In the previous section we saw how to calculate the distance between two points for the following four cases:
Case 1: The two given points lie on the x-axis
Case 2:The two given points lie on a line parallel to the x-axis
Case 3: The two given points lie on the y-axis
Case 4: The two given points lie on a line parallel to the y-axis
• In this section we will see Case 5:
The two given points can lie any where in the Cartesian plane
1. Fig.31.37 below shows two such points P and Q.
• P has coordinates: (x1,y1)
• Q has coordinates: (x2,y2)
2. We want the distance between P and Q
• For that, draw a horizontal dashed line through P
• Draw a vertical dashed line through Q
3. These lines intersect at R
• We want the coordinates of R
• The vertical dashed line passes through Q (x2,y2)
♦ So any point on this vertical dashed line will have x coordinate as x2
• The horizontal dashed line passes through P (x1,y1)
♦ So any point on this vertical dashed line will have y coordinate as y1
• So the coordinates of R are: (x2,y1)
4. Now, P and R lies on a line parallel to the x-axis.
• So this is Case 2. We can easily find the distance PR
• PR = |(x2-x1)|
5. Similarly, Q and R lies on a line parallel to the y-axis.
• So this is Case 4. We can easily find the distance QR
• QR = |(y2-y1)|
6. Since the dashed lines are horizontal and vertical, the triangle PRQ is right angled
• We can apply Pythagoras theorem:
PQ2 = PR2+ QR 2 ⟹ PQ2 = [|(x2-x1)|]2+[|(y2-y1)|]2
7. But [|(x2-x1)|]2 = (x2-x1)2 (∵ square of a number = square of it's absolute value)
• Also [|(y2-y1)|]2 = (y2-y1)2
8. So from (6) we get:
PQ2 = [(x2-x1)2 + (y2-y1)2]
From this we get:
• This formula is easy to remember because it involves:
♦ Difference between x coordinates
♦ Difference between y coordinates
• Since the differences are squared, we need not worry about the order. That is., we can take either P or Q as the first point.
■ The following fig.31.38 shows three examples.
• The reader may check them by doing the necessary calculations
We can add this also to our list as 'Case 5'. So the final list is:
Calculation of distance between two given points
Case 1: The two given points lie on the x-axis
• When the points lie on the x-axis, their general form can be written as: (x1,0) and (x2,0)
♦ Distance between them = |(x2-x1)|
Case 2: The two given points lie on a line parallel to the x-axis
• When the points lie on a line parallel to the x-axis, their general form can be written as: (x1,y) and (x2,y)
['y' is same for both points because, all points on a line parallel to the x-axis will have the same y coordinate]
♦ Distance between them = |(x2-x1)|
Case 3: The two given points lie on the y-axis
• When the points lie on the y-axis, their general form can be written as: (0,y1) and (0,y2)
♦ Distance between them = |(y2-y1)|
Case 4: The two given points lie on a line parallel to the y-axis
• When the points lie on a line parallel to the y-axis, their general form can be written as: (0,y1) and (0,y2)
['x' is same for both points because, all points on a line parallel to the y-axis will have the same x coordinate]
♦ Distance between them = |(y2-y1)|
Case 5: The two given points have different x and y coordinates
• When the points have different x and y coordinates, their general form can be written as: (x1,y1) and (x2,y2)
♦ Distance between them = √[(x2-x1)2 + (y2-y1)2]
This formula is known as the Distance formula.
■ 'Case 5' can be used for 'any two points'. So the first four cases would be included in Case 5. Let us prove it:
1. What if the two given points are on the x-axis?
• Then P will be (x1,0) and Q will be (x2,0)
• PQ = √[(x2-x1)2 + (y2-y1)2] = √[(x2-x1)2 + (0-0)2] = √[(x2-x1)2 ] = |(x2-x1)|.
• So this is same as Case 1
2. What if the two given points are on a line parallel to the x-axis?
• Then P will be (x1,y) and Q will be (x2,y)
• PQ = √[(x2-x1)2 + (y-y)2] = √[(x2-x1)2 + (0)2] = √[(x2-x1)2 ] = |(x2-x1)|.
• So this is same as Case 2
3. What if the two given points are on the y-axis?
• Then P will be (0,y1) and Q will be (0,y2)
• PQ = √[(x2-x1)2 + (y2-y1)2] = √[(0-0)2 + (y2-y1)2] = √[(y2-y1)2 ] = |(y2-y1)|.
• So this is same as Case 3
4. What if the two given points are on a line parallel to the y-axis?
• Then P will be (x,y1) and Q will be (x,y2)
• PQ = √[(x2-x1)2 + (y2-y1)2] = √[(x-x)2 + (y2-y1)2] = √[(y2-y1)2 ] = |(y2-y1)|.
• So this is same as Case 4
■ Thus we find that, the first four cases are included in Case 5
Now we will see a very interesting result:
Given any two points, we are now able to find the distance between them. What if one of the two given points is the origin O?
• Then we have:
♦ First point (x1,y1) = (0,0)
♦ Second point (x2,y2) = (x,y)
• So OP = √[(x2-x1)2 + (y2-y1)2] = √[(x-0)2 + (y-0)2] = √[x2 + y2]
■ Thus we can write:
Distance of any point P [with coordinates (x,y)] from the origin O is : √[x2 + y2]
Now we will see some solved examples
Solved example 31.16
Find whether the points with coordinates (-1,2), (3,5), (9,-3) lie on a line
Solution:
1. Consider any three points '1', '2' and '3. Let us first assume that they lie on a line. Then we will get a line as shown in fig.31.39 below:
2. From the fig., we can write the following:
• Among the three points, those which are farthest apart are 1 and 3
• So 'Distance from 1 to 3' must be equal to the sum of the following:
♦ Distance from 1 to 2
♦ Distance from 2 to 3
• So we need three distances:
♦ Distance from 1 to 3
♦ Distance from 1 to 2
♦ Distance from 2 to 3
3. In short, we need all the possible distances between the three given points. So we will calculate them:
• Let the given points be A, B and C
♦ Let coordinates of A be (-1,2)
♦ Let coordinates of B be (3,5)
♦ Let coordinates of C be (9,-3)
4. So we get:
AB = √[(x2-x1)2 + (y2-y1)2] = √[(3-(-1))2 + (5-2)2] = √[42 + 32] = √[16 + 9] = √[25] = 5
BC = √[(x2-x1)2 + (y2-y1)2] = √[(9-3)2 + (-3-5)2] = √[62 + (-8)2] = √[36 + 64] = √[100] = 10
AC = √[(x2-x1)2 + (y2-y1)2] = √[(9-(-1))2 + (-3-2)2] = √[102 + (-5)2] = √[100 + 25]
= √[125] = √[25×5] = √[25] × √[5] = 5√5 = 5 2.23 = 11.18
5. The longest length is AC
• But (AB + BC) = (5+10) = 15 cm. It is not equal to AC
• So A, B and C do not lie on a line.
Solved example 31.17
Do the points (3,2), (-2,-3) and (2,3) form a triangle? If they do form a triangle, what type of triangle will it be?
Solution:
1. Three points will always form a triangle unless they lie on the same line.
2. If they do not lie on the same line, they will satisfy one condition:
• The sum of any two sides will be greater than the third side (see details here)
3. So we will find the sides:
(a) Side 1 formed by the first and second points
= √[(3-(-2))2 + (2-(-3))2] = √[52 + 52] = √[25 + 25] = √[50] = 7.071
(b) Side 2 formed by the first and last points
= √[(3-2)2 + (2-3)2] = √[12 + (-1)2] = √[1 + 1] = √[2] = 1.4142
(c) Side 3 formed by the second and last points
= √[(-2-2)2 + (-3-3)2] = √[(-4)2 + (-6)2] = √[16 + 36] = √[52] = 7.211
4. Now we can check the conditions:
• (side 1 + side 2) = 7.071 + 1.4142 = 8.4852
This is greater than side 3
• (side 2 + side 3) = 1.4142 + 7.211 = 8.6252
This is greater than side 1
• (side 1 + side 3) = 7.071 + 7.211 = 14.282
This is greater than side 2
5. So all three conditions are satisfied. The given points will definitely form a triangle. It is shown in the fig.31.40 below:
6. Now we have to find what type of triangle is this.
(a) Consider (side 1)2+(side 2)2. We get: (√[50])2+(√[2])2 = 50 + 2 = 52
(b) Now consider (side 3)2. We get: (√[52])2 = 52
(c) We get: (side 1)2+(side 2)2 = (side 3)2.
• So Pythagoras theorem is applicable to this triangle. Thus it is a right triangle
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.40 is given for a better understanding of the problem
Solved example 31.18
Show that the points (1,7), (4,2), (-1,-1) and (-4,4) are the vertices of a square
Solution:
1. First, we will find the sides:
(a) Side 1 formed by the first and second points
= √[(1-4)2 + (7-2)2] = √[9 + 25] = √[34]
(b) Side 2 formed by the second and third points
= √[(4-(-1))2 + (2-(-1))2] = √[25 + 9] = √[34]
(c) Side 3 formed by the third and fourth points
= √[(-1-(-4))2 + (-1-4)2] = √[9 + 25] = √[34]
(d) Side 4 formed by the first and fourth points
= √[(1-(-4))2 + (7-4)2] = √[25 + 9] = √[34]
(e) Side 5 formed by the first and third points
= √[(1-(-1))2 + (7-(-1))2] = √[4 + 64] = √[68]
(f) Side 6 formed by the second and fourth points
= √[(4-(-4))2 + (2-4)2] = √[64 + 4] = √[68]
• Note that, these six are the only possible combinations
2. So four sides are equal
• Also two diagonals are equal.
■ This proves that the given points form a square. This is shown in the fig.31.41 below:
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.41 is given for a better understanding of the problem
Case 1: The two given points lie on the x-axis
Case 2:The two given points lie on a line parallel to the x-axis
Case 3: The two given points lie on the y-axis
Case 4: The two given points lie on a line parallel to the y-axis
• In this section we will see Case 5:
The two given points can lie any where in the Cartesian plane
1. Fig.31.37 below shows two such points P and Q.
 |
| Fig.31.37 |
• Q has coordinates: (x2,y2)
2. We want the distance between P and Q
• For that, draw a horizontal dashed line through P
• Draw a vertical dashed line through Q
3. These lines intersect at R
• We want the coordinates of R
• The vertical dashed line passes through Q (x2,y2)
♦ So any point on this vertical dashed line will have x coordinate as x2
• The horizontal dashed line passes through P (x1,y1)
♦ So any point on this vertical dashed line will have y coordinate as y1
• So the coordinates of R are: (x2,y1)
4. Now, P and R lies on a line parallel to the x-axis.
• So this is Case 2. We can easily find the distance PR
• PR = |(x2-x1)|
5. Similarly, Q and R lies on a line parallel to the y-axis.
• So this is Case 4. We can easily find the distance QR
• QR = |(y2-y1)|
6. Since the dashed lines are horizontal and vertical, the triangle PRQ is right angled
• We can apply Pythagoras theorem:
PQ2 = PR2
7. But [|(x2-x1)|]2 = (x2-x1)2 (∵ square of a number = square of it's absolute value)
• Also [|(y2-y1)|]2 = (y2-y1)2
8. So from (6) we get:
PQ2 = [(x2-x1)2 + (y2-y1)2]
From this we get:
• This formula is easy to remember because it involves:
♦ Difference between x coordinates
♦ Difference between y coordinates
• Since the differences are squared, we need not worry about the order. That is., we can take either P or Q as the first point.
■ The following fig.31.38 shows three examples.
 |
| Fig.31.38 |
We can add this also to our list as 'Case 5'. So the final list is:
Calculation of distance between two given points
Case 1: The two given points lie on the x-axis
• When the points lie on the x-axis, their general form can be written as: (x1,0) and (x2,0)
♦ Distance between them = |(x2-x1)|
Case 2: The two given points lie on a line parallel to the x-axis
• When the points lie on a line parallel to the x-axis, their general form can be written as: (x1,y) and (x2,y)
['y' is same for both points because, all points on a line parallel to the x-axis will have the same y coordinate]
♦ Distance between them = |(x2-x1)|
Case 3: The two given points lie on the y-axis
• When the points lie on the y-axis, their general form can be written as: (0,y1) and (0,y2)
♦ Distance between them = |(y2-y1)|
Case 4: The two given points lie on a line parallel to the y-axis
• When the points lie on a line parallel to the y-axis, their general form can be written as: (0,y1) and (0,y2)
['x' is same for both points because, all points on a line parallel to the y-axis will have the same x coordinate]
♦ Distance between them = |(y2-y1)|
Case 5: The two given points have different x and y coordinates
• When the points have different x and y coordinates, their general form can be written as: (x1,y1) and (x2,y2)
♦ Distance between them = √[(x2-x1)2 + (y2-y1)2]
This formula is known as the Distance formula.
■ 'Case 5' can be used for 'any two points'. So the first four cases would be included in Case 5. Let us prove it:
1. What if the two given points are on the x-axis?
• Then P will be (x1,0) and Q will be (x2,0)
• PQ = √[(x2-x1)2 + (y2-y1)2] = √[(x2-x1)2 + (0-0)2] = √[(x2-x1)2 ] = |(x2-x1)|.
• So this is same as Case 1
2. What if the two given points are on a line parallel to the x-axis?
• Then P will be (x1,y) and Q will be (x2,y)
• PQ = √[(x2-x1)2 + (y-y)2] = √[(x2-x1)2 + (0)2] = √[(x2-x1)2 ] = |(x2-x1)|.
• So this is same as Case 2
3. What if the two given points are on the y-axis?
• Then P will be (0,y1) and Q will be (0,y2)
• PQ = √[(x2-x1)2 + (y2-y1)2] = √[(0-0)2 + (y2-y1)2] = √[(y2-y1)2 ] = |(y2-y1)|.
• So this is same as Case 3
4. What if the two given points are on a line parallel to the y-axis?
• Then P will be (x,y1) and Q will be (x,y2)
• PQ = √[(x2-x1)2 + (y2-y1)2] = √[(x-x)2 + (y2-y1)2] = √[(y2-y1)2 ] = |(y2-y1)|.
• So this is same as Case 4
■ Thus we find that, the first four cases are included in Case 5
Now we will see a very interesting result:
Given any two points, we are now able to find the distance between them. What if one of the two given points is the origin O?
• Then we have:
♦ First point (x1,y1) = (0,0)
♦ Second point (x2,y2) = (x,y)
• So OP = √[(x2-x1)2 + (y2-y1)2] = √[(x-0)2 + (y-0)2] = √[x2 + y2]
■ Thus we can write:
Distance of any point P [with coordinates (x,y)] from the origin O is : √[x2 + y2]
Now we will see some solved examples
Solved example 31.16
Find whether the points with coordinates (-1,2), (3,5), (9,-3) lie on a line
Solution:
1. Consider any three points '1', '2' and '3. Let us first assume that they lie on a line. Then we will get a line as shown in fig.31.39 below:
 |
| Fig.31.39 |
• Among the three points, those which are farthest apart are 1 and 3
• So 'Distance from 1 to 3' must be equal to the sum of the following:
♦ Distance from 1 to 2
♦ Distance from 2 to 3
• So we need three distances:
♦ Distance from 1 to 3
♦ Distance from 1 to 2
♦ Distance from 2 to 3
3. In short, we need all the possible distances between the three given points. So we will calculate them:
• Let the given points be A, B and C
♦ Let coordinates of A be (-1,2)
♦ Let coordinates of B be (3,5)
♦ Let coordinates of C be (9,-3)
4. So we get:
AB = √[(x2-x1)2 + (y2-y1)2] = √[(3-(-1))2 + (5-2)2] = √[42 + 32] = √[16 + 9] = √[25] = 5
BC = √[(x2-x1)2 + (y2-y1)2] = √[(9-3)2 + (-3-5)2] = √[62 + (-8)2] = √[36 + 64] = √[100] = 10
AC = √[(x2-x1)2 + (y2-y1)2] = √[(9-(-1))2 + (-3-2)2] = √[102 + (-5)2] = √[100 + 25]
= √[125] = √[25×5] = √[25] × √[5] = 5√5 = 5 2.23 = 11.18
5. The longest length is AC
• But (AB + BC) = (5+10) = 15 cm. It is not equal to AC
• So A, B and C do not lie on a line.
Solved example 31.17
Do the points (3,2), (-2,-3) and (2,3) form a triangle? If they do form a triangle, what type of triangle will it be?
Solution:
1. Three points will always form a triangle unless they lie on the same line.
2. If they do not lie on the same line, they will satisfy one condition:
• The sum of any two sides will be greater than the third side (see details here)
3. So we will find the sides:
(a) Side 1 formed by the first and second points
= √[(3-(-2))2 + (2-(-3))2] = √[52 + 52] = √[25 + 25] = √[50] = 7.071
(b) Side 2 formed by the first and last points
= √[(3-2)2 + (2-3)2] = √[12 + (-1)2] = √[1 + 1] = √[2] = 1.4142
(c) Side 3 formed by the second and last points
= √[(-2-2)2 + (-3-3)2] = √[(-4)2 + (-6)2] = √[16 + 36] = √[52] = 7.211
4. Now we can check the conditions:
• (side 1 + side 2) = 7.071 + 1.4142 = 8.4852
This is greater than side 3
• (side 2 + side 3) = 1.4142 + 7.211 = 8.6252
This is greater than side 1
• (side 1 + side 3) = 7.071 + 7.211 = 14.282
This is greater than side 2
5. So all three conditions are satisfied. The given points will definitely form a triangle. It is shown in the fig.31.40 below:
 |
| Fig.31.40 |
(a) Consider (side 1)2+(side 2)2. We get: (√[50])2+(√[2])2 = 50 + 2 = 52
(b) Now consider (side 3)2. We get: (√[52])2 = 52
(c) We get: (side 1)2+(side 2)2 = (side 3)2.
• So Pythagoras theorem is applicable to this triangle. Thus it is a right triangle
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.40 is given for a better understanding of the problem
Solved example 31.18
Show that the points (1,7), (4,2), (-1,-1) and (-4,4) are the vertices of a square
Solution:
1. First, we will find the sides:
(a) Side 1 formed by the first and second points
= √[(1-4)2 + (7-2)2] = √[9 + 25] = √[34]
(b) Side 2 formed by the second and third points
= √[(4-(-1))2 + (2-(-1))2] = √[25 + 9] = √[34]
(c) Side 3 formed by the third and fourth points
= √[(-1-(-4))2 + (-1-4)2] = √[9 + 25] = √[34]
(d) Side 4 formed by the first and fourth points
= √[(1-(-4))2 + (7-4)2] = √[25 + 9] = √[34]
(e) Side 5 formed by the first and third points
= √[(1-(-1))2 + (7-(-1))2] = √[4 + 64] = √[68]
(f) Side 6 formed by the second and fourth points
= √[(4-(-4))2 + (2-4)2] = √[64 + 4] = √[68]
• Note that, these six are the only possible combinations
2. So four sides are equal
• Also two diagonals are equal.
■ This proves that the given points form a square. This is shown in the fig.31.41 below:
 |
| Fig.31.41 |
In the next section we will see a few more solved examples.
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