In the previous section we saw the Distance formula. We also saw some solved examples. In this section we will see a few more solved examples.
Solved example 31.19
Find whether the points with coordinates (3,1), (6,4), (8,6) 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.31.42 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 (3,1)
♦ Let coordinates of B be (6,4)
♦ Let coordinates of C be (8,6)
4. So we get:
• AB = √[(x2-x1)2 + (y2-y1)2] = √[(6-3)2 + (4-1)2] = √[32 + 32] = √[9 + 9]
= √[18] = √[9×2] = √[9] × √[2] = 3√2
• BC = √[(x2-x1)2 + (y2-y1)2] = √[(8-6)2 + (6-4)2] = √[22 + 22] = √[4 + 4]
= √[8] = √[4×2] = √[4] × √[2] = 2√2
• AC = √[(x2-x1)2 + (y2-y1)2] = √[(8-3)2 + (6-1)2] = √[52 + 52] = √[25 + 25]
= √[50] = √[25×2] = √[25] × √[2] = 5√2
5. The longest length is AC
• (AB + BC) = (3√2+2√2) = 5√2 cm. It is equal to AC
• So A, B and C lie on a line.
• This is shown in fig.31.43 below. The reader may draw it on a graph paper by taking appropriate scales along the x and y axes.
Solved example 31.20
Find the distance between the following pairs of points:
(i) (2,3), (4,1) (ii) (-5,7), (-1,3) (iii) (a,b), (-a,-b)
Solution:
We have the Distance formula:
■ Distance between any two points with coordinates (x1,x2) and (y1,y2) = √[(x2-x1)2 + (y2-y1)2]
(i) √[(4-2)2 + (1-3)2] = √[22 + (-2)2] = √[4 + 4]
= √[8] = √[4×2] = √[4] × √[2] = 2√2
(ii) √[(-1-(-5))2 + (3-7)2] = √[42 + (-4)2] = √[16 + 16]
= √[32] = √[16×2] = √[16] × √[2] = 4√2
(iii) √[(-a-a)2 + (-b-b)2] = √[(-2a)2 + (-2b)2] = √[4a2 + 4b2] = √[4(a2 + b2)] = 2√[(a2 + b2)]
Solved example 31.21
Find the distance between (0,0) and (36,15)
Solution:
We have the 'special case of the Distance formula' for distance of any point from the origin:
■ Distance of any point with coordinates (x,y) from the origin = √[x2 + y2]
So we get:
• Distance of the point [with coordinates (36,15)] from the origin
= √[362 + 152] = √[1296 + 225] = √[1521] = 39
Solved example 31.22
Find whether the points with coordinates (1,5), (2,3), (-2,-11) lie on a line
Solution:
1. If 3 points are to be collinear, the following condition should be satisfied:
• The Longest distance among them should be equal to the sum of the other two distances
2. So let us calculate the distances:
(i) Distance between first and second points
= √[(x2-x1)2 + (y2-y1)2] = √[(2-1)2 + (3-5)2] = √[12 + (-2)2] = √[1 + 4] = √[5] = 2.236
(ii) Distance between second and third points
= √[(-2-2)2 + (-11-3)2] = √[(-4)2 + (-14)2] = √[16 + 196] = √[212] = 14.560
(iii) Distance between first and third points
= √[(-2-1))2 + (-11-5)2] = √[(-3)2 + (-16)2] = √[9 + 196] = √[204] = 14.283
3. The longest distance is the one between the second and third points. It is 14.560 units
• The sum of other two = 14.283 + 2.236 = 16.519
• The above two quantities are not equal. So the given points are not collinear.
• The actual plot is shown in the fig.31.44
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.44 is given for a better understanding of the problem
Solved example 31.23
Check whether the points with coordinates (5,-2), (6,4), (7,-2) are the vertices of an isosceles triangle
Solution:
1. When 3 points are given, there will be three possible distances between them
2. So let us calculate the distances:
(i) Distance between first and second points
= √[(x2-x1)2 + (y2-y1)2] = √[(6-5)2 + (4-(-2))2] = √[12 + 62] = √[1 + 36] = √[37]
(ii) Distance between second and third points
= √[(7-6)2 + (-2-4)2] = √[12 + (-6)2] = √[1 + 36] = √[37]
(iii) Distance between first and third points
= √[(7-5))2 + (-2-(-2))2] = √[22 + 02] = √[4 + 0] = √[4] = 2
3. Two distances are equal. So it is an isosceles triangle. The actual plot is shown in fig.31.45 below:
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.45 is given for a better understanding of the problem
Solved example 31.24
Show that the points (6,1), (9,4), (6,7) and (3,4) are the vertices of a square
Solution:
1. First, we will find the distances:
(a) Distance 1 formed by the first and second points
= √[(9-6)2 + (4-1)2] = √[9 + 9] = √[18]
(b) Distance 2 formed by the second and third points
= √[(6-9)2 + (7-4)2] = √[9 + 9] = √[18]
(c) Distance 3 formed by the third and fourth points
= √[(3-6)2 + (4-7)2] = √[9 + 9] = √[18]
(d) Distance 4 formed by the first and fourth points
= √[(3-6)2 + (4-1)2] = √[9 + 9] = √[18]
(e) Distance 5 formed by the first and third points
= √[(6-6)2 + (7-1)2] = √[0 + 36] = √[36] = 6
(f) Distance 6 formed by the second and fourth points
= √[(3-9)2 + (4-4)2] = √[36 + 0] = √[36] = 6
• 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.
The actual plot is shown in fig.31.46 below:
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.46 is given for a better understanding of the problem
Solved example 31.25
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (-1,-2), (1,0), (-1,2), (-3,0)
(ii) (-3,5), (3,1), (0,3), (-1,-4)
(iii) (4,5), (7,6), (4,3), (1,2)
Solution:
Case (i):
1. There are 4 points. It is possible that, they are vertices of a quadrilateral. 4 points will give a maximum of 6 distances. We will find all those 6 distances:
(a) Distance 1 formed by the first and second points
= √[(1-(-1))2 + (0-(-2))2] = √[4 + 4] = √[8]
(b) Distance 2 formed by the second and third points
= √[(-1-1)2 + (2-0)2] = √[4 + 4] = √[8]
(c) Distance 3 formed by the third and fourth points
= √[(-3-(-1))2 + (0-2)2] = √[4 + 4] = √[8]
(d) Distance 4 formed by the first and fourth points
= √[(-3-(-1))2 + (0-(-2))2] = √[4 + 4] = √[8]
(e) Distance 5 formed by the first and third points
= √[(-1-(-1))2 + (2-(-2))2] = √[0 + 16] = √[16] = 4
(f) Distance 6 formed by the second and fourth points
= √[(-3-1)2 + (0-0)2] = √[16 + 0] = √[16] = 4
2. So four sides are equal
• Also two diagonals are equal.
■ This proves that the given points form a square.
The actual plot is shown as (i) in fig.31.47 below:
Case (ii):
1. There are 4 points. It is possible that, they are vertices of a quadrilateral. 4 points will give a maximum of 6 distances. We will find all those 6 distances:
(a) Distance 1 formed by the first and second points
= √[(3-(-3))2 + (1-5)2] = √[36 + 16] = √[52]
= √[13×4] = √[13] × √[4] = 2√13
(b) Distance 2 formed by the second and third points
= √[(0-3)2 + (3-1)2] = √[9 + 4] = √[13]
(c) Distance 3 formed by the third and fourth points
= √[(-1-0)2 + (-4-3)2] = √[1 + 49] = √[50]
(d) Distance 4 formed by the first and fourth points
= √[(-1-(-3))2 + (-4-5)2] = √[4 + 81] = √[84]
(e) Distance 5 formed by the first and third points
= √[(0-(-3))2 + (3-5))2] = √[9 + 4] = √[13]
(f) Distance 6 formed by the second and fourth points
= √[(-3-1)2 + (0-0)2] = √[16 + 0] = √[16] = 4
2. Consider (b) and (e). We have:
• (b) Distance 2 formed by the second and third points = √13
• (e) Distance 5 formed by the first and third points = √13
3. Sum of the two distances = (√13 + √13) = 2√13
4. But from (a) we have:
• Distance 1 formed by the first and second points = 2√13
5. So the given third point lies midway between the first and second points. Also those three points are collinear. So a quadrilateral will not be formed. This is shown as (ii) in fig.31.47 above
Case (iii):
1. There are 4 points. It is possible that, they are vertices of a quadrilateral. 4 points will give a maximum of 6 distances. We will find all those 6 distances:
(a) Distance 1 formed by the first and second points
= √[(7-4)2 + (6-5)2] = √[9 + 1] = √[10]
(b) Distance 2 formed by the second and third points
= √[(4-7)2 + (3-6)2] = √[9 + 9] = √[18]
(c) Distance 3 formed by the third and fourth points
= √[(1-4)2 + (2-3)2] = √[9 + 1] = √[10]
(d) Distance 4 formed by the first and fourth points
= √[(1-4)2 + (2-5)2] = √[9 + 9] = √[18]
(e) Distance 5 formed by the first and third points
= √[(4-4)2 + (3-5)2] = √[0 + 4] = √[4] = 2
(f) Distance 6 formed by the second and fourth points
= √[(1-7)2 + (2-6)2] = √[36 + 16] = √[52]
2. So opposite sides are equal
• But two diagonals are not equal.
• So it is a parallelogram. This is shown as (iii) in fig.31.47 above.
Solved example 31.26
Find the point on the x-axis which is equidistant from (2,-5) and (-2,9)
Solution:
1. Coordinates of any point on the x axis will have a general form: (x,0). Here, 'x' can take any value.
• We have to find a particular point. For that point, 'x' will be a constant. So we will put 'a' in the place of 'x'. So our required point is (a,0). We have to find the value of 'a'
2. Distance between (a,0) and (2,-5) = √[(2-a)2 + (-5-0)2] = √[4 - 4a + a2 + 25] = √[29 - 4a + a2]
• Distance between (a,0) and (-2,9) = √[(-2-a)2 + (9-0)2] = √[4 + 4a + a2 + 81] = √[85 + 4a + a2]
3. These two distances are equal. So we can write:
√[29 - 4a + a2] = √[85 + 4a + a2]
• Squaring both sides we get: [29 - 4a + a2] = [85 + 4a + a2]
⟹ 8a = (29-85) = -56 ⟹ a = -7
• So the required point is: (-7,0)
4. Check:
• Distance between (-7,0) and (2,-5) = √[(2-(-7))2 + (-5-0)2] = √[81 + 25] = √[106]
• Distance between (-7,0) and (-2,9) = √[(-2-(-7))2 + (9-0)2] = √[25 + 81] = √[106]
• The points are shown in fig.31.48 below:
Solved example 31.27
Find the value of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units
Solution:
1. Distance between P(2,-3) and Q(10,y) = √[(10-2)2 + (y-(-3))2] = √[82 + (y+3)2]
= √[64 + y2 + 6y + 9] = √[73 + y2 + 6y]
2. So we can write: √[73 + y2 + 6y] = 10
• Squaring both sides, we get: [73 + y2 + 6y] = 100
⟹ y2 + 6y - 27 = 0
Solved example 31.19
Find whether the points with coordinates (3,1), (6,4), (8,6) 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.31.42 below:
 |
| Fig.31.42 |
• 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 (3,1)
♦ Let coordinates of B be (6,4)
♦ Let coordinates of C be (8,6)
4. So we get:
• AB = √[(x2-x1)2 + (y2-y1)2] = √[(6-3)2 + (4-1)2] = √[32 + 32] = √[9 + 9]
= √[18] = √[9×2] = √[9] × √[2] = 3√2
• BC = √[(x2-x1)2 + (y2-y1)2] = √[(8-6)2 + (6-4)2] = √[22 + 22] = √[4 + 4]
= √[8] = √[4×2] = √[4] × √[2] = 2√2
• AC = √[(x2-x1)2 + (y2-y1)2] = √[(8-3)2 + (6-1)2] = √[52 + 52] = √[25 + 25]
= √[50] = √[25×2] = √[25] × √[2] = 5√2
5. The longest length is AC
• (AB + BC) = (3√2+2√2) = 5√2 cm. It is equal to AC
• So A, B and C lie on a line.
• This is shown in fig.31.43 below. The reader may draw it on a graph paper by taking appropriate scales along the x and y axes.
 |
| Fig.31.43 |
Find the distance between the following pairs of points:
(i) (2,3), (4,1) (ii) (-5,7), (-1,3) (iii) (a,b), (-a,-b)
Solution:
We have the Distance formula:
■ Distance between any two points with coordinates (x1,x2) and (y1,y2) = √[(x2-x1)2 + (y2-y1)2]
(i) √[(4-2)2 + (1-3)2] = √[22 + (-2)2] = √[4 + 4]
= √[8] = √[4×2] = √[4] × √[2] = 2√2
(ii) √[(-1-(-5))2 + (3-7)2] = √[42 + (-4)2] = √[16 + 16]
= √[32] = √[16×2] = √[16] × √[2] = 4√2
(iii) √[(-a-a)2 + (-b-b)2] = √[(-2a)2 + (-2b)2] = √[4a2 + 4b2] = √[4(a2 + b2)] = 2√[(a2 + b2)]
Solved example 31.21
Find the distance between (0,0) and (36,15)
Solution:
We have the 'special case of the Distance formula' for distance of any point from the origin:
■ Distance of any point with coordinates (x,y) from the origin = √[x2 + y2]
So we get:
• Distance of the point [with coordinates (36,15)] from the origin
= √[362 + 152] = √[1296 + 225] = √[1521] = 39
Solved example 31.22
Find whether the points with coordinates (1,5), (2,3), (-2,-11) lie on a line
Solution:
 |
| Fig.31.44 |
1. If 3 points are to be collinear, the following condition should be satisfied:
• The Longest distance among them should be equal to the sum of the other two distances
2. So let us calculate the distances:
(i) Distance between first and second points
= √[(x2-x1)2 + (y2-y1)2] = √[(2-1)2 + (3-5)2] = √[12 + (-2)2] = √[1 + 4] = √[5] = 2.236
(ii) Distance between second and third points
= √[(-2-2)2 + (-11-3)2] = √[(-4)2 + (-14)2] = √[16 + 196] = √[212] = 14.560
(iii) Distance between first and third points
= √[(-2-1))2 + (-11-5)2] = √[(-3)2 + (-16)2] = √[9 + 196] = √[204] = 14.283
3. The longest distance is the one between the second and third points. It is 14.560 units
• The sum of other two = 14.283 + 2.236 = 16.519
• The above two quantities are not equal. So the given points are not collinear.
• The actual plot is shown in the fig.31.44
• In this problem, we are not asked to do the actual plotting of the points. Fig.31.44 is given for a better understanding of the problem
Solved example 31.23
Check whether the points with coordinates (5,-2), (6,4), (7,-2) are the vertices of an isosceles triangle
Solution:
1. When 3 points are given, there will be three possible distances between them
2. So let us calculate the distances:
(i) Distance between first and second points
= √[(x2-x1)2 + (y2-y1)2] = √[(6-5)2 + (4-(-2))2] = √[12 + 62] = √[1 + 36] = √[37]
(ii) Distance between second and third points
= √[(7-6)2 + (-2-4)2] = √[12 + (-6)2] = √[1 + 36] = √[37]
(iii) Distance between first and third points
= √[(7-5))2 + (-2-(-2))2] = √[22 + 02] = √[4 + 0] = √[4] = 2
3. Two distances are equal. So it is an isosceles triangle. The actual plot is shown in fig.31.45 below:
 |
| Fig.31.45 |
Solved example 31.24
Show that the points (6,1), (9,4), (6,7) and (3,4) are the vertices of a square
Solution:
(a) Distance 1 formed by the first and second points
= √[(9-6)2 + (4-1)2] = √[9 + 9] = √[18]
(b) Distance 2 formed by the second and third points
= √[(6-9)2 + (7-4)2] = √[9 + 9] = √[18]
(c) Distance 3 formed by the third and fourth points
= √[(3-6)2 + (4-7)2] = √[9 + 9] = √[18]
(d) Distance 4 formed by the first and fourth points
= √[(3-6)2 + (4-1)2] = √[9 + 9] = √[18]
(e) Distance 5 formed by the first and third points
= √[(6-6)2 + (7-1)2] = √[0 + 36] = √[36] = 6
(f) Distance 6 formed by the second and fourth points
= √[(3-9)2 + (4-4)2] = √[36 + 0] = √[36] = 6
• 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.
The actual plot is shown in fig.31.46 below:
 |
| Fig.31.46 |
Solved example 31.25
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (-1,-2), (1,0), (-1,2), (-3,0)
(ii) (-3,5), (3,1), (0,3), (-1,-4)
(iii) (4,5), (7,6), (4,3), (1,2)
Solution:
Case (i):
1. There are 4 points. It is possible that, they are vertices of a quadrilateral. 4 points will give a maximum of 6 distances. We will find all those 6 distances:
(a) Distance 1 formed by the first and second points
= √[(1-(-1))2 + (0-(-2))2] = √[4 + 4] = √[8]
(b) Distance 2 formed by the second and third points
= √[(-1-1)2 + (2-0)2] = √[4 + 4] = √[8]
(c) Distance 3 formed by the third and fourth points
= √[(-3-(-1))2 + (0-2)2] = √[4 + 4] = √[8]
(d) Distance 4 formed by the first and fourth points
= √[(-3-(-1))2 + (0-(-2))2] = √[4 + 4] = √[8]
(e) Distance 5 formed by the first and third points
= √[(-1-(-1))2 + (2-(-2))2] = √[0 + 16] = √[16] = 4
(f) Distance 6 formed by the second and fourth points
= √[(-3-1)2 + (0-0)2] = √[16 + 0] = √[16] = 4
2. So four sides are equal
• Also two diagonals are equal.
■ This proves that the given points form a square.
The actual plot is shown as (i) in fig.31.47 below:
 |
| Fig.31.47 |
1. There are 4 points. It is possible that, they are vertices of a quadrilateral. 4 points will give a maximum of 6 distances. We will find all those 6 distances:
(a) Distance 1 formed by the first and second points
= √[(3-(-3))2 + (1-5)2] = √[36 + 16] = √[52]
= √[13×4] = √[13] × √[4] = 2√13
(b) Distance 2 formed by the second and third points
= √[(0-3)2 + (3-1)2] = √[9 + 4] = √[13]
(c) Distance 3 formed by the third and fourth points
= √[(-1-0)2 + (-4-3)2] = √[1 + 49] = √[50]
(d) Distance 4 formed by the first and fourth points
= √[(-1-(-3))2 + (-4-5)2] = √[4 + 81] = √[84]
(e) Distance 5 formed by the first and third points
= √[(0-(-3))2 + (3-5))2] = √[9 + 4] = √[13]
(f) Distance 6 formed by the second and fourth points
= √[(-3-1)2 + (0-0)2] = √[16 + 0] = √[16] = 4
2. Consider (b) and (e). We have:
• (b) Distance 2 formed by the second and third points = √13
• (e) Distance 5 formed by the first and third points = √13
3. Sum of the two distances = (√13 + √13) = 2√13
4. But from (a) we have:
• Distance 1 formed by the first and second points = 2√13
5. So the given third point lies midway between the first and second points. Also those three points are collinear. So a quadrilateral will not be formed. This is shown as (ii) in fig.31.47 above
Case (iii):
1. There are 4 points. It is possible that, they are vertices of a quadrilateral. 4 points will give a maximum of 6 distances. We will find all those 6 distances:
(a) Distance 1 formed by the first and second points
= √[(7-4)2 + (6-5)2] = √[9 + 1] = √[10]
(b) Distance 2 formed by the second and third points
= √[(4-7)2 + (3-6)2] = √[9 + 9] = √[18]
(c) Distance 3 formed by the third and fourth points
= √[(1-4)2 + (2-3)2] = √[9 + 1] = √[10]
(d) Distance 4 formed by the first and fourth points
= √[(1-4)2 + (2-5)2] = √[9 + 9] = √[18]
(e) Distance 5 formed by the first and third points
= √[(4-4)2 + (3-5)2] = √[0 + 4] = √[4] = 2
(f) Distance 6 formed by the second and fourth points
= √[(1-7)2 + (2-6)2] = √[36 + 16] = √[52]
2. So opposite sides are equal
• But two diagonals are not equal.
• So it is a parallelogram. This is shown as (iii) in fig.31.47 above.
Solved example 31.26
Find the point on the x-axis which is equidistant from (2,-5) and (-2,9)
Solution:
1. Coordinates of any point on the x axis will have a general form: (x,0). Here, 'x' can take any value.
• We have to find a particular point. For that point, 'x' will be a constant. So we will put 'a' in the place of 'x'. So our required point is (a,0). We have to find the value of 'a'
2. Distance between (a,0) and (2,-5) = √[(2-a)2 + (-5-0)2] = √[4 - 4a + a2 + 25] = √[29 - 4a + a2]
• Distance between (a,0) and (-2,9) = √[(-2-a)2 + (9-0)2] = √[4 + 4a + a2 + 81] = √[85 + 4a + a2]
3. These two distances are equal. So we can write:
√[29 - 4a + a2] = √[85 + 4a + a2]
• Squaring both sides we get: [29 - 4a + a2] = [85 + 4a + a2]
⟹ 8a = (29-85) = -56 ⟹ a = -7
• So the required point is: (-7,0)
4. Check:
• Distance between (-7,0) and (2,-5) = √[(2-(-7))2 + (-5-0)2] = √[81 + 25] = √[106]
• Distance between (-7,0) and (-2,9) = √[(-2-(-7))2 + (9-0)2] = √[25 + 81] = √[106]
• The points are shown in fig.31.48 below:
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| Fig.31.48 |
Find the value of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units
Solution:
1. Distance between P(2,-3) and Q(10,y) = √[(10-2)2 + (y-(-3))2] = √[82 + (y+3)2]
= √[64 + y2 + 6y + 9] = √[73 + y2 + 6y]
2. So we can write: √[73 + y2 + 6y] = 10
• Squaring both sides, we get: [73 + y2 + 6y] = 100
⟹ y2 + 6y - 27 = 0
3. This is of the form ax2 + bx + c = 0 (Details here)
Where: a = 1, b = 6 and c = -27
4. So we can use the general formula to solve the equation
5. b2-4ac = 62-4×1×(-27) = 36 + 108 = 144
• So √[b2-4ac] = √144 = ±12
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(6)±12 = 6 AND -18
• The denominator = 2a = 2×1 = 2
• Thus y = 6⁄2 = 3 AND y = -18⁄2 = -9
6. Thus we can write:
Distance from (2-3) to (10,3) is 10 units
Distance from (2-3) to (10,-9) is 10 units
7. Check:
• Distance between P(2,-3) and Q(10,3) = √[(10-2)2 + (3-(-3))2] = √[82 + 62]
= √[64 + 36] = √[100] = 10
• Distance between P(2,-3) and Q(10,-9) = √[(10-2)2 + (-9-(-3))2] = √[82 + (-6)2]
= √[64 + 36] = √[100] = 10
8. This is shown in the fig.31.49 below:
Solved example 31.28
If Q(0,1) is equidistant from P(5,-3) and R(x,6), find the values of x. Also find the distances QR and PR
Solution:
1. Distance between P(5,-3) and Q(0,1) = √[(0-5)2 + (1-(-3))2] = √[52 + 42]
= √[25 + 16] = √[41]
2. Distance between R(x,6) and Q(0,1) = √[(0-x)2 + (1-6)2] = √[x2 + 25]
3. These two distances are equal. So we can write:
√[41] = √[x2 + 25]
• Squaring both sides we get: 41 = x2 + 25
⟹ 16 = x2 ⟹ x = ±4
5. So the two possible positions of R are: (4,6) and (-4,6)
We will call them R and R'
6. Distance between R(4,6) and Q(0,1) = √[(0-4)2 + (1-6)2] = √[16 + 25] = √41
• Distance between R'(-4,6) and Q(0,1) = √[(0-(-4))2 + (1-6)2] = √[16 + 25] = √41
• Distance between R(4,6) and P(5,-3) = √[(5-4)2 + (-3-6)2] = √[1 + 81] = √82
7. All the points are shown in fig.31.50 below:
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(6)±12 = 6 AND -18
• The denominator = 2a = 2×1 = 2
• Thus y = 6⁄2 = 3 AND y = -18⁄2 = -9
6. Thus we can write:
Distance from (2-3) to (10,3) is 10 units
Distance from (2-3) to (10,-9) is 10 units
7. Check:
• Distance between P(2,-3) and Q(10,3) = √[(10-2)2 + (3-(-3))2] = √[82 + 62]
= √[64 + 36] = √[100] = 10
• Distance between P(2,-3) and Q(10,-9) = √[(10-2)2 + (-9-(-3))2] = √[82 + (-6)2]
= √[64 + 36] = √[100] = 10
8. This is shown in the fig.31.49 below:
 |
| Fig.31.49 |
If Q(0,1) is equidistant from P(5,-3) and R(x,6), find the values of x. Also find the distances QR and PR
Solution:
1. Distance between P(5,-3) and Q(0,1) = √[(0-5)2 + (1-(-3))2] = √[52 + 42]
= √[25 + 16] = √[41]
2. Distance between R(x,6) and Q(0,1) = √[(0-x)2 + (1-6)2] = √[x2 + 25]
3. These two distances are equal. So we can write:
√[41] = √[x2 + 25]
• Squaring both sides we get: 41 = x2 + 25
⟹ 16 = x2 ⟹ x = ±4
5. So the two possible positions of R are: (4,6) and (-4,6)
We will call them R and R'
6. Distance between R(4,6) and Q(0,1) = √[(0-4)2 + (1-6)2] = √[16 + 25] = √41
• Distance between R'(-4,6) and Q(0,1) = √[(0-(-4))2 + (1-6)2] = √[16 + 25] = √41
• Distance between R(4,6) and P(5,-3) = √[(5-4)2 + (-3-6)2] = √[1 + 81] = √82
7. All the points are shown in fig.31.50 below:
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| Fig.31.50 |
In the next section we will see a few more solved examples.
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