7.2 Distance between two points and Section Formula:
7.2.1
Distance between two points: We have learnt how to plot points on a plane
in a graph sheet. There are many instances where we need to find the distance
between two points (length of the line segment joining two points).
We
know that any point can be represented in terms of coordinates of x and y. Let
P (x_{1},y_{1}) and Q (x_{2},y_{2}) be the
two points. We
are required to find the length of the line PQ. Draw
PA and QB perpendicular to x axis from P and Q respectively. Note
that OA = x_{1} and Draw
PC and QD horizontal to Y axis from P and Q respectively. Note
that OC = y_{1} and OD = y_{2} Produce
CP to meet BQ at R. PR = OBOA = x_{2}x_{1} QR = ODOC = y_{2}y_{1} Since
_{}PRQ is a right angled triangle, by Pythagoras theorem PQ^{2} = PR^{2}+RQ^{2}=
(x_{2}x_{1})^{2}+ (y_{2}y_{1})^{2} _{} PQ = _{} {(x_{2}x_{1})^{2}+ (y_{2}y_{1})^{2}} This formula is called 'distance
formula' 

Corollary: What if one point is the origin (0,0) ?
The distance of a point P (x,y) from the origin
O(0,0) is OP = _{} (x^{2}+ y^{2})
7.2 Problem 1: Find the value of k if P(0,2) is equidistant from
Q(3,k) and R(k,5)
Solution:
PQ = _{} {(30)^{2}+
(k2)^{2}} = _{} (9 +k^{2}4k+4) PR = _{} {(k0)^{2}+
(52)^{2}} = _{} ( k^{2}+9) Since
PQ=PR it follows that 9
+k^{2}4k+4 = k^{2}+9 On
simplification we get k = 1 The point P(0,2) is equidistant
from Q(3,1) and R(1,5). 

7.2 Problem 2: State the special property of the triangle formed
by three points A(10,18), B(3,6) and C(5,2)
Solution:
AB = _{} {(310)^{2}+
(6(18))^{2}} = _{} (49+ 576) = _{} (625) =25 AC = _{} {(510)^{2}+
(2(18))^{2}} = _{} (225+ 400) = _{} (625) =25 BC = _{} {(53)^{2}+
(26)^{2}} = _{} (64+ 16) = _{} (80) Since AB=AC it is clear that the
triangle formed by the given three points is an isosceles triangle 

7.2 Problem 3: Using distance formula show that points A(2,5),
B(1,2) and C(4,7) are collinear
Hint: Show that BA+AC = BC
(Verify by plotting points that they are collinear) 

7.2 Problem 4: Find the
coordinates of the Circumcenter of a triangle ABC whose vertices are A(4,6),B(0,4)
and C(6,2)
Hint: Let O(x,y) be the
Circumcenter. Then OA= Solution: OA^{2}
= (x4)^{2}+(y6)^{2}=x^{2}8x+16+y^{2}12y+36 OB^{2}
= (x0)^{2}+(y4)^{2}=x^{2}+y^{2}8y+16 OC^{2}
= (x6)^{2}+(y2)^{2}=x^{2}12x+36+y^{2}4y+4 Equating
OA^{2} = OB^{2} We get 2x+y =9 Equating
OA^{2} = OC^{2} We get x2y = 3 By
solving the above two equations we get x=3 and y=3. Hence
O(3,3) is the Circumcenter of _{}ABC. 

7.2.2
Dividing the line in the given ratio:
Section
Formula:
This section is about finding a point on a line
such that the point divides the line in the given ratio.
Let
AB be the line joining the point A (x_{1}, y_{1}) and B(x_{2},
y_{2}). We
are required to find a point P(x, y) on the line AB such that it divides AB in
the given ratio of m_{1}:m_{2}. From
A, P and B draw perpendiculars to the xaxis and let these perpendiculars
meet x axis at C,Q and D respectively. From
A and P draw parallel lines to x axis to meet PQ at E and BD at R. If
P is a point on AB dividing it in the given ratio of m_{1}:m_{2},
then AP/PB = m_{1}/m_{2} It
is clear from the adjacent figure that, _{} AEP and _{}PRB are similar (AAA Postulate). Hence AE/PR = PE/BR=AP/PB = m_{1}/m_{2} à(1) AE = OQOC = xx_{1}_{ }: PR = ODOQ = x_{2}x_{ }PE = QPQE(=CA) = yy_{1}_{ }BR = DBDR = y_{2}y_{} Thus
by substituting values in (1) we get AE/PR = (xx_{1})/(x_{2}x) = PE/PR = (yy_{1})/
(y_{2}y) = m_{1}/m_{2} à(2) By
taking first and last expression in (2), we have (xx_{1})/(x_{2}x)
= m_{1}/m_{2} _{} m_{2}(xx_{1}) = m_{1}(x_{2}x) (By cross multiplication)_{ }_{} m_{2}x  m_{2}x_{1} = m_{1}x_{2} m_{1}x (By expansion)_{} _{} x(m_{2}+m_{1}) = m_{1}x_{2}+ m_{2}x_{1}(By transposition)_{ }_{} x = (m_{1}x_{2}+ m_{2}x_{1})/(m_{2}+m_{1})(By division)_{} Similarly,
by taking second and last expression in (2) we get y = (m_{1}y_{2}+ m_{2}y_{1})/(m_{2}+m_{1}) Thus
the coordinates of point P, which divides the line joining points A(x_{1},
y_{1}) and B(x_{2}, y_{2}) in the ratio of m_{1}:m_{2
}are given by: {(m_{1}x_{2}+ m_{2}x_{1})/(m_{1}+m_{2}), (m_{1}y_{2}+ m_{2}y_{1})/(m_{1}+m_{2}) } This
is known as ‘section formula’. 

1. What are the coordinates of midpoint of AB
(i.e. when m_{1}:m_{2} = 1:1)?
It is {(x_{2}+x_{1})/2),
(y_{2}+ y_{1})/2}: (Mid point formula)
Note: The above formula can be used to show that the quadrilateral formed by
joining the midpoints of adjacent sides of a quadrilateral is a parallelogram.
2. What are the coordinates of the point which
divides the line in the ratio of k:1?
It is {(kx_{2}+x_{1})/(k+1),
(ky_{2}+ y_{1})/(k+1)}
7.2 Problem 5: Find the ratio
in which the point P(11,15) divides the line segment joining the points A(15,5)
and B(9,20)
Solution:
Let
P divide the line AB in the ratio of k:1 x_{1}=15,
y_{1}=5, x_{2}=9, y_{2}=20,x=11, y=15 We
have seen above that the coordinates of the point which divides a line in
the ratio of k:1 is {(kx_{2}+x_{1})/(k+1),
(ky_{2}+ y_{1})/(k+1)} _{} x = (kx_{2}+x_{1})/(k+1),
y = (ky_{2}+ y_{1})/(k+1) _{} x = 9k+15/(k+1) _{}11 = 9k+15/(k+1)(_{} it is given that x=11) _{} 11k+11 = 9k+15 _{}2k=4 or k=2 Hence
the point divides the given line in the ratio of 2:1 

7.2 Problem 6: Find the
coordinates of points which trisects the line joining A(6,2) and B(8,10)
Hint:
We
are required to find the coordinates of P and Q such that AP=PQ=QB (1:1:1) This
problem needs to be solved in two steps: 1. Find the coordinates of P(x_{1},y_{1})
such that AP:PB = 1:2. 2. Find the coordinates of Q(x_{2},y_{2})
such that AQ:QB = 2:1 They
are P (4/3,2) and Q (10/3,6) 

7.2 Problem 7: In triangle
ABC, D(2,5) is mid point of AB, E(2,4) is mid point of BC and F(1,2) is mid
point of AC.
Find the coordinates of A,B and C.
Hint: Let
A =(x_{1},y_{1}), B=(x_{2},y_{2}) and C=(x_{3},y_{3}) Since
D is mid point of AB, (x_{1}+x_{2})/2 = 2 and (y_{1}+y_{2})/2
= 5 Since
E is mid point of BC, (x_{2}+x_{3})/2 = 2 and (y_{2}+y_{3})/2
= 4 Since
F is mid point of AC, (x_{1}+x_{3})/2 = 1 and (y_{1}+y_{3})/2
= 2 By
solving these three equations we get x_{1}=
5, x_{2}=1, x_{3}= 3 y_{1}=
3, y_{2}=7, y_{3}= 1 Thus
the vertices are A(5,3), B(1,7) and C(3,1) 

7.2 Summary of
learning
No 
Points to remember 
1 
The
distance between points P (x_{1},y_{1})
and Q (x_{2},y_{2}) is
= _{} {(x_{2}x_{1})^{2}+ (y_{2}y_{1})^{2}} 
2 
The coordinates of the point which divides A(x_{1},y_{1})
and B (x_{2},y_{2}) in the given ratio of m_{1}:m_{2
} are 
Coordinates
of centroid:
We
have learnt that the centroid divides the median in the ratio of 2:1 (Refer
to Section 6.4) Given
the three vertices of a triangle let us calculate the coordinates of the
centroid. In
the adjoining figure A (x_{1}, y_{1}), B(x_{2}, y_{2})
and C(x_{3}, y_{3}) AD
is one median and G is the centroid which divides AD in the ratio 2:1. We
are required to find the coordinates of G. Since
AD is median BD = DC _{}The coordinates of D are {(x_{2}+x_{3})/2), (y_{2}+ y_{3})/2} Since
G(x,y) divides AD in 2:1 (m_{1}=2 and m_{2}=1) We have x
= {2(x_{2}+x_{3})/2)+1.
x_{1}}/(2+1) = (x_{2}+x_{3}+x_{1})/3 y
= {2(y_{2}+y_{3})/2)+1.
y_{1}}/(2+1) = (y_{2}+y_{3}+y_{1})/3 Thus
the coordinates of centroid are 

7.2 Problem 8: Find the third vertex of a triangle if two of its
vertices are at A(3,1) and B(0,2)
and the centroid is at the origin.
Solution:
Let C(x_{3}, y_{3}) be the third
vertex
We know that the coordinates of centroid are
x = (x_{1}+x_{2}+x_{3})/3
and
y = (y_{1}+y_{2}+y_{3})/3.
Since the coordinates of origin are (0,0)
It follows that
x_{1}+x_{2}+x_{3 }= 0 and y_{1}+y_{2}+y_{3 }=
0
By substituting the values for co–ordinates we have
x_{1}+x_{2}+x_{3}= 3+0+ x_{3}=0
_{}x_{3 }= 3
y_{1}+y_{2}+y_{3}= 12+ y_{3}=0
_{}y_{3 }= 1
Thus (3,1) is the third vertex.
Area of a
triangle given the coordinates of a triangle
As
in the adjoining figure let A (x_{1}, y_{1}), B(x_{2},
y_{2}) and C(x_{3}, y_{3}) be the three vertices of a
triangle. We are required to find the area of triangle ABC. Let
BL, AM and CN be perpendiculars from the vertices B, A and C to xaxis. _{} OL = x_{2}, From
the figure Area of Triangle ABC = =Area
of trapezium BLMA + Area of trapezium AMNC  Area of trapezium BLNC =
1/2(BL+AM)LM + 1/2(AM+NC)MN  1/2(BL+NC)LN =
1/2(y_{2}+ y_{1}) (x_{1} x_{2}) + 1/2(y_{1}+
y_{3}) (x_{3} x_{1})  1/2(y_{2}+ y_{3})
(x_{3} x_{2}) =
1/2[x_{1}(y_{2} y_{3})
+ x_{2}(y_{3} y_{1}) +x_{3}(y_{1} y_{2})]  (By rearranging the terms) Note
that if A B and C are collinear then area is zero. 

7.2 Problem 8: If D(3,1), E(2,6), F(5,7) are
the mid points of the sides of _{}ABC, find the area of _{}ABC.
Let us calculate the area of _{}DEF first Area of _{}DEF = 1/2[3(67)+2(7(1))+(5)(16)] =1/2[3+16+35] =1/2(48) = 24 sq units Since the area of _{}ABC is four times the area of _{}DEF (By BPT theorem – Refer 6.13) Area of _{}ABC = 4*24 = 96 sq units 
