Section Formula
Suppose a telephone company wants to position a relay tower at P between A and B is such a way that the distance of the tower from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1 : 2(see Fig. 7.9). If we take A as the origin O and 1 km as one unit on both the axis, the coordinates of B will be (36, 15). In order to know the position of the tower, we must know the
coordinates of P. How do we find these coordinates.
 Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the xaxis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by the AA similarity criterion, studied in Chapter 6, Δ POD and Δ BPC are similar.
 Therefore ,`(OD)/(PC)` = `(OP)/( PB)` = `1/2`
 and `(PD)/(BC)` = `(OP)/(PB)` = `1/2`
 So,` (X)/(36X)= 1/2` and `(y)/(15y)= 1/2` .
 These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB = 1 : 2.
 Consider any two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) and assume that P (x, y) divides AB internally in the ratio m_{1} : m_{2}, i.e.,
`(PA)/(PB)= m_1/m_2`( Fig. 7.10).
 Draw AR, PS and BT perpendicular to the xaxis. Draw AQ and PC parallel to
the xaxis. Then, by the AA similarity criterion,
 Δ PAQ ~ Δ BPC
Therefore, `(PA)/(BP) = (AQ)/(PC) = (PQ)/(BC)` (1)
Now, AQ = RS = OS – OR = x – x_{1}
PC = ST = OT – OS = x_{2} – x
PQ = PS – QS = PS – AR = y – y_{1}
BC = BT– CT = BT – PS = y_{2} – y
 Substituting these values in (1), we get
`m_1/m_2 = ((xx_1)/(x_2x))=((yy_1)/(y_2y))`
`m_1/m_2 = ((xx_1)/(x_2x))`, we get `x = (m_1x_2+ m_2 x_1)/(m_1+m_2)`
`m_1/m_2 =((yy_1)/(y_2y))`, we get `y = (m_1y_2+ m_2 y_1)/(m_1+m_2)`
 So, the coordinates of the point P(x, y) which divides the line segment joining the
points A(x_{1}, y_{1}) and B(x_{2}, y_{2}), internally, in the ratio m_{1} : m_{2} are
`[(m_1 x_2+ m_2 x_1)/(m_1+m_2),(m_1y_2+ m_2 y_1)/(m_1+m_2)]`
This is known as the section formula.
Example 6 :
Find the coordinates of the point which divides the line segment joining the points
(4, –3) and (8, 5) in the ratio 3 : 1 internally.
Solution :
Let P(x, y) be the required point. Using the section formula, we get
x =`{3(8)+ 1(4)}/{3+1}`=7
y =`{3(5)+1(–3)}/{3+1}`=3
Therefore, (7, 3) is the required point.
Example 7 :
In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?
Solution : Let (– 4, 6) divide AB internally in the ratio m_{1} : m_{2}. Using the section formula, we get
`(– 4, 6) = [(3m_16m_2)/(m_1+m_2),(8m_1+10m_2)/(m_1+m_2)]`
Recall that if (x, y) = (a, b) then x = a and y = b.
So, `– 4 = (3m_16m_2)/(m_1+m_2) and 6 = (8m_1+10m_2)/(m_1+m_2)`
`4 =(3m_16m_2)/(m_1+m_2) `gives us
`4m_1 – 4m_2 = 3m_1 – 6m_2`
`2m_2 = 7m_1`
i.e., m_{1} : m_{2} = 2 : 7
You should verify that the ratio satisfies the y coordinate also.
Now, ` (8m_1+10m_2)/(m_1+m_2)`
` = (8(m_1/m_2)+10)/((m_1/m_2)+1) ` (Dividing throughout by m_{2} )
= `[8(2/7)+10]/[(2/7)+1]` `[here (m_1/m_2)=2/7]`
=`[[16+70]/7] /[9/7]`
= `(54)/9`
=6
 Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and
B(3, – 8) in the ratio 2 : 7.
Alternatively : The ratio m_{1} : m_{2} can also be written as `(m_1)/(m_2)`:1 ,or k : 1. Let (– 4, 6)
divide AB internally in the ratio k : 1. Using the section formula, we get
 `(– 4, 6) = ((3k 6) /(k+18),(8k+10)/(k+1))`(2)
`4=(3k6)/(k+1)`
So, `– 4 = (3k  6)/(k+1`)
i.e., – 4k – 4 = 3k – 6
i.e., 7k = 2
i.e., k : 1 = 2 : 7
 You can check for the ycoordinate also.
 So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and
B(3, – 8) in the ratio 2 : 7.
Note :
You can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear.
Example 8 :
Find the coordinates of the points of trisection (i.e., points dividing in
three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).
Solution : Let P and Q be the points of
trisection of AB i.e., AP = PQ = QB
(see Fig. 7.11).
 Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by
applying the section formula, are
 `[(1( 7)+ 2(2))/(1+2), (1(4)+ 2( 2))/(1+2)]`, i.e(I,0)
 Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
`[(2( 7)+ 1(2)) /(2+1) ,(2(4) +1( 2)) / (2+1)]`, i.e., (– 4, 2)
 Therefore, the coordinates of the points of trisection of the line segment joining A and
B are (–1, 0) and (– 4, 2).
Note :
We could also have obtained Q by noting that it is the midpoint of PB. So, we could have obtained its coordinates using the midpoint formula.
Example 9 :
Find the ratio in which the yaxis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.
Solution:
Let the ratio be k : 1. Then by the section formula, the coordinates of the
point which divides AB in the ratio k : 1 are ` [(k+5)/(k+1) ,(4k6)/(k+1)]`
 This point lies on the yaxis, and we know that on the yaxis the abscissa is 0.
 Therefore, `[(k+5)/(k+1)]=0`
So, k =5
 That is, the ratio is 5 : 1.Putting the value of k = 5, we get the point of intersection as
{ 0,`(13)/3`.}
Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a
parallelogram, taken in order, find the value of p.
Solution :
We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the midpoint of AC = coordinates of the midpoint of BD
i.e.,
 `[(6+ 9)/2, (1+ 4)/2] =[(8+p)/2,(2+3)/2]`
 `[(15)/2,5/2] =[(8+p)/2,(2+3)/2]`
 `(15)/2 =(8+p)/2`
i.e., p =7
Area of a Triangle
 The area of a triangle when its base and corresponding height (altitude) are given. You have used the formula :
Area of a triangle =`1/2` × base × altitude
 you have also studied Heron’s formula to find the area of a triangle.Now, if the coordinates of the vertices of a triangle are given, Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula. But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.
 Let ABC be any triangle whose vertices are A(x _{1}, y _{1}), B(x_{2}, y_{2}) and C(x _{3}, y_{3}). Draw AP, BQ and CR perpendiculars from A, B and C,respectively, to the xaxis. Clearly ABQP, APRC and BQRC are all trapezia(see Fig. 7.13).
 Now, from Fig. 7.13, it is clear that area of Δ ABC = area of trapezium ABQP + area of trapezium APRC– area of trapezium BQRC.
 Area of a trapezium = `1/2` (sum of parallel sides)(distance between them)
Therefore,Area of Δ `ABC =1/2 (BQ + AP) QP +1/2 (AP + CR) PR – 1/2 (BQ + CR) QR`
`=1/2 (y_2+y_1)(x_1x_2)+1/2(y_1+y_2)(x_3x_1)1/2(y_2+y_3)(x_3x_2)`
 =`1/2[x_1(y_2y_3)+x_2(y_3y_1)+x_3(y_1y_3)]`
Thus, the area of Δ ABC is the numerical value of the expression
`1/2[x_1(y_2y_3)+x_2(y_3y_1)+x_3(y_1y_3)]`
Example 11 :
Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5).
Solution :
The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and
C (–3, –5), by using the formula above, is given by
1/2[ (6+ 5)+ ( 4) ( 5+ 1)+ ( 3) ( 1 6) ]
= 1 (11+ 16+ 21)= 24
So, the area of the triangle is 24 square units.
Example 12 :
Find the area of a triangle formed by the points A(5, 2), B(4, 7) and
C (7, – 4).
solution
–is by= `1/2[5(7+4)+4(42)+7(27)]`
=`1/2(552435)=4/2=2`
 Since area is a measure, which cannot be negative, we will take the numerical value
of – 2, i.e., 2. Therefore, the area of the triangle = 2 square units.
Example 13 :
Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4).
Solution:
The area of the triangle formed by the given points is equal to
If the area of a triangle is 0 square units, then its vertices will be collinear.
`=1/2[1.5( 2 4)+ 6(4 3)+ ( 3)(3+ 2)]`
`= 1/2 (9+ 615)= 0`
If the area of a triangle is 0 square units, then its vertices will be collinear.
Example 14 :
Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.
Solution :
Since the given points are collinear, the area of the triangle formed by them must be 0,
i.e.,`1/2[(k+ 3)+4 ( 3 3)+ 6(3k )]=0`

`1/2(4k)=0`
 Therefore, k = 0
 area of Δ `ABC = 1/2[ 2(0+ 3) +4 ( 3 3)+ 6 (3 0)]= 0`
Example 15 :
If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral,
find the area of the quadrilateral ABCD.
Solution :
By joining B to D, you will get two triangles ABD and BCD.
 Now the area of Δ ABD = `1/2[ 5( 5 5)+( 4)(5 7)+ 4 (7+ 5)]`
= `1/2`[ 50+8+48) = `106/2` = 53 square units
 The area of Δ BCD = `1/2[4( 6 5) – 1(5+ 5) + 4( 5+ 6)]`
= `1/2`(4410+4) = 19 square units
So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.
Note :
To find the area of a polygon, we divide it into triangular regions, which have no common area, and add the areas of these regions.
Example 16 : If two vertices of an quadrilateral triangle be(0, 0), (3,√3),find the third vertex.
Solution:
O(0,0) and A(3,3) be the given points and let B(x,y) be the third vertex of equilateral triangleOAD.then,
 OA=0B=AB
 OA^{2 }=OB^{2 }=AB^{2 }
WE have OA=(30)^{2 }+(`sqrt3`0)^{2 }=12
0B^{2 }=X^{2 }+Y^{2 }
 AB^{2}=(X3)^{2 }+(Y`sqrt3`)^{2 }
 AB^{2 }=X^{2 }+Y^{2 }6X2`sqrt3`Y+12
OA^{2 }=OB^{2 } AND OB^{2 }=AB^{2 }
X^{2 }+y^{2 }=12
X^{2 }+y^{2 }=x^{2 }+y^{2 }6x2√3Y+12
X^{2 }+y^{2 }=12 and 6x+2√3Y=12
[3x+`sqrt3`Y=6 therefore `y=(63x)/sqrt3`]
`X^2+[(63x)/sqrt3]^2=12`
`[X^2+(63x)^2]/3=12`
`3x^2+(63x)^2 =36`
`12x^236x=0`
`X=0,3`
 X=0; `sqrt3Y=6`
`Y=(6/sqrt3)=2/sqrt3`
 X=3;`9+sqrt3y=6`
`y=((69)/sqrt3)=sqrt3`
Hence the coordinate of the third vertex B are`(0,2sqrt3) or,(3,sqrt3)`.