In the previous section we saw how to find the unknown coordinates of a vertex in a parallelogram. We also saw some solved examples. In this section we will see coordinates of midpoint.
Consider the line AB in fig.34.7(a) below:
• Coordinates of A and B are known. We have to find the coordinates of the midpoint C
2. In fig.b, green lines are drawn.
• These green lines have a special property: They are all parallel to the axes
♦ The horizontal green line is parallel to the x axis
♦ The vertical green lines are parallel to the y axis
3. The two vertical green lines intersect the horizontal green line at C' and B'.
• We know that the angle between the axes will always be 90o
• Since the green lines are parallel to the axes, the angle at B' and C' will also be 90o
• So triangles AB'B and AC'C are right triangles
4. Now we will see the relation between the two right triangles:
• Let ∠BAB' be α.
• Let ∠ABB' be β.
• CC' and BB' are two parallel lines cut by a transversal AB
♦ ∠ACC' and ∠ABB' are corresponding angles
♦ So we get: ∠ACC' = ∠ABB' = β.
• Now consider the two right triangles: ⊿AB'B and ⊿AC'C
• Both of them have the same angles: αo, βo and 90o.
■ So they are similar triangles
5. Now we apply a special property that is applicable to any two similar triangles (Details here):
= side opposite B in ΔAB'B⁄side opposite C in ΔAC'C
= side opposite B' in ΔAB'B⁄side opposite C' in ΔAC'C
8. So we get: BB'⁄CC' = AB'⁄AC' = AB⁄AC
• Since C is the midpoint, we have:
AB = 2AC ⟹ AB⁄AC = 2
• So we can write: BB'⁄CC' = AB'⁄AC' = 2
9. Take the first and the third ratios. We get:
BB'⁄CC' = 2 ⟹ BB' = 2CC'
• Take the second and the third ratios. We get:
AB'⁄AC' = 2 ⟹ AB' = 2AC'
10. The results obtained in (9) are very useful. They tell us that:
• 'Altitude of the smaller triangle' is exact half of 'altitude of larger triangle'
• 'Base of the smaller triangle' is exact half of 'base of larger triangle'
11. Consider a person travelling from A to B
First he has to travel 6 units horizontally to the right to reach B' (∵ 8-2 =6)
Then he has to travel 3 units vertically upwards to reach B (∵ 6-3 =3)
12. Consider a person travelling from A to C
• First he has to travel horizontally to the right to reach C'
♦ But how much distance?
♦ We have seen in (9) that, AC' is half of AB'
♦ So the distance he has to travel to reach C' is (AB'⁄2 = 6⁄2 ) = 3 units
♦ So the coordinates of C' are (5,3)
• Then he has to travel vertically upwards to reach C
♦ But how much distance?
♦ We have seen in (9) that, CC' is half of BB'
♦ So the distance he has to travel to reach C is (BB'⁄2 = 3⁄2 ) = 1.5 units
♦ So the coordinates of C' are (5,4.5). [Adding 1.5 to the y coordinate of C']
So we obtained the coordinates of C. Let us write a summary:
• We are given the coordinates of two points. We are asked to find the mid point. The following 4 steps can be used:
1. Assume a travel from the first point to the second point
2. Note the 'horizontal travel distance' and the 'vertical travel distance'
3. Calculate half of the two distances in (2)
4. Add or subtract them appropriately to the coordinates of the first point
An example:
Find the midpoint A of P(2,1) and Q(6,3)
Solution:
• 6 >2 and 3 >1. So Q is at the right side top of P
1. Assume a travel fro P to Q
2. The horizontal distance is (6-2) = 4
The vertical distance is (3-1) = 2
3. Half of horizontal distance is 2
Half of vertical distance is 1
4. Add the half distances to the coordinates of P:
• x coordinate = (2+2) = 4
• y coordinate = (1+1) = 2
■ So the coordinates of the midpoint A are: (4,2).
[See the side PQ and it's midpoint A in fig.34.4(e) in the previous section]
• In the above discussion, we were able to find the midpoint because the bases are half. The altitudes are also half. We proved it using the principles of similar triangles.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.7(c) above, consider the right triangle AB'B. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = BB'⁄AB.
cos α = adjacent side⁄hypotenuse = AB'⁄AB.
2. Again in fig.34.7(c) above, consider the right triangle AC'C. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = CC'⁄AC.
cos α = adjacent side⁄hypotenuse = AC'⁄AC.
3. Now, the sine in ⊿AB'B can be equated to the sine in ⊿AC'C. Because, both are taken for the same angle α
4. So we can write: BB'⁄AB = CC'⁄AC ⟹ BB'⁄CC' = AB⁄AC = 2
So we get: BB' = 2CC'
■ That is., altitude of smaller triangle is half of the altitude of the larger triangle
5. The same is applicable to cosine also:
• Equating the cosines, we get:
AB'⁄AB = AC'⁄AC ⟹ AB'⁄AC' = AB⁄AC = 2
So we get: AB' = 2AC'
■ That is., base of smaller triangle is half of the base of the larger triangle
• The discussions so far in this section helps us to understand the 'basics about finding the midpoint'.
• But while solving problems, it is convenient to use a formula. So we will derive it:
1. Let the coordinates of A and B be (x1,y1) and (x2,y2) respectively. This is shown in fig.34.8 below
• We have to find the coordinates of the midpoint C
2. We have seen that AC' is half of AB'
• But AB' = (x2-x1)
• So AC' = (x2-x1)⁄2.
• So the x coordinate of C' will be:
x coordinate of A + AC'
= [x1 + (x2-x1)⁄2] = (x1+x2)⁄2.
• Then coordinates of C' are [(x1+x2)⁄2,y1]
3. We have seen that CC' is half of BB'
• But BB' = (y2-y1)
• So CC' = (y2-y1)⁄2.
• So the y coordinate of C' will be:
y coordinate of C' + CC'
= [y1 + (y2-y1)⁄2] = (y1+y2)⁄2
■ Thus coordinates of midpoint C are [(x1+x2)⁄2,(y1+y2)⁄2]
An example:
Find the midpoint of the line joining (-3,5) and (7,3)
Solution:
1. x coordinate of the midpoint = (x1+x2)⁄2 = (-3+7)⁄2 = 4⁄2 = 2
2. y coordinate of the midpoint = (y1+y2)⁄2 = (5+3)⁄2 = 8⁄2 = 4
3. So the coordinates of the midpoint are: (2,4)
The actual positions of the points in the Cartesian plane are shown in the fig.34.9 below:
• This example shows us that the formula is applicable even if the end points of the line lie in different quadrants.
In the next section, we will see Section formula.
Consider the line AB in fig.34.7(a) below:
 |
| Fig.34.7 |
2. In fig.b, green lines are drawn.
• These green lines have a special property: They are all parallel to the axes
♦ The horizontal green line is parallel to the x axis
♦ The vertical green lines are parallel to the y axis
3. The two vertical green lines intersect the horizontal green line at C' and B'.
• We know that the angle between the axes will always be 90o
• Since the green lines are parallel to the axes, the angle at B' and C' will also be 90o
• So triangles AB'B and AC'C are right triangles
4. Now we will see the relation between the two right triangles:
• Let ∠BAB' be α.
• Let ∠ABB' be β.
• CC' and BB' are two parallel lines cut by a transversal AB
♦ ∠ACC' and ∠ABB' are corresponding angles
♦ So we get: ∠ACC' = ∠ABB' = β.
• Now consider the two right triangles: ⊿AB'B and ⊿AC'C
• Both of them have the same angles: αo, βo and 90o.
■ So they are similar triangles
5. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔAB'B⁄side opposite smallest angle in ΔAC'C
= side opposite medium angle in ΔAB'B⁄side opposite medium angle in ΔAC'C
= side opposite largest angle in ΔAB'B⁄side opposite largest angle in ΔAC'C
6. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔAB'B = Smallest angle in ΔAC'C
• Medium angle in ΔAB'B = Medium angle in ΔAC'C
• Largest angle in ΔAB'B = Largest angle in ΔAC'C
7. So we can write this:
7. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite A in ΔAB'B⁄side opposite A in ΔAC'C= side opposite B in ΔAB'B⁄side opposite C in ΔAC'C
= side opposite B' in ΔAB'B⁄side opposite C' in ΔAC'C
8. So we get: BB'⁄CC' = AB'⁄AC' = AB⁄AC
• Since C is the midpoint, we have:
AB = 2AC ⟹ AB⁄AC = 2
• So we can write: BB'⁄CC' = AB'⁄AC' = 2
9. Take the first and the third ratios. We get:
BB'⁄CC' = 2 ⟹ BB' = 2CC'
• Take the second and the third ratios. We get:
AB'⁄AC' = 2 ⟹ AB' = 2AC'
10. The results obtained in (9) are very useful. They tell us that:
• 'Altitude of the smaller triangle' is exact half of 'altitude of larger triangle'
• 'Base of the smaller triangle' is exact half of 'base of larger triangle'
11. Consider a person travelling from A to B
First he has to travel 6 units horizontally to the right to reach B' (∵ 8-2 =6)
Then he has to travel 3 units vertically upwards to reach B (∵ 6-3 =3)
12. Consider a person travelling from A to C
• First he has to travel horizontally to the right to reach C'
♦ But how much distance?
♦ We have seen in (9) that, AC' is half of AB'
♦ So the distance he has to travel to reach C' is (AB'⁄2 = 6⁄2 ) = 3 units
♦ So the coordinates of C' are (5,3)
• Then he has to travel vertically upwards to reach C
♦ But how much distance?
♦ We have seen in (9) that, CC' is half of BB'
♦ So the distance he has to travel to reach C is (BB'⁄2 = 3⁄2 ) = 1.5 units
♦ So the coordinates of C' are (5,4.5). [Adding 1.5 to the y coordinate of C']
So we obtained the coordinates of C. Let us write a summary:
• We are given the coordinates of two points. We are asked to find the mid point. The following 4 steps can be used:
1. Assume a travel from the first point to the second point
2. Note the 'horizontal travel distance' and the 'vertical travel distance'
3. Calculate half of the two distances in (2)
4. Add or subtract them appropriately to the coordinates of the first point
An example:
Find the midpoint A of P(2,1) and Q(6,3)
Solution:
• 6 >2 and 3 >1. So Q is at the right side top of P
1. Assume a travel fro P to Q
2. The horizontal distance is (6-2) = 4
The vertical distance is (3-1) = 2
3. Half of horizontal distance is 2
Half of vertical distance is 1
4. Add the half distances to the coordinates of P:
• x coordinate = (2+2) = 4
• y coordinate = (1+1) = 2
■ So the coordinates of the midpoint A are: (4,2).
[See the side PQ and it's midpoint A in fig.34.4(e) in the previous section]
• In the above discussion, we were able to find the midpoint because the bases are half. The altitudes are also half. We proved it using the principles of similar triangles.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.7(c) above, consider the right triangle AB'B. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = BB'⁄AB.
cos α = adjacent side⁄hypotenuse = AB'⁄AB.
2. Again in fig.34.7(c) above, consider the right triangle AC'C. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = CC'⁄AC.
cos α = adjacent side⁄hypotenuse = AC'⁄AC.
3. Now, the sine in ⊿AB'B can be equated to the sine in ⊿AC'C. Because, both are taken for the same angle α
4. So we can write: BB'⁄AB = CC'⁄AC ⟹ BB'⁄CC' = AB⁄AC = 2
So we get: BB' = 2CC'
■ That is., altitude of smaller triangle is half of the altitude of the larger triangle
5. The same is applicable to cosine also:
• Equating the cosines, we get:
AB'⁄AB = AC'⁄AC ⟹ AB'⁄AC' = AB⁄AC = 2
So we get: AB' = 2AC'
■ That is., base of smaller triangle is half of the base of the larger triangle
• The discussions so far in this section helps us to understand the 'basics about finding the midpoint'.
• But while solving problems, it is convenient to use a formula. So we will derive it:
1. Let the coordinates of A and B be (x1,y1) and (x2,y2) respectively. This is shown in fig.34.8 below
 |
| Fig.34.8 |
2. We have seen that AC' is half of AB'
• But AB' = (x2-x1)
• So AC' = (x2-x1)⁄2.
• So the x coordinate of C' will be:
x coordinate of A + AC'
= [x1 + (x2-x1)⁄2] = (x1+x2)⁄2.
• Then coordinates of C' are [(x1+x2)⁄2,y1]
3. We have seen that CC' is half of BB'
• But BB' = (y2-y1)
• So CC' = (y2-y1)⁄2.
• So the y coordinate of C' will be:
y coordinate of C' + CC'
= [y1 + (y2-y1)⁄2] = (y1+y2)⁄2
■ Thus coordinates of midpoint C are [(x1+x2)⁄2,(y1+y2)⁄2]
An example:
Find the midpoint of the line joining (-3,5) and (7,3)
Solution:
1. x coordinate of the midpoint = (x1+x2)⁄2 = (-3+7)⁄2 = 4⁄2 = 2
2. y coordinate of the midpoint = (y1+y2)⁄2 = (5+3)⁄2 = 8⁄2 = 4
3. So the coordinates of the midpoint are: (2,4)
The actual positions of the points in the Cartesian plane are shown in the fig.34.9 below:
 |
| Fig.34.9 |
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