Question

To the nearest tenth, find the perimeter of triangle ABC with vertices A(3, 2), B(-2, 3), and C(2, 6)

14.2

B.
17.5

C.
18.2

D.
18.7

Find the midpoint between (-4, -2) and (-10, -4)

A.
(-16, -6)

B.
(-7, -3)

C.
(-3, -7)

D.
(3, 1)

Answer 1

The correct option is A). 14.2

The correct option is B). (-7,-3)

Explanation:

QA). The perimeter of triangle ABC with vertices A(3, 2), B(-2, 3), and C(2, 6)

Ans.

The distance between two points A(X1,Y1) and B(Y1,Y2) is

=
`\sqrt{(X1-X2)^(2)+(Y1-Y2)^(2)`

Now,

The distance between two points A(3, 2) and B(-2, 3) is

=
`\sqrt{(X1-X2)^(2)+(Y1-Y2)^(2)`

=
`\sqrt{(3-(-2))^(2)+(2-3)^(2)`

=
`√(25 +1)`

=
`√(26)`

The distance between two points B(-2, 3) and C(2, 6) is

=
`\sqrt{(X1-X2)^(2)+(Y1-Y2)^(2)`

=
`\sqrt{((-2)-(2))^(2)+(3-6)^(2)`

=
`√(16 +9)`

=
`√(25)`

=5

The distance between two points A(3, 2) and C(2, 6) is

=
`\sqrt{(X1-X2)^(2)+(Y1-Y2)^(2)`

=
`\sqrt{((3)-(2))^(2)+(2-6)^(2)`

=
`√(1 +16)`

=
`√(17)`

.

The perimeter of Triangle ABC is AB+BC+AC

=
`√(26)`+5+
`√(17)`

=5.099+5+4.1231

=14.2221

The correct option is A). 14.2

QB). Find the midpoint between (-4, -2) and (-10, -4)

Ans.

The midpoint of two points A(X1,Y1) and B(Y1,Y2) is

=
`(((X1+X2))/(2),((Y1+Y2))/(2))`

The midpoint of two points (-4, -2) and (-10, -4) IS

=
`(((X1+X2))/(2),((Y1+Y2))/(2))`

=
`((((-4)+(-10)))/(2),(((-2)+(-4)))/(2))`

=
`(((-14))/(2),((-6))/(2))`

=(-7,-3)

The correct option is B). (-7,-3)