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).
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We know
that any point can be represented in terms of co-ordinates of x and y. Let P
(x1,y1) and Q (x2,y2)
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 = x1 and Draw PC
and QD horizontal to x axis from P and Q respectively. Note
that OC = y1 and OD = y2 Produce
CP to meet BQ at R. PR =
OB-OA = x2-x1 QR =
OD-OC = y2-y1 Since PQ2 = PR2+RQ2= (x2-x1)2+
(y2-y1)2
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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
= SQRT (x2+ y2)
7.2
Problem 1: Find
the value of k if P(0,2) is equidistant from Q(3,k)
and R(k,5)
Solution:
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PQ = SQRT {(3-0)2+ (k-2)2} = SQRT(9 +k2-4k+4) PR = SQRT {(k-0)2+ (5-2)2} = SQRT( k2+9) Since
PQ=PR it follows that 9 +k2-4k+4
= k2+9 On
simplification we get k = 1 Thus the point P(0,2) is
equidistant from Q(3,1) and R(1,5). |
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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 = SQRT {(3-10)2+ (6-(-18))2} = SQRT (49+ 576) = SQRT (625) =25
AC = SQRT {(-5-10)2+ (2-(-18))2} = SQRT (225+ 400) = SQRT (625) =25
BC = SQRT {(-5-3)2+ (2-6)2} = SQRT
(64+ 16) = SQRT (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
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Hint: Show that BA+AC = BC (Verify by
plotting points that they are collinear) |
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7.2
Problem 4: Find the co-ordinates of the Circumcenter of a
triangle ABC whose vertices are A(4,6),B(0,4) and
C(6,2)
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Hint: Let O(x,y) be the Circumcenter. Then
OA= Solution: OA2
= (x-4)2+(y-6)2 OB2
= (x-0)2+(y-4)2 OC2
= (x-6)2+(y-2)2 Equating
OA2 = OB2 We get 2x+y =9 Equating
OA2 = OC2 We get x-2y = -3 By
solving the above two equations we get x=3 and y=3. Hence O(3,3) is the Circumcenter of |
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7.2.2 Section
Formula:
This section is about finding a point on a line such that
the point divides the line in the given ratio.
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Let AB
be the line joining the point A (x1, y1) and B(x2,
y2). We are required
to find a point P(x, y) on the line AB such that it divides AB in the given
ratio of m1:m2. From A,
P and B draw perpendiculars to the x-axis 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 m1:m2,
then AP/PB = m1/m2 It is
clear from the adjacent figure that, AE/PR = PE/BR=AP/PB = m1/m2
--------à(1) AE =
OQ-OC = x-x1 : PR =
OD-OQ = x2-x PE =
QP-QE(=CA) = y-y1 BR =
DB-DR = y2-y Thus by
substituting values in (1) we get AE/PR = (x-x1)/(x2-x) = PE/PR
= (y-y1)/ (y2-y) = m1/m2 --------à(2) By taking
first and last expression in (2), we have (x-x1)/(x2-x)
= m1/m2
Similarly,
by taking second and last expression in (2) we get y = (m1y2+ m2y1)/(m2+m1) Thus
the co-ordinates of point P, which divides the line joining points A(x1,
y1) and B(x2, y2) in the ratio of m1:m2
are given by: {(m1x2+ m2x1)/(m1+m2), (m1y2+ m2y1)/(m1+m2) } This is
known as ‘section formula’. |
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Dividing the line in
the given ratio:
1. What are the co-ordinates of midpoint of AB (i.e. when
m1:m2 = 1:1)?
It is {(x2+x1)/2),
(y2+ y1)/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 co-ordinates of the point which divides
the line in the ratio of k:1?
It is {(kx2+x1)/(k+1), (ky2+ y1)/(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:
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Let P divide
the line AB in the ratio of k:1 x1=15,
y1=5, x2=9, y2=20,x=11, y=15 We have
seen above that the co-ordinates of the point which divides a line in the
ratio of k:1 is {(kx2+x1)/(k+1),
(ky2+ y1)/(k+1)}
Hence the
point divides the given line in the ratio of 2:1 |
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it is given that x=11)
7.2
Problem 6: Find the co-ordinates of points which trisects
the line joining A(6,-2) and B(-8,10)
Hint:
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We are
required to find the co-ordinates of P and Q such that AP=PQ=QB (1:1:1) This
problem needs to be solved in two steps: 1. Find the co-ordinates of P(x1,y1) such that AP:PB = 1:2. 2. Find the co-ordinates of Q(x2,y2)
such that AQ:QB = 2:1 They
are P (4/3,2) and Q (-10/3,6) |
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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 co-ordinates of A,B and
C.
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Hint: Let A
=(x1,y1), B=(x2,y2) and C=(x3,y3) Since D
is mid point of AB, (x1+x2)/2 = -2 and (y1+y2)/2
= 5 Since E
is mid point of BC, (x2+x3)/2 = 2 and (y2+y3)/2
= 4 Since F
is mid point of AC, (x1+x3)/2 = -1 and (y1+y3)/2
= 2 By
solving these three equations we get x1=
-5, x2=1, x3= 3 y1=
3, y2=7, y3= 1 Thus
the vertices are A(-5,3), B(1,7) and C(3,1) |
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Co-ordinates of
centroid:
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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 co-ordinates of the
centroid. In the
adjoining figure A (x1, y1), B(x2, y2)
and C(x3, y3) AD is
one median and G is the centroid which divides AD in the ratio 2:1. We are
required to find the co-ordinates of G. Since
AD is median BD = DC
Since
G(x,y) divides AD in 2:1 (m1=2 and m2=1)
We have x = {2(x2+x3)/2)+1. x1}/(2+1) = (x2+x3+x1)/3 y = {2(y2+y3)/2)+1. y1}/(2+1) = (y2+y3+y1)/3 Thus
the co-ordinates of centroid are
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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(x3, y3) be the third vertex
We know that the co-ordinates of centroid are
x = (x1+x2+x3)/3 and
y = (y1+y2+y3)/3.
Since the co-ordinates of origin are (0,0)
It follows that
x1+x2+x3 =
0 and y1+y2+y3
= 0
By substituting the values for co–ordinates we have
x1+x2+x3=
-3+0+ x3=0
x3 = 3
y1+y2+y3= 1-2+ y3=0
y3 = 1
Thus (3,1) is the third vertex.
Area of a triangle
given the co-ordinates of a triangle
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As in
the adjoining figure let A (x1, y1), B(x2, y2)
and C(x3, y3) 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 x-axis.
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(y2+
y1) (x1- x2) + 1/2(y1+ y3)
(x3- x1) - 1/2(y2+ y3) (x3-
x2) = 1/2[x1(y2- y3) + x2(y3-
y1) +x3(y1- y2)] --------- (By
rearranging the terms) Note
that if A B and C are collinear then area is zero. |
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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.
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Let us calculate the area of Area of =1/2[-3+16+35] =1/2(48) = 24 sq units Since the area of Area of |
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7.2 Summary of learning
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No |
Points to remember |
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1 |
The
distance between points P (x1,y1)
and Q (x2,y2) is
= SQRT {(x2-x1)2+
(y2-y1)2} |
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2 |
The co-ordinates of the point which divides A(x1,y1)
and B (x2,y2) in the given ratio of m1:m2
are
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3 |
The
co-ordinates of centroid are
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4 |
Area of
triangle bound by A (x1, y1), B(x2, y2),C(x3,
y3)= 1/2[x1(y2- y3)
+ x2(y3- y1) +x3(y1- y2)] |
