In the previous section we saw how to find the midpoint of two given points. We also saw some examples. The midpoint of a line divides the line into two equal parts. In this section we will see the division of a line in a given ratio.
First of all, we will see what is meant by 'dividing a line in a ratio'.
We will analyse it using an example.
1. Consider the line AB in fig.34.10(a) below:
• A point P is marked on it. We are told that this P divides the line AB in the ratio 3:5.
2. If the coordinates of A and B are given, we must be able to find the coordinates of P.
• Before learning to find those coordinates, we will see 'how exactly P does the division'.
• In fig.b, the line AB is divided into 8 equal parts.
♦ This 8 comes from the 'given ratio 3:5'
♦ We simply add the portions on either sides of the ':' symbol
3. Now the point P is exactly at the end of the third (counting from end A) segment. Why is this so?
• Out of the 8 equal parts, 3 parts fall within AP. So the fraction representing AP is 3⁄8
♦ In other words, length of AP is 3⁄8 of the total length AB
• Out of the 8 equal parts, the other 5 parts fall within BP. So the fraction representing BP is 5⁄8
♦ In other words, length of BP is 5⁄8 of the total length AB
4. Let us take the ratio AP⁄BP :
AP⁄BP = Length of AP⁄Length of BP
= [(3⁄8 ×AB) ÷ (5⁄8 ×AB)]
= [(3⁄8) ÷ (5⁄8)] = [(3⁄8) × (8⁄5)] = 3⁄5
5. So we can write:
Length of AP⁄Length of BP = 3⁄5.
■ That is., P divides AB in the ratio 3:5
■ In general we can write:
If a point P divides a line AB in the ratio m:n, then:
• AP = {[m⁄(m+n)]×AB}
♦ If we put (m+n) = k, we get:
AP = {[m⁄k]×AB}
• BP = {[n⁄(m+n)]×AB}
♦ If we put (m+n) = k, we get:
BP = {[n⁄k]×AB}
• We will need these results in our further discussions
1. Consider the line AB in fig.34.11(a) below:
• The coordinates of A and B are (x1,y1) and (x2,y2) respectively
• The point P divides AB in the ratio m:n
• We have to find the coordinates of P
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 P' 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 P' will also be 90o
• So triangles AB'B and AP'P are right triangles
4. Now we will see the relation between the two right triangles:
• Let ∠BAB' be α.
• Let ∠ABB' be β.
• PP' and BB' are two parallel lines cut by a transversal AB
♦ ∠APP' and ∠ABB' are corresponding angles
♦ So we get: ∠APP' = ∠ABB' = β.
• Now consider the two right triangles: ⊿AB'B and ⊿AP'P
• Both of them have the same angles: αo, βo and 90o.
■ So they are similar triangles
5. To establish the relation between the two triangles, we can use two methods:
• Applying the principles of similar triangles
• Applying the principles of trigonometry
In the previous section where we calculated the midpoint, we demonstrated both the methods. In this section we will use trigonometry only. However readers are advised to write the steps using 'principles of similar triangles' in his/her own note books.
6. We have already seen the basics of trigonometry here.
• In fig.34.11(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
7. Again in fig.34.11(c) above, consider the right triangle AP'P. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = PP'⁄AP
cos α = adjacent side⁄hypotenuse = AP'⁄AP
8. Now, the sine in ⊿AB'B can be equated to the sine in ⊿AP'P. Because, both are taken for the same angle α
9. So we can write: BB'⁄AB = PP'⁄AP ⟹ BB'⁄PP' = AB⁄AP
10. But AP = {[m⁄k]×AB} ⟹ AB⁄AP = k⁄m
• So (9) becomes:
BB'⁄PP' = AB⁄AP = k⁄m .
• Thus we get:
PP' = {[m⁄k]×BB'}
■ That is., altitude of smaller triangle is (m⁄k) times the altitude of the larger triangle
11. The same is applicable to cosine also:
• Equating the cosines, we get:
AB'⁄AB = AP'⁄AP ⟹ AB'⁄AP' = AB⁄AP
12. But AP = {[m⁄k]×AB} ⟹ AB⁄AP = k⁄m
• So (11) becomes:
AB'⁄AP' = AB⁄AP = k⁄m .
• Thus we get:
AP' = {[m⁄k]×AB'}
■ That is., base of smaller triangle is (m⁄k) times the base of the larger triangle
13. Now we can write the coordinates of P':
• x coordinate of P'
= (x coordinate of A) + (AP')
= x1 + {[m⁄k]×AB'}
= x1 + {[m⁄k]×(x2-x1)}
• y coordinate of P'
= y coordinate of A
= y1
14. Based on the coordinates of P', we can write the coordinates of P:
• x coordinate of P
= (x coordinate of P')
= x1 + {[m⁄k]×(x2-x1)}
• y coordinate of P
= (y coordinate of P') + (PP')
= y1 + {[m⁄k]×BB'}
= y1 + {[m⁄k]×(y2-y1)}
The above two formulas can be written in another form also:
• x coordinate of P
= x1 + {[m⁄k]×(x2-x1)}
= x1 + {[m⁄(m+n)]×(x2-x1)} (∵ k = m+n)
= {x1(m+n) + m(x2-x1)} ÷ {m+n}
= {x1×m + x1×n + x2×m - x1×m} ÷ {m+n}
= (nx1+mx2)⁄(m+n)
• y coordinate of P
= y1 + {[m⁄k]×(y2-y1)}
= y1 + {[m⁄(m+n)]×(y2-y1)} (∵ k = m+n)
= {y1(m+n) + m(y2-y1)} ÷ {m+n}
= {y1×m + y1×n + y2×m - y1×m} ÷ {m+n}
= (ny1+my2)⁄(m+n)
■ This is known as the section formula
■ In both the forms of the formula, it is important to note the pattern:
• m corresponds to the segment near the (x1,y1)
• n corresponds to the segment near the (x2,y2)
An example:
Coordinates of A are (2,4). Coordinates of B are (8,7). Find the coordinates of the point P which divides AB in the ratio 3:5
Solution:
1. Let (x1,y1) be (2,4) and (x2,y2) be (8,7)
2. Given m:n = 3:5
• So m = 3, n = 5 and k = 8
3. Then x coordinate of P = x1 + {[m⁄k]×(x2-x1)}
= 2 + {[3⁄8]×(8-2)} = 2 + 9⁄4 = 17⁄4 = 41⁄4
4. y coordinate of P = y1 + {[m⁄k]×(y2-y1)}
= 4 + {[3⁄8]×(7-4)} = 4 + 9⁄8 = 41⁄8 = 51⁄8
■ Let us try the section formula also:
• x coordinate of P = (nx1+mx2)⁄(m+n) = (5×2+3×8)⁄(3+5) = 34⁄8 = 17⁄4 = 41⁄4
• y coordinate of P = (ny1+my2)⁄(m+n) = (5×4+3×7)⁄(3+5) = 41⁄8 = 51⁄8
The results are same as before
Solved example 34.5
The coordinates of two points A, B are (3,2) and 8,7)
(i) Calculate the coordinates of the point P on AB such that AP:PB = 2:3
(ii) Calculate the coordinates of the point Q on AB such that AQ:QB = 3:2
Solution:
Part (i):
1. Let (x1,y1) be (3,2) and (x2,y2) be (8,7)
2. Given m:n = 2:3
• So m = 2, n = 3 and k = 5
3. Then x coordinate of P = x1 + {[m⁄k]×(x2-x1)}
= 3 + {[2⁄5]×(8-3)} = 3 + 2 = 5
4. y coordinate of P = y1 + {[m⁄k]×(y2-y1)}
= 2 + {[2⁄5]×(7-2)} = 2 + 2 = 4
• So coordinates of P are (5,4)
Part (ii):
1. Given m:n = 3:2
• So m = 3, n = 2 and k = 5
2. Then x coordinate of P = x1 + {[m⁄k]×(x2-x1)}
= 3 + {[3⁄5]×(8-3)} = 3 + 3 = 6
3. y coordinate of P = y1 + {[m⁄k]×(y2-y1)}
= 2 + {[3⁄5]×(7-2)} = 2 + 3 = 5
• So coordinates of Q are (6,5)
■ Fig.34.12 below shows the actual positions in the Cartesian plane:
In the next section, we will see a few more solved examples.
First of all, we will see what is meant by 'dividing a line in a ratio'.
We will analyse it using an example.
1. Consider the line AB in fig.34.10(a) below:
 |
| Fig.34.10 |
2. If the coordinates of A and B are given, we must be able to find the coordinates of P.
• Before learning to find those coordinates, we will see 'how exactly P does the division'.
• In fig.b, the line AB is divided into 8 equal parts.
♦ This 8 comes from the 'given ratio 3:5'
♦ We simply add the portions on either sides of the ':' symbol
3. Now the point P is exactly at the end of the third (counting from end A) segment. Why is this so?
• Out of the 8 equal parts, 3 parts fall within AP. So the fraction representing AP is 3⁄8
♦ In other words, length of AP is 3⁄8 of the total length AB
• Out of the 8 equal parts, the other 5 parts fall within BP. So the fraction representing BP is 5⁄8
♦ In other words, length of BP is 5⁄8 of the total length AB
4. Let us take the ratio AP⁄BP :
AP⁄BP = Length of AP⁄Length of BP
= [(3⁄8 ×AB) ÷ (5⁄8 ×AB)]
= [(3⁄8) ÷ (5⁄8)] = [(3⁄8) × (8⁄5)] = 3⁄5
5. So we can write:
Length of AP⁄Length of BP = 3⁄5.
■ That is., P divides AB in the ratio 3:5
■ In general we can write:
If a point P divides a line AB in the ratio m:n, then:
• AP = {[m⁄(m+n)]×AB}
♦ If we put (m+n) = k, we get:
AP = {[m⁄k]×AB}
• BP = {[n⁄(m+n)]×AB}
♦ If we put (m+n) = k, we get:
BP = {[n⁄k]×AB}
• We will need these results in our further discussions
1. Consider the line AB in fig.34.11(a) below:
 |
| Fig.34.11 |
• The point P divides AB in the ratio m:n
• We have to find the coordinates of P
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 P' 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 P' will also be 90o
• So triangles AB'B and AP'P are right triangles
4. Now we will see the relation between the two right triangles:
• Let ∠BAB' be α.
• Let ∠ABB' be β.
• PP' and BB' are two parallel lines cut by a transversal AB
♦ ∠APP' and ∠ABB' are corresponding angles
♦ So we get: ∠APP' = ∠ABB' = β.
• Now consider the two right triangles: ⊿AB'B and ⊿AP'P
• Both of them have the same angles: αo, βo and 90o.
■ So they are similar triangles
5. To establish the relation between the two triangles, we can use two methods:
• Applying the principles of similar triangles
• Applying the principles of trigonometry
In the previous section where we calculated the midpoint, we demonstrated both the methods. In this section we will use trigonometry only. However readers are advised to write the steps using 'principles of similar triangles' in his/her own note books.
6. We have already seen the basics of trigonometry here.
• In fig.34.11(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
7. Again in fig.34.11(c) above, consider the right triangle AP'P. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = PP'⁄AP
cos α = adjacent side⁄hypotenuse = AP'⁄AP
8. Now, the sine in ⊿AB'B can be equated to the sine in ⊿AP'P. Because, both are taken for the same angle α
9. So we can write: BB'⁄AB = PP'⁄AP ⟹ BB'⁄PP' = AB⁄AP
10. But AP = {[m⁄k]×AB} ⟹ AB⁄AP = k⁄m
• So (9) becomes:
BB'⁄PP' = AB⁄AP = k⁄m .
• Thus we get:
PP' = {[m⁄k]×BB'}
■ That is., altitude of smaller triangle is (m⁄k) times the altitude of the larger triangle
11. The same is applicable to cosine also:
• Equating the cosines, we get:
AB'⁄AB = AP'⁄AP ⟹ AB'⁄AP' = AB⁄AP
12. But AP = {[m⁄k]×AB} ⟹ AB⁄AP = k⁄m
• So (11) becomes:
AB'⁄AP' = AB⁄AP = k⁄m .
• Thus we get:
AP' = {[m⁄k]×AB'}
■ That is., base of smaller triangle is (m⁄k) times the base of the larger triangle
13. Now we can write the coordinates of P':
• x coordinate of P'
= (x coordinate of A) + (AP')
= x1 + {[m⁄k]×AB'}
= x1 + {[m⁄k]×(x2-x1)}
• y coordinate of P'
= y coordinate of A
= y1
14. Based on the coordinates of P', we can write the coordinates of P:
• x coordinate of P
= (x coordinate of P')
= x1 + {[m⁄k]×(x2-x1)}
• y coordinate of P
= (y coordinate of P') + (PP')
= y1 + {[m⁄k]×BB'}
= y1 + {[m⁄k]×(y2-y1)}
The above two formulas can be written in another form also:
• x coordinate of P
= x1 + {[m⁄k]×(x2-x1)}
= x1 + {[m⁄(m+n)]×(x2-x1)} (∵ k = m+n)
= {x1(m+n) + m(x2-x1)} ÷ {m+n}
= {x1×m + x1×n + x2×m - x1×m} ÷ {m+n}
= (nx1+mx2)⁄(m+n)
• y coordinate of P
= y1 + {[m⁄k]×(y2-y1)}
= y1 + {[m⁄(m+n)]×(y2-y1)} (∵ k = m+n)
= {y1(m+n) + m(y2-y1)} ÷ {m+n}
= {y1×m + y1×n + y2×m - y1×m} ÷ {m+n}
= (ny1+my2)⁄(m+n)
■ This is known as the section formula
■ In both the forms of the formula, it is important to note the pattern:
• m corresponds to the segment near the (x1,y1)
• n corresponds to the segment near the (x2,y2)
An example:
Coordinates of A are (2,4). Coordinates of B are (8,7). Find the coordinates of the point P which divides AB in the ratio 3:5
Solution:
1. Let (x1,y1) be (2,4) and (x2,y2) be (8,7)
2. Given m:n = 3:5
• So m = 3, n = 5 and k = 8
3. Then x coordinate of P = x1 + {[m⁄k]×(x2-x1)}
= 2 + {[3⁄8]×(8-2)} = 2 + 9⁄4 = 17⁄4 = 41⁄4
4. y coordinate of P = y1 + {[m⁄k]×(y2-y1)}
= 4 + {[3⁄8]×(7-4)} = 4 + 9⁄8 = 41⁄8 = 51⁄8
■ Let us try the section formula also:
• x coordinate of P = (nx1+mx2)⁄(m+n) = (5×2+3×8)⁄(3+5) = 34⁄8 = 17⁄4 = 41⁄4
• y coordinate of P = (ny1+my2)⁄(m+n) = (5×4+3×7)⁄(3+5) = 41⁄8 = 51⁄8
The results are same as before
Solved example 34.5
The coordinates of two points A, B are (3,2) and 8,7)
(i) Calculate the coordinates of the point P on AB such that AP:PB = 2:3
(ii) Calculate the coordinates of the point Q on AB such that AQ:QB = 3:2
Solution:
Part (i):
1. Let (x1,y1) be (3,2) and (x2,y2) be (8,7)
2. Given m:n = 2:3
• So m = 2, n = 3 and k = 5
3. Then x coordinate of P = x1 + {[m⁄k]×(x2-x1)}
= 3 + {[2⁄5]×(8-3)} = 3 + 2 = 5
4. y coordinate of P = y1 + {[m⁄k]×(y2-y1)}
= 2 + {[2⁄5]×(7-2)} = 2 + 2 = 4
• So coordinates of P are (5,4)
Part (ii):
1. Given m:n = 3:2
• So m = 3, n = 2 and k = 5
2. Then x coordinate of P = x1 + {[m⁄k]×(x2-x1)}
= 3 + {[3⁄5]×(8-3)} = 3 + 3 = 6
3. y coordinate of P = y1 + {[m⁄k]×(y2-y1)}
= 2 + {[3⁄5]×(7-2)} = 2 + 3 = 5
• So coordinates of Q are (6,5)
■ Fig.34.12 below shows the actual positions in the Cartesian plane:
 |
| Fig.34.12 |
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