coordinate geometry - inyatrust€¦ · - danica mckellar coordinate geometry. 324 unit-13 know...
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This unit facilitates you in,
explaining the meaning of coordinate axes.
drawing coordinate axes.
locating the given points on the coordinate
plane.
identifying that the graph of the equation
ax + by = c represents a straight line.
defining, slope of a straight line.
calculating the slope of a given line.
explaining the meaning of x-intercept of a
line and y - intercept of a line.
deriving the slope-intercept form of the
equation of a straight line.
calculating the distance between two points
on a plane.
calculating the coordinates of a point which
divides the line in a given ratio.
Basic ideas - coordinate axis,
origin.
Writing the coordinates of a
point.
Graph of ax + by = c
[Linear graph].
Slope of a straight line.
Equation of a straight line in
slope-intercept form.
Distance between two points in
a plane.
Distance of a point in the plane
from origin.
Section Formula.
13
Rene Descartes
[1596 - 1650]
He was a french philosopher,
mathematician whose work
'La geometrie' includes his
application of Algebra to Geo-
metry from which we now have
'cartesian geometry'.
Mathematics is the only place where truth and
beauty mean the samething.
- Danica Mckellar
Coordinate Geometry
324 UNIT-13
Know this
The words ABSCISSA and
ORDINATE are Latin words
ABSCISSA means cut off
ORDINATE means keep it in
order.
We commonly observe that, the roads leading to the top
of a hill are always winding roads. There are no roads
laid straight upto the top of the hill. Why?
To answer the above question, the ideas learnt
about rectangular co-ordinate system would be useful.
This branch of mathematics was mainly developed
by French philosopher and mathematician Rene
Descartes [1596 - 1650]
Hence, Rectangular coordinate system has been named as 'Cartesian system' after
him.
This has lead to the branch of mathematics called COORDINATE GEOMETRY OR
ANALYTICAL GEOMETRY. This branch of mathematics treats geometry algebraically.
Descartes developed the idea of placing two number lines perpendicular to each other on
a plane and locate points on this plane.
Let us recall the terms related to cartesian system.
* The two perpendicular lines are called coordinate
axes.
* Horizontal line (XOX) is called x-axis and vertical
line (YOY) is called y-axis.
* The point of intersection of the two axes is called
the origin 'O'.
* Coordinate axes divides the plane into four
quadrants.
* For a point on the cartesian plane,
(i) its perpendicular distance from y-axis is called
the x-coordinate (or ABSCISSA) of the point.
(ii) its perpendicular distance from x-axis is called
the y-coordinate (or ORDINATE) of the point.
* (i) the coordinates of a point P is represented as
P(x,y).
(ii) coordinates of the origin are (0.0)
(iii) coordinates of any point on x-axis will be of the form (x,o)
(iv) coordinates of any point on y-axis will be of the form (o,y)
EXERCISE 13.1
I. Locate the following points on a graph sheet.
(i) P(4, –3) (ii) R(–1, –1) (iii) I(0, –5) (iv) X(–5, –2)
(v) Y(3, 2) (vi) Z(4, 0) (vii) E(0, 6) (viii) F(–2, 5)
XI X
Y
YI
O
I quadrant(–x, +y)
II quadrant(–x, +y)
IV quadrant(+x, –y)
III quadrant(–x, –y)
Coordinate Geometry 325
II. In which quadrants do these points lie?
(i) (4, –6) (ii) (3, 1) (iii) (–10, –2) (iv) x(–5, –2)
(v) (–5, –1) (vi) (5, –7) (vii) (9, 9) (viii) (–2, 7)
III. Plot the points on a cartesian plane whose
(i) Ordinate is 4, abscissa is 0 (ii) Ordinate is –3, abscissa is –1
(iii) Ordinate is 5, abscissa is 4 (iv) Ordiante is –1, abscissa is 7
Graphical representation of equations of the form ax + by = c :
You are familiar with graphs of linear equations. Observe the graphs the following
equations.
(1) 2x + y = 7
Reduce it to standard form y = mx + c
we get, y = 7 – 2x
0 1 2 3
7 5 3 1
x
y
(2) 3x + 2y = 5
Rewrite this equation as
y = 5 3
2
x
0 1 1 3
2.5 1 4 2
x
y
O
8
7
6
5
4
3
2
1
–1
–2
–3
–4
–5
–6
–1–2–3–4–5
(0, 7)
(1, 5)
(2, 3)
(3, 1)
XI
X
Y
YI
O
4
3
2
1
–1
–2
–3
–1–2–3
(3, –2)
(–1, 4)
(0, 2.5)
(1, 1)
XI X
Y
YI
326 UNIT-13
From the above examples, it is evident that, the graph of equations of the form
ax + by = c is always a straight line.
Note : 2 points are sufficient to draw the line graph. One or two more points are
taken for verification or confirmation
Inclination of a straight line
Observe the graphs obtained for the equations 2x + y = 7 and 3x + 2y = 5
The straight lines form an angle with the x-axis. This angle can be measured both
in clockwise and anti-clockwise directions.
The angle formed by the linear graph with the positive direction of x-axis is called
the INCLINATION of that line.
This angle is generally represented by ‘ ’.
The inclination of a straight line is the angle formed at the point of intersection with
the X-axis which is in positive direction. This angle is measured in anti-clockwise
direction from the X-axis to the line.
B
A
oXI X
Y
YI
P
Q
XI X
Y
YI
o
Q
XI X
Y
YI
o
M
N
Note: It is a convention to consider the inclination of a line to be positive when the
angle is measured in anti-clockwise direction from the axis to the line.
Slope of a line
Observe the following instances:
A ladder is placed against a wall such that its top touches the
wall at a certain height from the floor.
While loading trucks, labourers use a wooden plank placed in
an inclined position which helps them to load the truck easily.
Coordinate Geometry 327
The staircase to a building is constructed in a particular position
so that the strain of climbing the staircase is minimum.
In each of the above instances, the position in which the ladder/wooden plank /
stairs placed reduces the strain of climbing..
At what distance the ladder should be placed from the wall so that the strain is
minimised ?
Study the following three figures. Discuss in groups and try to find the solution
mathematically.
(1) (2) (3)C B
A
A
BC BC
A
This can be done by taking the ratio of two lengths i.e., the vertical height and
horizontal distance. Note that the length of ladder is same in all cases.
In the above figures, observe that ABBC and ABC is a right angled triangle.
Here, AB is called the vertical distance (height) and, CB is called the horizontal
distance.
Observe the following figures
A
BC4.7 cm
1.6 cm
A
BC3.5 cm
3.5
cm
A
BC2.6 cm
4.3
cm
Vertical distance
Horizontal distance
AB 1.6cm0.34
BC 4.7 cm
Vertical distance
Horizontal distance
AB 3.5cm1
BC 3.5 cm
Vertical distance
Horizontal distance
AB 4.3cm1.65
BC 2.6 cm
In each example, the ratio obtained is called the slope of the line AC.
328 UNIT-13
The ratio of the vertical distance to the horizontal distance is called slope.
Slope = Vertical distance
Horizontal distance
In the above examples,
Case (1), AB < BC Case (2), AB = BC Case (3), AB > BC.
We observe that,
* When vertical distance is less than the horizontal distance, slope is less than 1.
* When vertical distance is equal to the horizontal distance, slope is equal to 1.
* When vertical distance is more than the horizontal distance, slope is more than 1.
The strain of climbing will be least when the slope is less than 1.
Slope of a line can also be expressed in trigonometric ratio form. Study the following
example.
A
BCHorizontal distance
Vert
ical
dis
tance
The slope of a line is the tangent of the angle of its inclination. It is generally
denoted by 'm'.
m = tan
Gradient of a straight line
Gradient of a straight line is defined as the
increase in
increase in
y
x moving from one point on the line to another.
Gradient = BC
AC
y
x
Hence, Gradient = Vertical height
Horizontal distance
Gradient of a straight line is nothing but the slope of
a straight line.
With respect to ABC,
Slope of AC = AB
BC
AC forms an angle with BC.
With respect to angle ,
AB is the opposite side and BC is the adjacent side
Slope = Opposite side
Adjacent side
Slope = tan
A
B
C
OX
I
YI
Y
X
Coordinate Geometry 329
Note:
1. The X-axis subtends an angle of 0° with itself.
Hence, slope of X-axis is m = tan 0° = 0 [ = 0°]
2. The y-axis subtends an angle of 90° with the X-axis.
Hence, slope of Y-axis is m = tan 90° = not defined [ = 90°]
3. Value of angle of Value of slope
inclination
= 0° 0
0° < < 90° positive number
= 90° not defined
90° < < 180° negative number
Now you must be able to understand why roads leading up the mountains are not
laid straight upto the top.
Discuss with your friends and teacher.
Slope of a straight line passing through two given points
Let A(xI, yI) and B(x2, y
2) be any two fixed points on a
cartesian plane.
Draw AL X-axis, BM X-axis, AN BM
From the figure, OL = x1, OM = x
2
LM = x2 – x
1 = AN
AL = y1 = NM, BM = y
2
BN= y2 – y
1
In ABN, BAN =
with respect to , tan = BN
AN
slope = tan = 2 1
2 1
y y
x x
m = 2 1
2 1
y - y
x - x
Slope of a straight line passing through two given fixed points
= difference of ordinates of given points
difference of their abscissa
You have learnt to find the slope of a straight line. Let us find the slopes of two lines
in the same cartesian plane.
o
y1
( ,y )x2 2
A(,
)
xy 1
1
( – )x2 x1
x2
L x1 MX
IX
Y
YI
y 1
(–
)y
y2
1y
2
B
N ( – )x2 x1
330 UNIT-13
Parallel lines
Two parallel lines make the same inclination
with X-axis. Hence they have same slopes.
= [ corresponding angles]
tan = tan
slope of AB = slope of CD.
Parallel lines have equal slopes.
Conversely, if the slopes of two lines are equal,
the lines are parallel to each other.
Note: * Slope of X-axis is 0.
slope of any line parallel to X-axis is also 0.
* Slope of Y-axis is not defined.
slope of any line parallel to Y-axis is also not defined.
Perpendicular lines
If AB is perpendicular to CD and their inclinations are and respectively.
= 90 + [ exterior angle = sum of interior opposite angles]
tan = tan (90 + )
tan = – cot [ tan (90 + ) = – cot ]
tan = 1
tan θ
tan . tan = – 1
m1 . m
2= –1
where m1 slope of AB
m2 slope of CD.
m1
= 2
1
m or m
2 =
1
1
m
If two lines are mutually perpendicular,
then, the product of their slopes is –1.
Conversely, if the product of the slopes of two lines is –1,
then the lines are mutually perpendicular.
ILLUSTRATIVE EXAMPLES
1. Find the slope of the line whose inclination is 60°.
Sol. = 60°
Slope = m = tanm = tan 60°
m = 3
A
B
C
D
O
XI
X
Y
YI
A
B
C
DO
XI X
Y
YI
Coordinate Geometry 331
2. Find the angle of inclination of straight line whose slope is 0.
Sol. Slope = tan
0 = tan
we know that, tan 0º = 0
= 0°
3. Determine the slope of the line joining the points (3, –2) and (4, 5)
Sol. Let (x1, y
1) = (3, –2) and (x
2, y
2) = (4, 5)
Slope = 2 1
2 1
y y
x x =
5 ( 2)
4 3 =
5 2
1
Slope = 7
4. Prove that the line joining the points (4, 5) and (0, –2) is perpendicular to the
line passing through (2, –3) and (–5, 1).
Sol. For line 1 : Let (x1, y
1) = (4, 5) and (x
2, y
2)= (0, – 2)
Slope = 2 1
2 1
y y
x x =
2 5
0 4
Slope = 7
4 =
7
4 = m
1
For line 2 : (x1, y
1) = (2, –3), (x
2, y
2) = (–5, 1)
Slope =2 1
2 1
y y
x x =
1 ( 3)
5 2 =
1 3
7
Slope = 4
7 =
4
7 = m
2
m1 × m
2=
7 4
4 7= –1
The product of their slopes is –1
The two lines are perpendicular.
5. Find the slope of the line passing through the points P(7, 3) and Q(0, –4). Also
find slope of a line (i) parallel to PQ (ii) perpendicular to PQ.
Sol. Let (x1, y
1) = (7, 3) and (x
2, y
2) = (0, –4)
Slope = 2 1
2 1
y y
x x =
4 3
0 7 =
7
7
Slope = 1
Slope of PQ = 1
(i) Slope of a line parallel to PQ = 1 [ for parallel lines, m1 = m
2]
(ii) Slope of a line perpendicular to PQ = – 1 [ for perpendicular lines, m1 × m
2 = –1]
332 UNIT-13
EXERCISE 13.2
1. Find the slope of the line whose inclination is
(i) 90° (ii) 45° (iii) 30° (iv) 0°
2. Find the angles of inclination of straight lines whose slopes are
(i) 3 (ii) 1 (iii)1
3
3. Find the slope of the line joining the points
(i) (–4, 1) and (–5, 2) (ii) (4, –8) and (5, –2) (iii) (0, 0) and 3,3
(iv) (–5, 0) and (0, – 7) (v) (2a, 3b) and (a, –b)
4. Find whether the lines drawn through the two pairs of points are parallel or
perpendicular
(i) (5, 2), (0, 5) and (0, 0), (–5, 3) (ii) (3, 3), (4, 6) and (4, 1), (6, 7)
(iii) (4, 7), (3, 5) and (–1, 7), (1, 6) (iv) (–1, –2), (1, 6) and (–1, 1), (–2, –3)
5. Find the slope of the line perpedicular to the line joining the points.
(i) (1, 7) and (–4, 3) (ii) (2, –3) and (1, 4)
6. Find the slope of the line parallel to the line joining the points.
(i) (–4, 3) and (2, 5) (ii) (1, –5) and (7, 1)
7. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a)
and (11, 3). Find a.
8. A line passing through the points (1, 0) and (4, 3) is perpendicular to the line joining
(–2, –1) and (m, 0). Find the value of m.
Equation of a line in slope intercept form
Observe the linear graph. The straight line PQ is
intersecting the X-axis at P and Y-axis at Q.
The distance OP i.e., the distance from origin to the point
where the line intersects the X-axis is called the
x-intercept of that line.
The distance OQ i.e., the distance from the origin to the
point where the line intersects the Y-axis is called the
y - intercept of that line.
In the above example, x-intercept is 3 and y-intercept is 4.
Identify the x and y intercepts in the following cases.
O
4
3
2
1
–1
–2
–3
–1–2–3P
Q
XI X
Y
YI
Coordinate Geometry 333
Now, study the following graph.
Let line 'l ' meet the Y-axis at A(0,c).
OA is the y-intercept of the line l.
Let P be any point on line l.
Draw PQ OX and AB PQ,
Let 'l ' makes an angle with X-axis
PAB = (corresponding angles)
Slope = m = tan
m = Opposite side
Adjacent side
In PAB, with respect to ,
m = PB
AB [PB = PQ – BQ, AB = OQ]
m = cy
x
mx = y – c
y = mx + c
The equation of a l ine with slope 'm' and whose y-intercept is 'c'
is given by y = mx + c
ILLUSTRATIVE EXAMPLES
1. Find the equation of a line having slope 1
2 and y-intercept –3.
Sol. m = 1
2, y intercept = c = – 3
Equation : y = mx + c
y = 1
2x + (–3) =
1
2x – 3 y=
6
2
x
2y = x – 6 or x – 2y – 6 = 0
O
4
3
2
1
–1
–2
–3
–4
–1–2XI X
Y
YI
O
5
4
3
2
1
–1
–2
–3
–1–2–3X
IX
Y
YI
O
y
l
A B
Q x
c x
(–
)
yc
P(,
)x
y
(0, c)
XI
X
Y
YI
334 UNIT-13
2. Find the equation of a line having angle of inclination 45° and y-intercept 2.
Sol. = 45°, y-intercept = c = 2
Slope = m = tan m = tan 45°
m = 1
Equation : y = mx + c
y = 1(x) + 2 = x + 2
or x – y + 2 = 0
3. The equation of a line is 3x + 2y + 1 = 0. Find its slope and y-intercept.
Sol. Equation: 3x + 2y + 1 = 0
Rewrite in the standard form of y = mx + c
2y = –3x – 1
Dividing by 2, we get
y = 3 1
2 2x
compare with y = mx + c
m = 3
2, c =
1
2
Slope = 3
2
y - intercept = 1
2
EXERCISE 13.3
1. Find the equation of the line whose angle of inclination and y-intercept are given.
(i) = 60°, y-intercept is – 2 (ii) = 45°, y-intercept is 3.
2. Find the equation of the line whose slope and y-intercept are given.
(i) Slope = 2, y-intercept = – 4
(ii) Slope = 2
3, y-intercept =
1
2
(iii) Slope = –2, y-intercept = 3
3. Find the slope and y-intercept of the lines
(i) 2x + 3y = 4 (ii) 3x = y
(iii) x – y + 5 = 0 (iv) 3x – 4y = 5
4. Is the line x = 2y parallel to 2x – 4y + 7 = 0. [Hint : Parallel lines have same slopes]
5. Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each
other. [Hint : For perpendicular lines, m1m
2 = –1]
Coordinate Geometry 335
Distance between two points
Observe the points A and B. The distance between them is the length of the line
segment joining them.
A B A B
We use a ruler to measure the length of the line segment or the distance between
two points.
Is it possible to find the distance between any two points without actually
measuring ?
It is interesting to study that we can find the
distance between them, if they are located on a cartesian
plane.
If the points A and B lie on X-axis, we can count the
number of unit segments between A and B
AB= 4 cm or
Coordinates of A are (1, 0) and of B (5, 0)
AB= x2 – x
1 = 5 – 1
AB = 4 cm
In general, if the coordinates A and B respectively are
(x1, 0) and (x
2, 0) then AB = x
2 – x
1.
If two points P and Q lie on y-axis. We can count the number
of unit segments between P and Q.
PQ= 6 cm or
Coordinates of P = (0, 4) and of Q = (0, –2)
PQ = y2 – y
1 or PQ = y
1–y
2
= –2 – 4 = –6 = 4 – (–2) = 6
P Q = 6 cm = 6
In general, if the coordinates of P and Q respectively are
(0, y1) and (0, y
2) then PQ = y
2 – y
1
Length of a line parallel to x-axis
If AB is parallel to X-axis.
Draw, AA' X-axis and BB' X-axis
The figure formed AA'B'B is a rectangle
AB = A'B'
A'B' = x2 – x
1 = 6 – 1
A'B' = 5 AB = 5cm
Note: All points on a line parallel to x-axis have the same
y-coordinate.
Remember: Distance is
always a positive number.
O
3
2
1
–1
–2
–1–2XI
X
Y
YI
A B
O
5
4
3
2
1
–1
–2
–3
–1–2
P
Q
XI
X
Y
YI
XI X
YI
–2
–3
Y
O 1 2 3 4 5 6–1–2
–1
5
4
3
1
2
AI BI
336 UNIT-13
Length of a line parallel to Y-axis
If AB is parallel to Y-axis,
Draw AA' Y-axis
BB' Y-axis
The figure formed by AA'B'B is a rectangle
AB = A'B'
A'B' = y2 – y
1
= –3 – 4 = – 7
A'B' = 7 cm AB = 7 cm
Note : Points on a line parallel to Y-axis all have the
same x coordinate.
In all the above examples, where we found the
distance between two points, two conditions are noted,
* the points either lie on X-axis or Y-axis
* the points form a line which is parallel to either X-axis or
Y-axis
Now observe the points P and Q are on the cartesian plane.
Neither of these points lie either on X-axis or on Y-axis.
Let us join them. Line segment PQ is neither parallel to
X-axis nor to Y-axis.
Now, let us learn how to find the distance between 2 points
which neither lie on X or Y-axis nor form a line parallel to them.
Distance formula
If A(x1, y
1) and B(x
2, y
2) are the two given points, draw
BC Y-axis and AC X-axis the co-ordinates of C are (x2, y
1)
AC= x2 – x
1 and BC = y
2 – y
1
ABC is right angled triangle, By Pythagoras theorem,
AB2 = AC2 + BC2
AB2 = (x2 – x
1)2 + (y
2 – y
1)2
AB = 2 2
2 1 2 1x x y y
distance between two points (x1, y
1) and (x
2, y
2)
d = 2 2
2 1 2 1x - x + y - y
d = 2 2difference of absscissa difference of ordinates
Note : For the sake of convenience, we take points lying in I quadrant to derive formulae.
But, all the formulae remain true for points lying in any quadrant.
O
5
4
3
2
1
–1
–2
–3
–1–2
AI
BI
A(2, 4)
B(2, –3)
–3–4X
IX
Y
YI
P
Q
OX
I X
Y
YI
OX
IX
Y
YI
Coordinate Geometry 337
Distance of a point in a plane from the origin
Given A(x, y) and O(0, 0) are the two points,
By distance formula,
d = 2 2
2 1 2 1x x y y
d = 2 2
0 0x y = 2 2x y
ILLUSTRATIVE EXAMPLES
1. Find the distance between the points (2, 3) and (6, 6).
Sol. Let (x1, y
1) = (2, 3) and (x
2, y
2) = (6, 6)
d = 2 2
2 1 2 1x x y y = 2 2
6 2 6 3
d = 2 24 3 = 16 9 = 25
d = 5 units
2. Find the distance between the origin and the point (12, –5).
Sol. Distance between origin and (x, y) = 2 2x y
Here, (x, y) = (12, –5)
d = 2 212 ( 5) 144 25 169
d = 13 units
3. Find the co-ordinates of points on x-axis which are at a distance of 5 units fromthe point (6, –3).
Sol. Any point on X-axis has co-ordinates of form (x, 0)
(x1, y
1) = (6, –3) and (x
2, y
2) = (x, 0) and d = 5
d = 2 2
2 1 2 1x x y y
5 = 22( 6) 0 ( 3)x = 2 2( 6) 3x
Squaring on both sides 52= 2
26 9x
25 = x2 – 12x + 36 + 9
x2 – 12x + 45 – 25 = 0 x2 – 12x + 20 = 0 (x – 10)(x – 2) = 0
x – 10 = 0 or x – 2 = 0 x = 10 or x = 2
Required points on X - axis are (10, 0) and (2, 0)
O(0, 0)
A(x,y)
XI
X
Y
YI
338 UNIT-13
4. Find the value of x such that the distance between the points (2, 5) and
(x, –7) is 13.
Sol. d = 13, (x1, y
1) = (2, 5) and (x
2, x
2) = (x, –7)
d = 2 2
2 1 2 1x x y y
13 = 2 22 ( 7 5)x = 2( 2) 144x
Squaring both sides,132= x2 – 4x + 4 + 144
169 = x2 – 4x +148
x2 – 4x + 148 – 169 = 0
x2 – 4x – 21 = 0
(x – 7) (x + 3) = 0
x = 7 or x = – 3
5. Show that the triangle whose vertices are (8, –4), (9,5) and (0, 4) is an isosceles
triangle.
Sol. To show that the triangle is isosceles, we must show that the length of 2 sides of
the triangle are equal.
Let A(8, –4), B(9, 5), C(0, 4)
d = 2 2
2 1 2 1x x y y
AB = 2 2
9 8 5 ( 4) = 2 21 9 1 81 82
BC = 2 2 2 2(9 0) (4 5) 9 ( 1) 81 1 82
CA = 22 2 2(0 8) 4 ( 4) ( 8) 8 64 64 128
Since AB = BC, the triangle is isosceles.
6. Prove that the points A(–3, –2), B(5, –2), C(9, 3) and D(1, 3) are the vertices of a
parallelogram.
Sol. A quadrilateral is a parallelogram if its opposite sides are equal
d = 2 2
2 1 2 1x x y y
AB = 2 2
5 ( 3) 2 ( 2) = 2 28 0 64 8
CD = 2 2(9 1) (3 3) = 2 28 0 64 8
BC = 2 22 2(5 9) ( 2 3) 4 5 16 25 41
DA = 2 2 2 21 ( 3) 3 ( 2) 4 5 16 25 41
AB= CD and BC = DA
ABCD is a parallelogram.
B(9, 5)
C(0, 4) A(8, –4)
A(–3, –2) B(5, –2)
C(9, 3)
Coordinate Geometry 339
EXERCISE 13.4
1. Find the distance between the following pairs of points
(i) (8, 3) and (8, –7) (ii) (1, –3) and (–4, 7) (iii) (–4, 5) and (–12, 3)
(iv) (6, 5) and (4, 4) (v) (2, 0) and (0, 3) (vi) (2, 8) and (6, 8)
(viii) (a, b) and (c, b) (viii) (cos – sin) and (sin – cos )
2. Find the distance between the origin and the point
(i) (–6, 8) (ii) (5, 12) (iii) (–8, 15)
3. (i) The distance between the points (3, 1) and (0, x) is 5 units. Find x.
(ii) A point P(2, –1) is equidistant from the points (a, 7) and (–3, a). Find 'a'.
(iii) Find a point on y-axis which is equidistant from the points (5, 2) and (–4, 3)
4. Find the perimeter of the triangles whose vertices have the following coordinates
(i) (–2, 1), (4, 6), (6, –3) (ii) (3, 10), (5, 2), (14, 12)
5. Prove that the points A(1, –3), B(–3, 0) and C(4, 1) are the vertices of a right isosceles
triangle.
6. Find the radius of a circle whose centre is (–5, 4) and which passes through the
point (–7, 1).
7. Prove that each of the set of coordinates are the vertices of parallelograms.
(i) (–5, –3), (1, –11), (7, –6), (1, 2) (ii) (4, 0), (–2, –3), (3, 2), (–3, –1)
8. The coordinate of vertices of triangles are given. Identify the types of triangles.
(i) (2, 1) (10, 1) (6, 9) (ii) (1, 6) (3, 2) (10, 8)
(iii) (3, 5) (–1, 1) (6, 2) (iv) (3, –3) (3, 5) (11, –3)
Section Formula
Let AB be a line joining the points A(x1, y
1) and
B(x2, y
2) and point P divides the line segment AB in
the ratio m : n AP
PB=
m
n
Now, let us try to find the coordinates of point P.
Let the coordinates of P be (x, y)
Draw AL, PM and BN, perpendiculars on X-axis.
Also draw, AR PM and PQ BN
From the figure, AR = LM = OM – OL = x – x1
PR = PM – RM = PM – AL = y – y1
PQ = MN = ON – OM = x2 – x
BQ = BN – QN = BN – PM = y2 – y
Observe that, APR ~ PBQ
L M N X
Y
O
R
–
yy
2
Q
B (x ,y )22
(x y)1 1 P y– y1 x– x2
x– x1
Y
X
A
(x, y)
340 UNIT-13
AR PR AP
= =PQ BQ PB
[ corresponding sides of similar s are proportional]
Let us take, AR
PQ =
AP
PB
x - x1
x - x2
= m
n
nx + mx = mx2 + nx
1, x (m + n) = mx
2 + nx
1, x(m + n) = mx
2 + nx
1
x = 2 1mx nx
m n
Consider PR
BQ =
AP
PB
1
2
y y
y y =
m
n
ny – ny1
= my2 – my
ny + my = my2 + ny
1
y(m + n) = my2 + ny
1
y =2 1my ny
m n
coordinates of P = (x, y)
coordinates of P = 2 1 2 1mx + nx my + ny
,m + n m + n
If P is the midpoint of AB, m : n = 1 : 1 then coordinates of P = 2 1 2 1x + x y + y
,2 2
.
This is also called the mid point fomula.
ILLUSTRATIVE EXAMPLES
1. Find the coordinates of point P which divides the line joining A(4, –5) and
B(6, 3) in the ratio 2 : 5.
Sol. Let the coordinates of P be (x, y). Given: m : n = 2 : 5
(x1, y
1) = (4, –5) and (x
2, y
2) = (6, 3)
by section formula,
x = 2 1 2(6) 5(4) 12 20 32
2 5 7 7
mx nx
m n
y = 2 1 2(3) 5( 5) 6 25 19
2 (5) 7 7
my ny
m n coordinates of P =
32 19,
7 7
2. Find the ratio in which point (5, 4) divides the line joining points (2, 1) and (7, 6).
Sol. Let the required ratio be m : n
Given: (x1, y
1) = (2, 1), (x
2, y
2) = (7, 6) and (x, y) = (5, 4)
Coordinate Geometry 341
By section formula,
x = 2 1mx nx
m n or y =
2 1my ny
m n
5 = (7) (2)m n
m n4 =
(6) (1)m n
m n
5 = 7 2m n
m n4 =
6m n
m n
5m + 5n = 7m + 2n 4m + 4n = 6m + n
5m – 7m + 5n – 2n = 0 4m – 6m + 4n – n = 0
–2m + 3n = 0 –2m + 3n = 0
–2m = –3n –2m = –3n
m
n =
3
2
m
n =
3
2
m : n = 3 : 2 m : n = 3 : 2
3. Find the coordinates of the midpoint of the line segments joining the points
(2, 3) and (4, 7).
Sol. (x1, y
1) = (2, 3) and (x
2, y
2) = (4, 7)
by midpoint formula, coordinates of mid point= 1 2 1 2,
2 2
x x y y
= 2 4 3 7
,2 2
= 6 10
,2 2
= (3, 5)
EXERCISE 13.5
1) In what ratio does the point (-2, 3) divide the line segment joining the points
(-3, 5) and (4, -9) ?
2) In the point C(1,1) divides the line segment joining A(-2,7) and B in the ratio 3:2,
find the coordinates of B.
3) Find the ratio in which the point (-1, k) divides the line joining the points
(-3, 10) and (6, -8), and find the value of k.
4) Find the coordinates of the midpoint of the line joining the points
(-3, 10) and (6, -8).
5) Three consecutive vertices of a parallelogram are A(1,2), B(2,3) and C(8,5). Find
the fourth vertex. (Hint : diagonals of a parallelogram bisect each other).
342 UNIT-13
ANSWERS
EXERCISE 13.2
1] (i) ND (ii) 1 (iii) 1
3(iv) 0 2] (i) 60º (ii) 45º (iii) 30º
3] (i) –1 (ii) 6 (iii) 3 (iv)7
5(v)
4b
a4] (i) parallel (ii) parallel
(iii) perpendicular (iv) parallel 5] (i)5
4(ii)
1
76] (i)
1
3(ii) 1 7] 5 8] –3
EXERCISE 13.3
1] (i) y = 3 x – 2 (ii) y = x + 3 2] (i) y = 2x – 4 (ii) 6y = –4x + 3 (iii) y = –2x + 3
3] (i) m = 2
3, c =
4
3(ii) m = 3, c = 0 (iii) m = 1, c = 5 (iv) m =
3
4, c =
5
4
EXERCISE 13.4
1] (i) 10 units (ii) 5 5 units (iii) 2 17 units (vi) 5 units (v) 13 units
(vi) 4 units (vii) (c–a) units (viii) 2 (sin cos )units
2] (i) 10 units (ii) 13 units (iii) 17 units 3] (i) 5 or –3 (ii) 7 (iii) (0, –2)
4] (i) 61 80 85 units (ii) 68 125 181 units 6] 13 units
EXERCISE 13.5
1] 1:6 2] (3, –3) 3] 2:7 4] (3
2, 1) 5] (7, 4)
Drawing graph of equation ax+by=c
Coordinate Geometry
Distance between two points
Inclination of a straight line lying on
axisx-
lying onaxisy-
Length of lineparallel to
axisx-
Length of lineparallel to
axisy-
2 1
2 1
y ym =
x x
Distance Formula
2 2
2 1 2 1d ( ) ( y )x x y
Section Formula
p ,2 1 2 1mx +nx my +ny
m+n m+n
Section Formula
p ,2 1 2 1
2 2
x + x y + y
Slope of a straight
line, m = tan
s