sqrt{130}H=LM=

130

, P=MN=\sqrt{40}P=MN=

40

, B=LN=\sqrt{90}B=LN=

90

LHS, H^2=(\sqrt{130})^2=130H

2

=(

130

)

2

=130

RHS, B^2+P^2=(\sqrt{40})^2+(\sqrt{90})^2=40+90=130B

2

+P

2

=(

40

)

2

+(

90

)

2

=40+90=130

LHS=RHS.

Therefore, Yes the given points form a right angled triangle.