﻿ Distance and Midpoint Formula Worksheet - Page 2 | Problems & Solutions

# Distance and Midpoint Formula Worksheet - Page 2

Distance and Midpoint Formula Worksheet
• Page 2
11.
Find the distance between A(4, - 5) and B(5, - 8).
 a. 4.16 units b. 3.16 units c. 6.16 units d. 5.16 units

#### Solution:

Distance between two points d = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

d = (5 - 4)2+((- 8) - (- 5))2
[Replace (x2, y2) with (5, - 8) and (x1, y1) with (4, - 5).]

d = (1)2+(- 3)2
[Subtract.]

d = 10
[Simplify.]

d 3.16
[Find the positive square root.]

The distance between, A and B is 3.16 units.

12.
The distance between two points ($x$1, $y$1) and ($x$2, $y$2) is _____.
 a. $\sqrt{\left({x}_{2}-{x}_{1}\right)+\left({y}_{2}-{y}_{1}\right)}$ b. $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$ c. $\frac{{x}_{1}+{y}_{1}+{x}_{2}+{y}_{2}}{2}$ d. | (

#### Solution:

The distance between two points (x1, y1) and (x2, y2) is (x2-x1)2+(y2-y1)2.

13.
Find the distance between A(- 5, 5) and B(- 7, 3) and round the solution to the nearest tenth.
 a. 5.8 units b. 3.8 units c. 2.8 units d. 4.8 units

#### Solution:

The distance between (x1, y1) and (x2, y2) = d = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

d = (- 7 - (- 5))2 +(3 - 5)2
[Replace (x1, y1) with (- 5, 5) and (x2, y2) with (- 7, 3).]

d = (- 2)2+(- 2)2
[Subtract.]

d = 4 + 4 = 8

d 2.828

The distance between A and B is 2.8 units.
[Round the distance to the nearest tenth.]

14.
What is the distance between ($x$1, $y$1) and origin?
 a. $\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$ b. $\sqrt{{{x}_{1}}^{2}-{{y}_{1}}^{2}}$ c. d.

#### Solution:

The coordinates of origin are (0, 0).

Distance d = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

d = (0-x1)2+(0-y1)2
[Replace (x2, y2) with (0, 0).]

d = x12+y12
[Subtract.]

Distance between (x1, y1) and the origin is x12+y12.

15.
Which of the following ordered pairs is at a distance of 2 units from (5, 3)?
 a. (7, 4) b. (10, 6) c. (8, 7) d. (7, 3)

#### Solution:

Distance, d = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

Consider choice A, the distance between (5, 3) and (7, 4).

d = (7 - 5)2+(4 - 3)2 = 5 = 2, which is not true.
[Replace (x1, y1) with (5, 3) and (x2, y2) with (7, 4) and simplify.]

Consider choice B, the distance between (5, 3) and (10, 6).

d = (10 - 5)2+(6 - 3)2 = 34 = 2, which is not true.
[Replace (x1, y1) with (5, 3) and (x2, y2) with (10, 6).]

Consider choice C, the distance between (5, 3) and (8, 7).

d = (8 - 5)2+(7 - 3)2 = 25 = 2, which is not true.
[Replace (x1, y1) with (5, 3) and (x2, y2) with (8, 7).]

Consider choice D, the distance between (5, 3) and (7, 3).

d = (7 - 5)2+(3 - 3)2 = 4 = 2, which is true.
[Replace (x1, y1) with (5, 3) and (x2, y2) with (7, 3).]

Therefore, the point (7, 3) is at a distance of 2 units from (5, 3).

16.
Find the perimeter of the triangle ABC shown in the graph.

 a. 26.08 units b. 30 units c. 24 units d. 20.92 units

#### Solution:

From the figure, A is at (6, - 1), B is at (2, 5) and C is at (- 1, - 1).

Distance d = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

AB = (2-6)2+(5-(- 1))2
[Replace (x1, y1) with (6, - 1) and (x2, y2) with (2, 5).]

AB = (- 4)2+62

= 16+36 = 52
[Simplify.]

BC = (- 1-2)2+(- 1-5)2
[Replace (x1, y1) with (2, 5) and (x2, y2) with (- 1, - 1).]

BC = (- 3)2+(- 6)2

= 9+36 = 45
[Simplify.]

CA = (6-(- 1))2+((- 1)-(- 1))2
[Replace (x1, y1) with (- 1, - 1) and (x2, y2) with (6, - 1).]

CA = 72+02

= 49+0 = 49 = 7
[Simplify.]

Perimeter of ΔABC = AB + BC + CA

= 52+45 + 7
[Substitute the values.]

= 7.21 + 6.71 + 7 = 20.92

The perimeter of ΔABC = 20.92 units.

17.
Find the perimeter of quadrilateral ABCD shown in the graph.

 a. 13.24 units b. 22.62 units c. 24.78 units d. 19.98 units

#### Solution:

From the graph, the co-ordinates of A are (7, 7), B are (11, 3), C are (7, - 1) and D are (3, 3).

Distance d = (x2-x1)2+(y2-y1)2
[Use the distance formula.]

AB = (11 - 7)2+(3 - 7)2
[Replace (x1, y1) with (7, 7) and (x2, y2) with (11, 3).]

AB = 42+(- 4)2=16+16=32
[Simplify.]

BC = (7 - 11)2+(- 1 - 3)2
[Replace (x1, y1) with (11, 3) and (x2, y2) with (7, - 1).]

BC = (- 4)2+(- 4)2=16+16=32
[Simplify.]

CD = (3 - 7)2+(3 - (- 1))2
[Replace (x1, y1) with (7, - 1) and (x2, y2) with (3, 3).]

CD = (- 4)2+42=16+16=32

DA = (7 - 3)2+(7 - 3)2=16+16=32
[Replace (x1, y1) with (3, 3) and (x2, y2) with (7, 7).]

Perimeter of the quadrilateral ABCD = AB + BC + CD + DA

= 32+32+32+32
[Substitute the values.]

= 432 = 22.62

The perimeter of quadrilateral ABCD is 22.62 units.

18.
If AB is a line segment and P is the midpoint, then which of the following is true?
 a. BP = AB b. AP = AB c. AB = $\frac{AP}{2}$ d. AP = BP

#### Solution:

AB is the line segment and P is the midpoint.

P divides the line segment AB into two equal parts AP and BP.

AP = BP = AB / 2 .

19.
Midpoint divides the line-segment into _____.
 a. Two unequal parts b. Four equal parts c. Two equal parts d. Three equal parts

#### Solution:

Midpoint divides the line-segment into two equal parts.

20.
Find the midpoint of the line-segment AB in the graph.

 a. ($\frac{5}{2}$ , $\frac{1}{2}$) b. (2, 3) c. (4, 5) d. ($\frac{1}{2}$, $\frac{5}{2}$)

#### Solution:

From the graph, the coordinates of A are (- 3, - 1) and B are (4, 6).

Midpoint = (x1+x22, y1+y22)
[Use the midpoint formula.]

Midpoint of AB = (- 3 + 42, - 1 + 62)
[Replace (x1, y1) with (- 3, - 1) and (x2, y2) with (4, 6).]

Midpoint = (1 / 2 , 5 / 2)
[Simplify.]

Midpoint of the line segment AB = (1 / 2 , 5 / 2)