# Distance Formula Worksheet

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.

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.

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.

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.

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.

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.

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.

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).

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.

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.