Question

Find the perimeter of the triangle whose vertices are (-2,1),(4,6) and (6,3). ​

Answer 1

Answer:

We know that the distance between the two points (x

1

,y

1

) and (x

2

,y

2

) is

d=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

Let the given vertices be A=(−2,1), B=(4,6) and C=(6,−3)

We first find the distance between A=(−2,1) and B=(4,6) as follows:

AB=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(4−(−2))

2

+(6−1)

2

=

(4+2)

2

+5

2

=

6

2

+5

2

=

36+25

=

61

Similarly, the distance between B=(4,6) and C=(6,−3) is:

BC=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(6−4)

2

+(−3−6)

2

=

2

2

+(−9)

2

=

4+81

=

85

Now, the distance between C=(6,−3) and A=(−2,1) is:

CA=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(6−(−2))

2

+(−3−1)

2

=

(6+2)

2

+(−4)

2

=

8

2

+(−4)

2

=

64+16

=

80

Since the perimeter P of a triangle ABC is AB+BC+CA, therefore,

P=

61

+

85

+

80

Hence, the perimeter of the triangle is (

61

+

85

+

80

) units.

Answer 2

Given:

• A triangle formed by the coordinates (-2, 4), (4, 6) and (6, 3).

To find:

• The perimeter.

Answer:

To do so, we'll have to find the length of each side [distance between two points.]

Distance formula:

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

Let's first find the distance between the points (-2, 4) and (4, 6).

From those points,

$\tt x_1\ =\ -2\\\\x_2\ =\ 4\\\\y_1\ =\ 4\\\\y_2\ =\ 6$

Using them in the formula,

$\tt Distance\ =\ \sqrt{\Bigg(4\ +\ 2\Bigg)^2\ +\ \Bigg(6\ -\ 4\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(6\Bigg)^2\ +\ \Bigg(2\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(36\ +\ 4\Bigg)}\\\\\\\\\bf Distance\ =\ \sqrt{40}\ =\ 6.32\ units\ [approx.]$

Now, let's find the distance between the points (4, 6) and (6, 3).

From those points,

$\tt x_1\ =\ 4\\\\x_2\ =\ 6\\\\y_1\ =\ 6\\\\y_2\ =\ 3$

Using them in the formula,

$\tt Distance\ = \sqrt{\Bigg(6\ -\ 4\Bigg)^2\ +\ \Bigg(3\ -\ 6\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(2\Bigg)^2\ +\ \Bigg(-3\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(4\ +\ 9\Bigg)^2}\\\\\\\\\bf Distance\ =\ \sqrt{13}\ =\ 3.6\ units\ [approx.]$

Now, the distance between the points (6, 3) and (-2, 4).

From those points,

$\tt x_1\ =\ 6\\\\x_2\ =\ -2\\\\y_1\ =\ 3\\\\y_2\ =\ 4$

Using them in the formula,

$\tt Distance\ =\ \sqrt{\Bigg(-2\ -\ 6\Bigg)^2\ +\ \Bigg(4\ -\ 3\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(-8\Bigg)^2\ +\ \Bigg(1\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(64\ +\ 1\Bigg)}\\\\\\\\\bf Distance\ =\ \sqrt{65}\ =\ 8.06\ units\ [approx.]$

Now, to find the perimeter.

Perimeter = Sum of the lengths of all sides

Perimeter = 6.32 + 3.6 + 8.06

Perimeter = 17.98

Therefore, the perimeter of the triangle formed by the points (-2, 4), (4, 6) and (6, 3) = 17.98 units.