﻿ Distance Formula Worksheet | Problems & Solutions
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Distance Formula Worksheet
• Page 1
1.
Find the distance between the points (2, 3) and (5, 7). a. $\sqrt{7}$ units b. 6 units c. 25 units d. 17 units e. 5 units

#### Solution:

The distance between the points A(x1, y1) and B(x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Let A(x1, y1) = (2, 3) and B(x2, y2) = (5, 7)

AB = (5-2)2+(7-3)2
[Distance between two points.]

= (3)2+(4)2 = 9+16
[Simplify.]

= 25 = 5

So, the distance between the points is 5 units.

Correct answer : (5)
2.
Find the distance between two points A and B shown in the graph.  a. 4 units b. 34 units c. $\sqrt{2}$ units d. 5.83 units e. 6 units

#### Solution:

The distance between the points (x1, y1) and (x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Co - ordinates of A are (-1, 2) and B are (2, -3).
[Write the coordinates from the graph.]

Distance between the points AB = (2-(-1))2+(-3-2)2
[(x1, y1) = (-1, 2) and (x2, y2) = (2, - 3).]

= (3)2+(-5)2 = 9+25
[Simplify.]

= 34 5.83

So, the distance between two points A and B is approximately 5.83 units.

Correct answer : (4)
3.
The point (2, $p$) lies in the first quadrant, at a distance of 3 units form (2, 2). Find the possible value of $p$. a. 5 b. 1 c. 2 d. 3

#### Solution:

The distance between two points (x1, y1) and (x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Let (x1, y1) = (2, 2) and (x2, y2) = (2, p).

3 = (2-2)2+(p-2)2
[The distance between the points is 3 units.]

Þ 3 = 0+(p-2)2

± 3 = p - 2, so, p = 5 or p = - 1
[Solve for p.]

Since the point (2, p) lies in the first quadratnt, p must be 5.

So, the possible value of p is 5.

Correct answer : (1)
4.
Find the distance between the two points A and B shown in the graph.  a. 10 units b. 0 units c. $\sqrt{28}$ units d. 14 units e. 48 units

#### Solution:

The distance between the points A(x1, y1) and B(x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Co - ordiantes of A are (3, 4) and B are (- 3, - 4).

Distance between A and B = (-3-3)2+(-4-4)2
[(x1, y1) = (3, 4) and (x2, y2) = (- 3, - 4).]

= (-6)2+(-8)2

= 36+64 = 100 = 10
[Simplify.]

The distance between the points A and B is 10 units.

Correct answer : (1)
5.
Find the distance between the points (2.8, 3.3) and (-7.4, 1). a. 10 units b. 10.45 units c. 5.14 units d. 11 units e. 4.77 units

#### Solution:

The distance between the two points (x1, y1) and (x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Let (x1, y1) = (2.8, 3.3) and (x2, y2) = (-7.4, 1).

Distance between the points, d = (-7.4-2.8)2+(1-3.3)2
[Use distance formula.]

= (-10.2)2+(-2.3)2

= 104.04+5.29 10.45
[Simplify.]

The distance between the points is approximately 10.45 units.

Correct answer : (2)
6.
A line segment AB is of length 6 units, whose end points are A(0, 3) and B($a$, 3). Find the possible values of $a$. a. - 3 and 3 b. $\sqrt{6}$ c. - 6 and 6 d. 6 only e. - 36 and 36

#### Solution:

The distance between the points (x1, y1) and (x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Let A(x1, y1) = (0, 3) and B(x2, y2) = (a, 3)

6 = (a-0)2+(3-3)2
[The distance between the points is 6 units.]

6 = a2+0

a = ± 6
[Solve for a.]

So, the possible values of a are - 6 and 6.

Correct answer : (3)
7.
The distance from origin to a point A(3$q$, 5$q$) is 7$\sqrt{34}$ units. Find the possible values of $q$. a. $\sqrt{34}$ b. 7 only c. - 7 and 7 d. - 49 and 49

#### Solution:

The distance between two points (x1, y1) and (x2, y2) in the coordinate plane is, d = (x2-x1)2+(y2-y1)2.

The distance from origin to a point A(3q, 5q) is 734.

734 = (3q-0)2+(5q-0)2
[(x1, y1) = (0, 0) and (x2, y2) = (3q, 5q).]

734 = 9q2+25q2 = 34q2

49 × 34 = 34q2
[Square the both sides of the equation.]

49 = q2
[Cancel the common factors.]

± 7 = q

So, the possible values of q are - 7 and 7.

Correct answer : (4)
8.
Choose the coordinates of the point P, which is 10 units from the point Q(3, 4). a. (- 3, 4) b. (5, 5) c. (4, 0) d. (- 3, - 4)

#### Solution:

The distance between the two points in a co-ordinate plane is, d = (x2-x1)2+(y2-y1)2.

The distance between Q and P is 10 units.

Distance between (3, 4) and (5, 5) = (5-3)2+(5-4)2
[(x1, y1) = (3, 4) and (x2, y2) = (5, 5).]

= 4+1 = 5 ≠ 10
[Simplify.]

Distance between (3, 4) and (- 3, - 4) = (-3-3)2+(-4-4)2
[(x1, y1) = (3, 4) and (x2, y2) = (- 3, - 4).]

= 36+64 = 100 = 10
[Simplify.]

Distance between (3, 4) and (- 3, 4) = (-3-3)2+(4-4)2
[(x1, y1) = (3, 4) and (x2, y2) = (- 3, 4).]

= 36 = 6 ≠ 10
[Simplify.]

Distance between (3, 4) and (4, 0) = (4-3)2+(0-4)2
[(x1, y1) = (3, 4) and (x2, y2) = (4, 0).]

= 17 ≠ 10
[Simplify.]

The point (- 3, - 4) is at a distance of 10 units from Q.

So, the coordinates of P are (- 3, - 4).

Correct answer : (4)
9.
The point B is the reflection of the point A(4, - 7) through the origin. Find the length of the line segment AB. a. 0 units b. 16.12 units c. 15 units d. 17 units e. 11 units

#### Solution:

The point B is the reflection of the point A(4, - 7) through the origin.

So, the coordinates of B are (- 4, 7).

The distance between two points (x1, y1) and (x2, y2) in a coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Let (x1, y1) = (4, - 7) and (x2, y2) = (- 4, 7).

Distance between A and B = (-4-4)2+(7+7)2
[Use distance formula.]

= (-8)2+(14)2

= 64+196 = 260 16.12
[Simplify.]

The length of the line segment AB is approximately 16.12 units.

Correct answer : (2)
10.
The diagonals of the rectangle PQRS intersect each other at the point O, as shown in the figure. Find the length of the line segment OP.  a. 3.6 units b. $\sqrt{5}$ units c. 13 units d. 3 units e. 4 units

#### Solution:

From the graph, the coordinates of the points O and P are (5, 5) and (2, 3).

The distance between the two points (x1, y1) and (x2, y2) in a coordinate plane is, d = (x2-x1)2+(y2-y1)2.

Distance between the points O and P = (2-5)2+(3-5)2
[Take (x1, y1) = (5, 5) and (x2, y2) = (2, 3).]

= (-3)2+(-2)2 = 9+4
[Simplify.]

= 13 3.6
[Simplify.]

So, the length of the line segment OP is approximately 3.6 units.

Correct answer : (1)

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