Chapter 34.4 - Solved examples on Section formula

In the previous section we saw the section formula. We also saw some examples. In this section we will see a few more solved examples.

Solved example 34.6
The coordinates of the vertices of a quadrilateral are (2,1), (5,3), (8,7) and (4,9) in order.
(i) Find the coordinates of the midpoints of all sides
(ii) Prove that the quadrilateral with these midpoints as vertices is a parallelogram
Solution:
• Let us name the coordinates:
A(2,1), B(5,3), C(8,7), D(4,9)
1. Now we will calculate the midpoints:
(i) Midpoint of AB (Let us call it P):
[^(x1+x2)⁄₂,^(y1+y2)⁄₂] = [⁽²⁺⁵⁾⁄₂,⁽¹⁺³⁾⁄₂] = [⁷⁄₂,⁴⁄₂] = (3.5,2)
(ii) Midpoint of BC (Let us call it Q):
[⁽⁵⁺⁸⁾⁄₂,⁽³⁺⁷⁾⁄₂] = [¹³⁄₂,¹⁰⁄₂] = (6.5,5)
(iii) Midpoint of CD (Let us call it R):
[⁽⁸⁺⁴⁾⁄₂,⁽⁷⁺⁹⁾⁄₂] = [¹²⁄₂,¹⁶⁄₂] = (6,8)
(iv) Midpoint of DA (Let us call it S):
[⁽⁴⁺²⁾⁄₂,⁽⁹⁺¹⁾⁄₂] = [⁶⁄₂,¹⁰⁄₂] = (3,5)
2. We took the vertices A, B C and D in order.
• So P, Q, R and S are the midpoints taken in order:
P(3.5,2), Q(6.5,5), R(6,8), S(3,5)
• We have to prove that these points are vertices of a parallelogram
■ If both pairs of opposite sides are equal, then the quadrilateral is a parallelogram. 
3. In our case:
• One pair of opposite sides is [PQ, RS]
• The other pair of opposite sides is [QR,SP]
(A rough sketch of a quadrilateral will help to identify opposite sides)
4. We will use the distance formula to calculate the distances
• Length of PQ = √[(6.5-3.5)² + (5-2)²] = √[(3)² + (3)²] = √[2×(3)²] = 3√2
• Length of RS = √[(3-6)² + (5-8)²] = √[(-3)² + (-3)²] = √[2×(-3)²] = 3√2
• Length of QR = √[(6-6.5)² + (8-5)²] = √[(-0.5)² + (3)²] = √[0.25 + 9] = √9.25
• Length of SP = √[(3.5-3)² + (2-5)²] = √[(0.5)² + (-3)²] = √[0.25 + 9] = √9.25
■ Thus we find that PQ = RS AND QR = SP
So both pairs of opposite sides are equal. Thus it is a parallelogram
• The actual points in the Cartesian plane are shown in fig.34.13 below:

Fig.34.13

Solved example 34.7
In the fig.34.14(a) below, the midpoints of the sides of the large quadrilateral ABCD are joined to draw the small quadrilateral PQRS inside

Fig.34.14

(i) Find the coordinates of the fourth vertex P of the small quadrilateral
(ii) Find the coordinates of the other three vertices B, C and D of the large quadrilateral
Solution:
• We have seen in an earlier chapter that, the quadrilateral obtained by joining the midpoints of any other quadrilateral will always be a parallelogram
    ♦ See solved example 18.13 in section 18.7
• So PQRS is a parallelogram. We have seen the method to find the unknown vertex of a parallelogram when all the other three vertices are given
1.So the first step is to find the coordinates of P
(i) Group the vertices into two: [P,Q] and [R,S]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach S from R:
• First travel 3 units horizontally to the left [∵ (3-6) = -3]
• Then travel 3 unit vertically downwards [∵ (3-6) = -3]
(iii) The same procedure of travel must be followed for the travel from Q to P
• First travel 3 units horizontally to the left
    ♦ At the end of this travel, the coordinates will be (6,5) [∵ (9-3) = 6]
• Then travel 3 unit vertically downwards
    ♦ At the end of this final lap, the coordinates will be (6,2) [∵ (5-3) = 2]
■ So the coordinates of P are (6,2)
2. P is the midpoint of AB
(i) Let the coordinates of A be (x1,y1)
    ♦ They are already given as (2,1)
(ii) Let the coordinates of B be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint P = ^(x1+x2)⁄₂ = ^(2+x2)⁄₂ = 6
⟹ (2+x2) = 12 ⟹ x2 = 10
• y coordinate of the midpoint P = ^(y1+y2)⁄₂ = ^(1+y2)⁄₂ = 2
⟹ (1+y2) = 4 ⟹ y2 = 3
(iv) Thus coordinates of B are (10,3)
3. Q is the midpoint of BC
(i) Let the coordinates of B be (x1,y1)
    ♦ They are already obtained as (10,3)
(ii) Let the coordinates of C be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint Q = ^(x1+x2)⁄₂ = ^(10+x2)⁄₂ = 9
⟹ (10+x2) = 18 ⟹ x2 = 8
• y coordinate of the midpoint Q = ^(y1+y2)⁄₂ = ^(3+y2)⁄₂ = 5
⟹ (3+y2) = 10 ⟹ y2 = 7
(iv) Thus coordinates of C are (8,7)
4. R is the midpoint of CD
(i) Let the coordinates of C be (x1,y1)
    ♦ They are already obtained as (8,7)
(ii) Let the coordinates of D be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint R = ^(x1+x2)⁄₂ = ^(8+x2)⁄₂ = 6
⟹ (8+x2) = 12 ⟹ x2 = 4
• y coordinate of the midpoint R = ^(y1+y2)⁄₂ = ^(7+y2)⁄₂ = 6
⟹ (7+y2) = 12 ⟹ y2 = 5
(iv) Thus coordinates of D are (4,5)
• The actual points in the Cartesian plane are shown in fig.34.14(b) above

Solved example 34.8
The vertices of a triangle ABC are points with coordinates (3,5), (9,13) and (10,6). Prove that it is an isosceles triangle. Calculate it's area.
Solution:
1. Fig.34.15(a) below shows a rough sketch

Fig.34.15

• We will use the distance formula to calculate the distances
• Length of AB = √[(9-3)² + (13-5)²] = √[(6)² + (8)²] = √[36 + 64] = √[100] = 10
• Length of BC = √[(10-9)² + (6-13)²] = √[(1)² + (7)²] = √[1 + 49] = √[50] = √[2×(5)²] = 5√2
• Length of AC = √[(10-3)² + (6-5)²] = √[(7)² + (1)²] = √[49 + 1] = √[50] = √[2×(5)²] = 5√2
2. So two sides AC and BC are equal. It is an isosceles triangle
• The meeting point of the two equal sides will be the apex. 
• And the third side will be the base
• So C is the apex and AB is the base
3. In an isosceles triangle, if we drop a perpendicular from the apex to the base, then the foot of the perpendicular will be the midpoint of the base. 
• Let this midpoint be D
• So we must find the coordinates of D. We can use the midpoint formula:
• x coordinate of the midpoint D = ^(x1+x2)⁄₂ = ⁽³⁺⁹⁾⁄₂ = 6
• y coordinate of the midpoint D = ^(y1+y2)⁄₂ = ⁽⁵⁺¹³⁾⁄₂ = 9
Thus the coordinates of D are: (6,9)
4. Now, CD is the altitude. We can use the distance formula to find it's length:
• Length of CD = √[(6-10)² + (9-6)²] = √[(-4)² + (3)²] = √[16 + 9] = √[25] = 5
5. Area of the triangle = ¹⁄₂ × base × altitude 
= ¹⁄₂ × AB × CD = ¹⁄₂ × 10 × 5 = 25 square units
• The fig.34.15(b) above shows the actual positions of the points in the Cartesian plane

Solved example 34.9
The coordinates of the vertices of a triangle are (-1,5), (3,7), (3,1). Find the coordinates of it's centroid
Solution:
Fig.34.16(a) below shows a rough sketch.

Fig.34.16

• D is the midpoint of AB. This midpoint is joined to the opposite vertex C. So CD is a median
• E is the midpoint of AC. This midpoint is joined to the opposite vertex B. So BE is a median
• The point of intersection of the two medians will give the centroid of the triangle ABC (Details here)
• Now we can write the steps:
1. Coordinates of D using the midpoint formula:
• x coordinate = ^(x1+x2)⁄₂ = ^(-1+3)⁄₂ = 1
• y coordinate = ^(y1+y2)⁄₂ = ⁽⁵⁺¹⁾⁄₂ = 3
2. Coordinates of E using the midpoint formula:
• x coordinate = ^(x1+x2)⁄₂ = ^(-1+3)⁄₂ = 1
• y coordinate = ^(y1+y2)⁄₂ = ⁽⁷⁺⁵⁾⁄₂ = 6
3. The centroid will divide the median in the ratio 2:1 measured from the vertex
• Let us divide the median CD in the ratio 2:1
    ♦ This ratio 2:1 is measured from the vertex. 
    ♦ So the segment corresponding to 2 will lie near the vertex C of the triangle
    ♦ And the segment corresponding to 1 will lie near the base side AB
4. We will use the section formula:
• m:n is 2:1
• k = 2+1 = 3
• The segment corresponding to 'm' lies near (x1,y1)
    ♦ So the segment corresponding to '2' lies near (x1,y1)
    ♦ Thus coordinates of C must be taken as (x1,y1) and coordinates of D must be taken as (x2,y2)  
5. Then x coordinate of the centroid = x1 + {[^m⁄_k]×(x2-x1)} 
= 3 + {[²⁄₃]×(1-3)} 
= 3 + {[²⁄₃]×(-2)} 
= 3 - ⁴⁄₃ = ⁵⁄₃
6. y coordinate of the centroid = y1 + {[^m⁄_k]×(y2-y1)} 
= 7 + {[²⁄₃]×(3-7)}
= 7 + {[²⁄₃]×(-4)} 
= 7 - ⁸⁄₃ = ¹³⁄₃
• So coordinates of the centroid are (⁵⁄₃,¹³⁄₃)
Check:
We can check the result by using the other median BE:
• Let us divide the median BE in the ratio 2:1
    ♦ This ratio 2:1 is measured from the vertex. 
    ♦ So the segment corresponding to 2 will lie near the vertex B of the triangle
    ♦ And the segment corresponding to 1 will lie near the base side AC
1. We will use the section formula:
• m:n is 2:1
• k = 2+1 = 3
• The segment corresponding to 'm' lies near (x1,y1)
    ♦ So the segment corresponding to '2' lies near (x1,y1)
    ♦ Thus coordinates of B must be taken as (x1,y1) and coordinates of E must be taken as (x2,y2)  
2. Then x coordinate of the centroid = x1 + {[^m⁄_k]×(x2-x1)} 
= 3 + {[²⁄₃]×(1-3)} 
= 3 + {[²⁄₃]×(-2)} 
= 3 - ⁴⁄₃ = ⁵⁄₃
3. y coordinate of the centroid = y1 + {[^m⁄_k]×(y2-y1)} 
= 1 + {[²⁄₃]×(6-1)}
= 1 + {[²⁄₃]×(5)} 
= 1 + ¹⁰⁄₃ = ¹³⁄₃
• So coordinates of the centroid are (⁵⁄₃,¹³⁄₃)
• These are the same coordinates that we obtained earlier
• The actual positions of the points in the Cartesian plane are shown in fig.34.16(b) above 

Solved example 34.10
The centre of a circle is (1,2) and (3,2) is a point on it. Find the coordinates of the other end of the diameter through this point
Solution:
1. Imagine a small horizontal line 
• Horizontal because y coordinates are same 
2. Put O(1,2) at it's center
3. Put the coordinates B(3,2) at the right end 
• B is towards right of O because x coordinate of B is greater than that of O
4. Let the left end be A
• We have to find the coordinates of A
5. AB is the diameter and O is the centre
• So OA = OB
6. Since OB is horizontal, it's length can be calculated from x coordinates alone. 
• That is., OB = (3-1) = 2 units
7. So OA = 2 units.
• So A is 2 units to the left of O
• Thus it's x coordinate will be (1-2) = -1
• y coordinate will be same as that of O and B, which is 2
• So the coordinates of A are (-1,2)

Solved example 34.11
Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6)
Solution:
• Let A(x1,y1) = (-3,10)
• Let B(x2,y2) = (6,-8)
• Let P(-1,6) divide AB in the ratio m:n
1. Then x coordinate of P =  x1 + {[^m⁄_k]×(x2-x1)} = -1
⟹ -3 + {[^m⁄_k]×(6-(-3))} = -1
⟹ -3 + {[^m⁄_k]×(9)} = -1
⟹ {[^m⁄_k]×(9)} = 2 ⟹ {[^m⁄_k]} = ²⁄₉
⟹ {[^m⁄_(m+n)]} = ²⁄₉ ⟹ 9m = 2m + 2n ⟹ 7m = 2n
⟹ ^m⁄_n = ²⁄₇. That is., m:n = 2:7
Check:
Considering the y coordinates:
y coordinate of P =  y1 + {[^m⁄_k]×(y2-y1)} = 6
⟹ 10 + {[^m⁄_k]×(-8-10)} = 6
⟹ 10 + {[^m⁄_k]×(-18)} = 6
⟹ {[^m⁄_k]×(-18)} = -4 ⟹ {[^m⁄_k]} = ²⁄₉
⟹ {[^m⁄_(m+n)]} = ²⁄₉ ⟹ 9m = 2m + 2n ⟹ 7m = 2n
⟹ ^m⁄_n = ²⁄₇. That is., m:n = 2:7
• This is the same ratio as above.
• The actual positions of the points in the Cartesian plane are shown in the fig.34.17 below:

Fig.34.17

Another method of check:
1. Distance AB = √[(6-(-3))² + (-8-10)²] = √[(9)² + (-18)²] = √[81 + 324] = √[405] = √[5×9×9] = 9√5
2. Distance AP = √[(-1-(-3))² + (6-10)²] = √[(2)² + (-4)²] = √[4 + 16] = √[20] = √[5×2×2] = 2√5  
3. If AP is ²⁄₉ of AB, our calculations are correct
²⁄₉ of AB = ²⁄₉ × 9√5 = 2√5
4. From (2) we see that AP is indeed 2√5
• So our calculations are correct

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In the next section, we will see Lines.

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Chapter 34.3 - The section formula

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:

Fig.34.10

• 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

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■ 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

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1. Consider the line AB in fig.34.11(a) below:

Fig.34.11

• 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 90^o
• Since the green lines are parallel to the axes, the angle at B' and P' will also be 90^o
• 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: α^o, β^o and 90^o.
■ 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)}

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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}

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■ 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) 

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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

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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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In the next section, we will see a few more solved examples.

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