 Worksheet on Distance Formula is available on this page for free of cost. The distance formula will help you to find the distance between the two points on the plane. Students can find the distance between the points using the concept of the Pythagoras Theorem. Check out the problems given in the Distance Formula Worksheet PDF and enhance the math skills that help you to score good marks in the exams.

Distance Formula Problems Worksheet with Answers PDF

Example 1.
If the distance between the points (6, – 2) and (1, a) is 5, find the values of a.
Solution:
We know that, the distance between (x1, y1) and (x2, y2)
is √(x1−x2)²+(y1−y2)²
Here, the distance = 5,
x1 = 6,
x2 = 1,
y1 = -2 and
y2 = a
Therefore, 5 = √(6−1)²+(−2−a)²
Squaring on both sides
25 = 25 + (2 + a)²
(2 + a)² = 25 – 25
(2 + a)² = 0
Taking square root, 2 + a = 0
a = (2,0)

Example 2.
Find the value of y for which the distance between the points P(2,3) and Q(10,y) is 10 units.
Solution:
Given that
Distance between (2,3) and (10,y) is 10
Using the distance formula
PQ = √(10 – 2)² + (y – 3)²
PQ = √8² + (y -3)²
Since PQ = 10
= √(8)² + (y-3)² = 10
Simplify the above equation and find the value of y
Squaring on both sides.
64 + (y – 3)² = 100
(y – 3)² = 36
y – 3 = ±6
y – 3 = 6
y = 9
Or
y – 3 = -6
y = 3
y = (9,3)

Example 3.
Find the distance between the points (0,0) and (4,8) can you now find the distance between the two towns A and B
Solution:
Given that
Let us consider town A at point (0,0).
Therefore
Town B will be at point (4,8)
Distance between the points (0,0) and (4,8) is
AB = √(4-0)² + (8-0)²
= √4² + 8²
= √80
The distance between A and B is √80.

Example 4.
Find the relation between x and y such that the points (x,y) are equidistant from the point (3,6) and (-3,4).
Solution:
Point (x,y) is equidistant from (3,6) and (-3,4).
We know that the distance formula is √(x2 – x1)² + (y2 – y1)²
= √(x – 3)² + (y – 6)² = (x + 3)2 + (y – 4)²
= √(x – 3)² + (y – 6)² = √(x + 3)² + (y – 4)²
Squaring on both sides
(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
x² + 9 -6x + y² + 36 – 12y = x² + 9 + 6x + y² + 16 – 8y
36 – 16 = 6x + 6x + 12y – 8y
20 = 12x + 4y
3x + y = 5
3x + y – 5 = 0

Example 5.
Find the distance between the two points (2, 3) and (1, 1).
Solution:
Given that
(x1,y1) = (2,3)
(x2, y2) = (1,1)
We know that
Distance formula = √(x2 – x1)² + (y2 – y1)²
= √(1−2)² + (1−3)²
= √(−1)²+(−1)²
= √1 + 1
= √2
Therefore the distance is √2 units.

Example 6.
Which point on the y-axis is equidistant from the points (5, 3) and (-5, 4)?
Solution:
Let the required point be on the y-axis (0, y).
Given (0, y) is equidistant from (5, 3) and (-5, 4)
Therefore, distance between (0, y) and (5, 3) = distance between (0, y) and (-5, 4)
√(5−0)²+(3−y)²= √(−5−0)²+(4−y)²
25 + 9 + y² – 6y = 25 + 16 + y² – 8y
34 + y² – 6y = 41 + y² – 8y
34 – 6y – 41 + 8y = 0
2y = -7
y = -7/2
y = -7/2
Therefore, the required point on the y-axis = (0, -7/2).

Example 7.
Find the values of a such that PQ = QR, where P, Q and R are the points whose coordinates are (4, – 1), (1, 3) and (a, 8) respectively.
Solution:
By using the distance formula
PQ = √(4−1)²+(−1−3)²
= √3²+(−4)²
= √9+16
= √25
QR = √(1−a)²+(3−8)²
= √(1−a)²+(−5)²
= √(1−a)²+25
Therefore, PQ = QR
√25= √(1−a)²+25
25 = (1 – a)²+ 25
(1 – a)² = 25 – 25
(1 – a)² = 0
1 – a = 0
-a = 0 – 1
a= 1
a = (1,0)

Example 8.
Find the distance between the two points (1,2) and (3,4).
Solution:
Given that
(x1,y1) = (1,1)
(x2, y2) = (3,4)
We know that
Distance formula = √(x2 – x1)² + (y2 – y1)²
= √(3−1)² + (4−1)²
= √(2)²+(3)²
= √4+ 9
= √13
Therefore the distance is √13 units.

Example 9.
If the distance between the points (4, – 2) and (1, a) is 6, find the values of a.
Solution:
We know that, the distance between (x1, y1) and (x2, y2)
is √(x1−x2)²+(y1−y2)²
Here, the distance = 6,
x1 = 4,
x2 = 1,
y1 = -3 and
y2 = a
Therefore, 6 = √(4−1)²+(−3−a)²
Squaring on both sides
36 = 9 + (3 + a)²
(3 + a)² = 36 – 9
(2 + a)² = 27
Taking square root, 2 + a = ±27
2 + a = -27
a = -29
or
2 + a = 27
a = 27 – 2
a = 25
a = (-29,25)