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Page 1
100
• Distance formula • Section formula • Slope of a line
Let’s recall.
We know how to find the distance between any two points on a number line.
If co-ordinates of points P,Q and R are -1,-5 and 4 respectively then find the
length of seg PQ, seg QR.
If x
1
and x
2
are the co-ordinates of points A and B and x
2
> x
1
then length of
seg AB = d(A,B) = x
2
- x
1
As shown in the figure, co-ordinates of points P,Q and R are -1,-5 and 4
respectively.
\ d(P, Q) = (-1)-(-5) = -1 + 5 = 4
and d(Q, R) = 4 - (-5) = 4 + 5 = 9
Using the same concept we can find the distance between two points on the
same axis in XY-plane.
Let’s learn.
(1) To find distance between any two points on an axis .
Two points on an axis are like two points on the number line. Note
that points on the X-axis have co-ordinates such as (2, 0), (
-5
2
, 0),
(8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0,
17
2
),
(0, -3). Part of the X-axis which shows negative co-ordinates is OX¢
and part of the Y-axis which shows negative co-ordinates is OY¢.
Fig. 5.1
P Q
-2 1 5 -3 2 0 -4 3
O
-5 4
R
-1
5
Co-ordinate Geometry
Let’s study.
Page 2
100
• Distance formula • Section formula • Slope of a line
Let’s recall.
We know how to find the distance between any two points on a number line.
If co-ordinates of points P,Q and R are -1,-5 and 4 respectively then find the
length of seg PQ, seg QR.
If x
1
and x
2
are the co-ordinates of points A and B and x
2
> x
1
then length of
seg AB = d(A,B) = x
2
- x
1
As shown in the figure, co-ordinates of points P,Q and R are -1,-5 and 4
respectively.
\ d(P, Q) = (-1)-(-5) = -1 + 5 = 4
and d(Q, R) = 4 - (-5) = 4 + 5 = 9
Using the same concept we can find the distance between two points on the
same axis in XY-plane.
Let’s learn.
(1) To find distance between any two points on an axis .
Two points on an axis are like two points on the number line. Note
that points on the X-axis have co-ordinates such as (2, 0), (
-5
2
, 0),
(8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0,
17
2
),
(0, -3). Part of the X-axis which shows negative co-ordinates is OX¢
and part of the Y-axis which shows negative co-ordinates is OY¢.
Fig. 5.1
P Q
-2 1 5 -3 2 0 -4 3
O
-5 4
R
-1
5
Co-ordinate Geometry
Let’s study.
101
i) To find distance between two
points on X-axis.
ii) To find distance between two
points on Y-axis.
(2) To find the distance between two points if the segment joining these points is
parallel to any axis in the XY plane.
i) In the figure, seg AB is parallel to
X- axis.
\ y co-ordinates of points A and B
are equal
Draw seg AL and seg BM
perpendicular to X-axis
\ c ABML is a rectangle.
\ AB = LM
But, LM = x
2
- x
1
\ d(A,B) = x
2
- x
1
ii) In the figure seg PQ is parallel to
Y- axis.
\ x co-ordinates of points P and Q
are equal
Draw seg PR and seg QS
perpendicular to Y-axis.
\ c PQSR is a rectangle
\ PQ = RS
But, RS = y
2
- y
1
\ d(P,Q) = y
2
- y
1
In the above figure, points
A(x
1
, 0 ) and B(x
2
, 0) are on
X- axis such that, x
2
> x
1
\ d(A, B) = x
2
- x
1
In the above figure, points
P(0, y
1
) and Q(0, y
2
) are on
Y- axis such that, y
2
> y
1
\ d(P,Q) = y
2
- y
1
Fig. 5.4 Fig. 5.5
Fig. 5.2
A (x
1
, 0) B (x
2
, 0)
X` X
Y
Y`
O
Fig. 5.3
(0, y
2
) Q
(0, y
1
) P
X X`
Y
Y`
O
A(x
1
, y
1
)
L(x
1
, 0)
B(x
2
, y
1
)
M(x
2
, 0)
X` X
Y`
Y
O
P (x
1
, y
2
)
(0, y
2
) R
Q (x
1
, y
1
) (0, y
1
) S
X` X
Y`
Y
O
Page 3
100
• Distance formula • Section formula • Slope of a line
Let’s recall.
We know how to find the distance between any two points on a number line.
If co-ordinates of points P,Q and R are -1,-5 and 4 respectively then find the
length of seg PQ, seg QR.
If x
1
and x
2
are the co-ordinates of points A and B and x
2
> x
1
then length of
seg AB = d(A,B) = x
2
- x
1
As shown in the figure, co-ordinates of points P,Q and R are -1,-5 and 4
respectively.
\ d(P, Q) = (-1)-(-5) = -1 + 5 = 4
and d(Q, R) = 4 - (-5) = 4 + 5 = 9
Using the same concept we can find the distance between two points on the
same axis in XY-plane.
Let’s learn.
(1) To find distance between any two points on an axis .
Two points on an axis are like two points on the number line. Note
that points on the X-axis have co-ordinates such as (2, 0), (
-5
2
, 0),
(8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0,
17
2
),
(0, -3). Part of the X-axis which shows negative co-ordinates is OX¢
and part of the Y-axis which shows negative co-ordinates is OY¢.
Fig. 5.1
P Q
-2 1 5 -3 2 0 -4 3
O
-5 4
R
-1
5
Co-ordinate Geometry
Let’s study.
101
i) To find distance between two
points on X-axis.
ii) To find distance between two
points on Y-axis.
(2) To find the distance between two points if the segment joining these points is
parallel to any axis in the XY plane.
i) In the figure, seg AB is parallel to
X- axis.
\ y co-ordinates of points A and B
are equal
Draw seg AL and seg BM
perpendicular to X-axis
\ c ABML is a rectangle.
\ AB = LM
But, LM = x
2
- x
1
\ d(A,B) = x
2
- x
1
ii) In the figure seg PQ is parallel to
Y- axis.
\ x co-ordinates of points P and Q
are equal
Draw seg PR and seg QS
perpendicular to Y-axis.
\ c PQSR is a rectangle
\ PQ = RS
But, RS = y
2
- y
1
\ d(P,Q) = y
2
- y
1
In the above figure, points
A(x
1
, 0 ) and B(x
2
, 0) are on
X- axis such that, x
2
> x
1
\ d(A, B) = x
2
- x
1
In the above figure, points
P(0, y
1
) and Q(0, y
2
) are on
Y- axis such that, y
2
> y
1
\ d(P,Q) = y
2
- y
1
Fig. 5.4 Fig. 5.5
Fig. 5.2
A (x
1
, 0) B (x
2
, 0)
X` X
Y
Y`
O
Fig. 5.3
(0, y
2
) Q
(0, y
1
) P
X X`
Y
Y`
O
A(x
1
, y
1
)
L(x
1
, 0)
B(x
2
, y
1
)
M(x
2
, 0)
X` X
Y`
Y
O
P (x
1
, y
2
)
(0, y
2
) R
Q (x
1
, y
1
) (0, y
1
) S
X` X
Y`
Y
O
102
Activity:
In the figure, seg AB || Y-axis and
seg CB || X-axis. Co-ordinates of
points A and C are given.
To find AC, fill in the boxes given below.
D ABC is a right angled triangle.
According to Pythagoras theorem,
(AB)
2
+ (BC)
2
=
We will find co-ordinates of point B
to find the lengths AB and BC,
CB || X- axis \ y co-ordinate of B =
BA || Y- axis \ x co-ordinate of B =
AB =
3
- = BC = - =
4
\ AC
2
= + = \ AC =
17
Let’s learn.
Distance formula
In the figure 5.7, A(x
1
, y
1
) and B(x
2
, y
2
) are any
two points in the XY plane.
From point B draw perpendicular BP on X-axis.
Similarly from point A draw perpendicular AD
on seg BP.
seg BP is parallel to Y-axis.
\ the x co-ordinate of point D is x
2
.
seg AD is parallel to X-axis.
\ the y co-ordinate of point D is y
1
.
\ AD =d(A, D) = x
2
- x
1
; BD= d(B, D) = y
2
- y
1
In right angled triangle D ABD,
AB =AD+BD
22 2
= xx yy
21
2
21
2
- () +- ()
\ AB =
xx yy
21
2
21
2
- () +- ()
This is known as distance formula.
Fig. 5.6
X X`
B
Y
Y`
A(2, 3)
C(-2, 2)
4
1
-1
3
-3 2
-2
2
-2 3
-3
1
-1
0
Fig. 5.7
X
P
X`
Y
Y`
A
(x
1
, y
1
)
B(x
2
, y
2
)
D(x
2
, y
1
)
O
Page 4
100
• Distance formula • Section formula • Slope of a line
Let’s recall.
We know how to find the distance between any two points on a number line.
If co-ordinates of points P,Q and R are -1,-5 and 4 respectively then find the
length of seg PQ, seg QR.
If x
1
and x
2
are the co-ordinates of points A and B and x
2
> x
1
then length of
seg AB = d(A,B) = x
2
- x
1
As shown in the figure, co-ordinates of points P,Q and R are -1,-5 and 4
respectively.
\ d(P, Q) = (-1)-(-5) = -1 + 5 = 4
and d(Q, R) = 4 - (-5) = 4 + 5 = 9
Using the same concept we can find the distance between two points on the
same axis in XY-plane.
Let’s learn.
(1) To find distance between any two points on an axis .
Two points on an axis are like two points on the number line. Note
that points on the X-axis have co-ordinates such as (2, 0), (
-5
2
, 0),
(8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0,
17
2
),
(0, -3). Part of the X-axis which shows negative co-ordinates is OX¢
and part of the Y-axis which shows negative co-ordinates is OY¢.
Fig. 5.1
P Q
-2 1 5 -3 2 0 -4 3
O
-5 4
R
-1
5
Co-ordinate Geometry
Let’s study.
101
i) To find distance between two
points on X-axis.
ii) To find distance between two
points on Y-axis.
(2) To find the distance between two points if the segment joining these points is
parallel to any axis in the XY plane.
i) In the figure, seg AB is parallel to
X- axis.
\ y co-ordinates of points A and B
are equal
Draw seg AL and seg BM
perpendicular to X-axis
\ c ABML is a rectangle.
\ AB = LM
But, LM = x
2
- x
1
\ d(A,B) = x
2
- x
1
ii) In the figure seg PQ is parallel to
Y- axis.
\ x co-ordinates of points P and Q
are equal
Draw seg PR and seg QS
perpendicular to Y-axis.
\ c PQSR is a rectangle
\ PQ = RS
But, RS = y
2
- y
1
\ d(P,Q) = y
2
- y
1
In the above figure, points
A(x
1
, 0 ) and B(x
2
, 0) are on
X- axis such that, x
2
> x
1
\ d(A, B) = x
2
- x
1
In the above figure, points
P(0, y
1
) and Q(0, y
2
) are on
Y- axis such that, y
2
> y
1
\ d(P,Q) = y
2
- y
1
Fig. 5.4 Fig. 5.5
Fig. 5.2
A (x
1
, 0) B (x
2
, 0)
X` X
Y
Y`
O
Fig. 5.3
(0, y
2
) Q
(0, y
1
) P
X X`
Y
Y`
O
A(x
1
, y
1
)
L(x
1
, 0)
B(x
2
, y
1
)
M(x
2
, 0)
X` X
Y`
Y
O
P (x
1
, y
2
)
(0, y
2
) R
Q (x
1
, y
1
) (0, y
1
) S
X` X
Y`
Y
O
102
Activity:
In the figure, seg AB || Y-axis and
seg CB || X-axis. Co-ordinates of
points A and C are given.
To find AC, fill in the boxes given below.
D ABC is a right angled triangle.
According to Pythagoras theorem,
(AB)
2
+ (BC)
2
=
We will find co-ordinates of point B
to find the lengths AB and BC,
CB || X- axis \ y co-ordinate of B =
BA || Y- axis \ x co-ordinate of B =
AB =
3
- = BC = - =
4
\ AC
2
= + = \ AC =
17
Let’s learn.
Distance formula
In the figure 5.7, A(x
1
, y
1
) and B(x
2
, y
2
) are any
two points in the XY plane.
From point B draw perpendicular BP on X-axis.
Similarly from point A draw perpendicular AD
on seg BP.
seg BP is parallel to Y-axis.
\ the x co-ordinate of point D is x
2
.
seg AD is parallel to X-axis.
\ the y co-ordinate of point D is y
1
.
\ AD =d(A, D) = x
2
- x
1
; BD= d(B, D) = y
2
- y
1
In right angled triangle D ABD,
AB =AD+BD
22 2
= xx yy
21
2
21
2
- () +- ()
\ AB =
xx yy
21
2
21
2
- () +- ()
This is known as distance formula.
Fig. 5.6
X X`
B
Y
Y`
A(2, 3)
C(-2, 2)
4
1
-1
3
-3 2
-2
2
-2 3
-3
1
-1
0
Fig. 5.7
X
P
X`
Y
Y`
A
(x
1
, y
1
)
B(x
2
, y
2
)
D(x
2
, y
1
)
O
103
Note that,
xx yy xx yy
21
2
21
2
12
2
12
2
- () +- () =- () +- ()
In the previous activity, we found the lengths of seg AB and seg AC and then
used Pythagoras theorem to find the length of seg AC.
Now we will use distance formula to find AC.
A(2, 3) and C(-2, 2) is given
Let A(x
1
, y
1
) and C(x
2
, y
2
).
x
1
= 2, y
1
= 3, x
2
= -2, y
2
= 2
AC =
xx yy
21
2
21
2
- () +- ()
= -- () +- () 22 23
22
=
- () +- () 41
22
= 16 1 +
=
17
seg AB || Y-axis and seg BC || X-axis.
\ co-ordinates of point B are (2, 2).
\ AB = xx yy
21
2
21
2
- () +- ()
= 22 23
22
- () +- ()
=
01 +
= 1
BC = xx yy
21
2
21
2
- () +- ()
=
-- () +- () 22 22
22
=
- () + 40
2
= 4
In the Figure 5.1, distance between points P and Q is found as (-1) - (-5) = 4. In
XY- plane co-ordinates of these points are (-1, 0) and (-5, 0). Verify that, using the
distance formula we get the same answer.
X X`
B
Y
Y`
A(2, 3)
C(-2, 2)
O
Fig. 5.8
• Co-ordinates of origin are (0, 0). Hence if co-ordinates of point P are
(x, y) then d(O, P) =
xy
22
+
.
• If points P(x
1
, y
1
), Q(x
2
, y
2
) lie on the XY plane then
d(P, Q) =
xx yy
21
2
21
2
- () +- ()
that is, PQ
2
= xx yy
21
2
21
2
- () +- () = xx yy
12
2
12
2
- () +- ()
Remember this!
Page 5
100
• Distance formula • Section formula • Slope of a line
Let’s recall.
We know how to find the distance between any two points on a number line.
If co-ordinates of points P,Q and R are -1,-5 and 4 respectively then find the
length of seg PQ, seg QR.
If x
1
and x
2
are the co-ordinates of points A and B and x
2
> x
1
then length of
seg AB = d(A,B) = x
2
- x
1
As shown in the figure, co-ordinates of points P,Q and R are -1,-5 and 4
respectively.
\ d(P, Q) = (-1)-(-5) = -1 + 5 = 4
and d(Q, R) = 4 - (-5) = 4 + 5 = 9
Using the same concept we can find the distance between two points on the
same axis in XY-plane.
Let’s learn.
(1) To find distance between any two points on an axis .
Two points on an axis are like two points on the number line. Note
that points on the X-axis have co-ordinates such as (2, 0), (
-5
2
, 0),
(8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0,
17
2
),
(0, -3). Part of the X-axis which shows negative co-ordinates is OX¢
and part of the Y-axis which shows negative co-ordinates is OY¢.
Fig. 5.1
P Q
-2 1 5 -3 2 0 -4 3
O
-5 4
R
-1
5
Co-ordinate Geometry
Let’s study.
101
i) To find distance between two
points on X-axis.
ii) To find distance between two
points on Y-axis.
(2) To find the distance between two points if the segment joining these points is
parallel to any axis in the XY plane.
i) In the figure, seg AB is parallel to
X- axis.
\ y co-ordinates of points A and B
are equal
Draw seg AL and seg BM
perpendicular to X-axis
\ c ABML is a rectangle.
\ AB = LM
But, LM = x
2
- x
1
\ d(A,B) = x
2
- x
1
ii) In the figure seg PQ is parallel to
Y- axis.
\ x co-ordinates of points P and Q
are equal
Draw seg PR and seg QS
perpendicular to Y-axis.
\ c PQSR is a rectangle
\ PQ = RS
But, RS = y
2
- y
1
\ d(P,Q) = y
2
- y
1
In the above figure, points
A(x
1
, 0 ) and B(x
2
, 0) are on
X- axis such that, x
2
> x
1
\ d(A, B) = x
2
- x
1
In the above figure, points
P(0, y
1
) and Q(0, y
2
) are on
Y- axis such that, y
2
> y
1
\ d(P,Q) = y
2
- y
1
Fig. 5.4 Fig. 5.5
Fig. 5.2
A (x
1
, 0) B (x
2
, 0)
X` X
Y
Y`
O
Fig. 5.3
(0, y
2
) Q
(0, y
1
) P
X X`
Y
Y`
O
A(x
1
, y
1
)
L(x
1
, 0)
B(x
2
, y
1
)
M(x
2
, 0)
X` X
Y`
Y
O
P (x
1
, y
2
)
(0, y
2
) R
Q (x
1
, y
1
) (0, y
1
) S
X` X
Y`
Y
O
102
Activity:
In the figure, seg AB || Y-axis and
seg CB || X-axis. Co-ordinates of
points A and C are given.
To find AC, fill in the boxes given below.
D ABC is a right angled triangle.
According to Pythagoras theorem,
(AB)
2
+ (BC)
2
=
We will find co-ordinates of point B
to find the lengths AB and BC,
CB || X- axis \ y co-ordinate of B =
BA || Y- axis \ x co-ordinate of B =
AB =
3
- = BC = - =
4
\ AC
2
= + = \ AC =
17
Let’s learn.
Distance formula
In the figure 5.7, A(x
1
, y
1
) and B(x
2
, y
2
) are any
two points in the XY plane.
From point B draw perpendicular BP on X-axis.
Similarly from point A draw perpendicular AD
on seg BP.
seg BP is parallel to Y-axis.
\ the x co-ordinate of point D is x
2
.
seg AD is parallel to X-axis.
\ the y co-ordinate of point D is y
1
.
\ AD =d(A, D) = x
2
- x
1
; BD= d(B, D) = y
2
- y
1
In right angled triangle D ABD,
AB =AD+BD
22 2
= xx yy
21
2
21
2
- () +- ()
\ AB =
xx yy
21
2
21
2
- () +- ()
This is known as distance formula.
Fig. 5.6
X X`
B
Y
Y`
A(2, 3)
C(-2, 2)
4
1
-1
3
-3 2
-2
2
-2 3
-3
1
-1
0
Fig. 5.7
X
P
X`
Y
Y`
A
(x
1
, y
1
)
B(x
2
, y
2
)
D(x
2
, y
1
)
O
103
Note that,
xx yy xx yy
21
2
21
2
12
2
12
2
- () +- () =- () +- ()
In the previous activity, we found the lengths of seg AB and seg AC and then
used Pythagoras theorem to find the length of seg AC.
Now we will use distance formula to find AC.
A(2, 3) and C(-2, 2) is given
Let A(x
1
, y
1
) and C(x
2
, y
2
).
x
1
= 2, y
1
= 3, x
2
= -2, y
2
= 2
AC =
xx yy
21
2
21
2
- () +- ()
= -- () +- () 22 23
22
=
- () +- () 41
22
= 16 1 +
=
17
seg AB || Y-axis and seg BC || X-axis.
\ co-ordinates of point B are (2, 2).
\ AB = xx yy
21
2
21
2
- () +- ()
= 22 23
22
- () +- ()
=
01 +
= 1
BC = xx yy
21
2
21
2
- () +- ()
=
-- () +- () 22 22
22
=
- () + 40
2
= 4
In the Figure 5.1, distance between points P and Q is found as (-1) - (-5) = 4. In
XY- plane co-ordinates of these points are (-1, 0) and (-5, 0). Verify that, using the
distance formula we get the same answer.
X X`
B
Y
Y`
A(2, 3)
C(-2, 2)
O
Fig. 5.8
• Co-ordinates of origin are (0, 0). Hence if co-ordinates of point P are
(x, y) then d(O, P) =
xy
22
+
.
• If points P(x
1
, y
1
), Q(x
2
, y
2
) lie on the XY plane then
d(P, Q) =
xx yy
21
2
21
2
- () +- ()
that is, PQ
2
= xx yy
21
2
21
2
- () +- () = xx yy
12
2
12
2
- () +- ()
Remember this!
104
?????????????? Solved Examples ?????????????
Ex. (1) Find the distance between the points P(-1, 1) and Q (5,-7) .
Solution ?Suppose co-ordinates of point P are (x
1
, y
1
) and of point Q are (x
2
, y
2
).
x
1
= -1, y
1
= 1, x
2
= 5, y
2
= -7
According to distance formula, d(P, Q) =
xx yy
21
2
21
2
- () +- ()
=
51 71
22
-- () ?
?
?
?
+- ()- ?
?
?
?
=
68
22
() +- ()
=
36 64 +
d(P, Q) =
100
= 10
\ distance between points P and Q is 10.
Ex. (2) Show that points A(-3, 2), B(1, -2) and C(9, -10) are collinear.
Solution ? If the sum of any two distances out of d(A, B), d(B, C) and d(A, C) is
equal to the third , then the three points A, B and C are collinear.
\ we will find d(A, B), d(B, C) and d(A, C).
Co-ordinates of A Co-ordinates of B Distance formula
(-3, 2) (1, -2) d(A,B) =
xx yy
21
2
21
2
- () +- ()
(x
1
, y
1
) (x
2
, y
2
)
\ d(A, B) = 13 22
22
-- () ?
?
?
?
+- ()- ?
?
?
?
.......... from distance formula
= 13 4
22
+ () +- ()
=
16 16 +
= 32 = 42 ..........(I)
d(B, C) =
91 10 2
22
- () +- + ()
= 64 64 + =
82
..........(II)
and d(A, C) =
93 10 2
22
+ () +- - ()
=
144 144 +
=
12 2
..........(III)
\ from(I), (II) and (III)
42
+
82
=
12 2
\ d(A, B) + d(B, C) = d(A, C)
\ Points A, B, C are collinear.
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FAQs on Textbook: Co-ordinate Geometry - Mathematics Class 10 (Maharashtra SSC Board)
| 1. What is the significance of the coordinate system in geometry? |  |
Ans. The coordinate system is fundamental in geometry as it provides a framework for locating points in a plane. It allows for the representation of geometric figures algebraically and visually, enabling the analysis of shapes, distances, and relationships between points. The most commonly used system is the Cartesian coordinate system, defined by two perpendicular axes (x and y), where each point can be identified by an ordered pair (x, y).
| 2. How do you find the distance between two points in a coordinate plane? | |
Ans. The distance between two points (x1, y1) and (x2, y2) in a coordinate plane can be calculated using the distance formula: Distance = √[(x2 - x1)² + (y2 - y1)²]. This formula derives from the Pythagorean theorem, providing a straightforward method to determine the length of the line segment connecting the two points.
| 3. What is the midpoint of a line segment in coordinate geometry? | |
Ans. The midpoint of a line segment connecting two points (x1, y1) and (x2, y2) is the point that divides the segment into two equal parts. It can be calculated using the midpoint formula: Midpoint = ((x1 + x2)/2, (y1 + y2)/2). This result gives the coordinates of the midpoint, which is useful in various geometric applications.
| 4. How can you determine the slope of a line given two points? | |
Ans. The slope of a line passing through two points (x1, y1) and (x2, y2) is determined by the slope formula: Slope (m) = (y2 - y1) / (x2 - x1). The slope indicates the steepness of the line and the direction of its incline (positive, negative, or zero), which is essential for understanding the line's behavior in the coordinate plane.
| 5. What is the equation of a straight line in slope-intercept form? | |
Ans. The equation of a straight line in slope-intercept form is expressed as y = mx + b, where m represents the slope of the line and b is the y-intercept, the point where the line crosses the y-axis. This form is particularly useful for graphing linear equations, allowing for easy identification of the slope and y-intercept.
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