Coordinate Geometry
Class10Mathematics2Chapter5Maharashtra Board
Notes
Topics to be learn

Introduction :
How to find the distance between any two points on a number line :
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}
Example :
If co—ordinates of points P, Q and R are —1,—5 and 4 respectively then find the length of seg PQ, seg QR.
As shown in the figure, co—ordinates of points P, Q and R are —1,—5 and 4 respectively.
∴ length of seg PQ = d(P, Q) = (—1)—(—5) = —1 + 5 = 4
and length of seg QR = d(Q, R) = 4 — (—5) = 4 + 5 = 9
 Using the same concept, we will understand to find the distance between two points in the XY—plane.
(1) To find distance between any two points on any axis :
(i) To find the distance between any two points on X—axis.
In the ﬁgure, points P(x_{1}, 0) and Q(x_{2}, 0) lie on X—axis.
Q lies to the right side of P, ∴ x_{2} > x_{1.}
∴ d(P, Q) = x_{2} — x_{1}.
(ii) To ﬁnd distance between any two points on Y—axis.
In the ﬁgure, points P(0, y_{l}) and Q(0, y_{2}) lie on Y—axis.
The points Q lie above point P, ∴ y_{2} > y_{1.}
∴ d(P,Q) = y_{2} — y_{1}
(2) To find the distance between the 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—coordinates of points A and B are equal.
We draw seg AL and seg BM perpendiculér to X—axis.
∴ c ALMB is a rectangle.
∴ AB = LM …(Opposite sides of a rectangle)
as LM = x_{2} — x_{1}, AB = x_{2} — x_{1}
(ii) In the ﬁgure, seg PQ is parallel to Y—axis.
x—coordinates of points P and Q are equal.
Draw seg PR and seg QS perpendicular to Y axis.
∴ c PQSR is a rectangle
∴ PQ = RS ….(Opposite sides of a rectangle)
But, RS = y_{2} — y_{1}
∴ d(P,Q) = y_{2} — y_{1}
Example :
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.
Δ 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 = [ \(\sqrt{17}\) ]
Solution :
(AB)^{2} + (BC)^{2} = [ (AC)^{2 }]
We will find co—ordinates of point B
to find the lengths AB and BC,
CB  X— axis ∴ y co—ordinate of B = [ 2 ]
BA  Y— axis ∴ x co—ordinate of B = [ 2 ]
AB = [3] — [ 2 ] = [ 1 ], BC = [ 2 ] — [ —2 ] = [4]
∴ AC^{2} = [ 1^{2} ] + [ 4^{2} ] = [ 17 ] ∴ AC = [ \(\sqrt{17}\) ]
Distance formula :
In the below figure, 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 Δ ABD,
AB^{2} = AD^{2} + BD^{2} = ( x_{2} — x_{1} )^{2} + ( y_{2} — y_{1} )^{ 2}
∴ AB = \(\sqrt{(x_2x_1)^2+(y_2y_1)^2}\)
This is known as distance formula.
Note that, \(\sqrt{(x_2x_1)^2+(y_2y_1)^2}\) = \(\sqrt{(x_1x_2)^2+(y_1y_2)^2}\)
Example :
(i) In the figure, seg AB  Y—axis and seg CB  X—axis. Co—ordinates of points A and C are given. Find lengths of seg AB seg AC and seg BC.
Solution :
Seg AB  Y—axis and seg BC  X—axis.
∴ Co—ordinates of point B are (2, 2).
Co—ordinates of point A(2, 3) and C(—2, 2) is given
(a) Find length of seg AB :
Co—ordinates of point A(2, 3) and B(2, 2)
By distance formula
AB = \(\sqrt{(x_2x_1)^2+(y_2y_1)^2}\)
∴ AB = \(\sqrt{(22)^2+(23)^2}\) = \(\sqrt{(0)^2+(1)^2}=\sqrt{1}\) = 1
(b) Find length of seg AC :
Co—ordinates of point A(2, 3) and C(—2, 2)
By distance formula
AC = \(\sqrt{(x_2x_1)^2+(y_2y_1)^2}\)
∴ AC = \(\sqrt{(22)^2+(23)^2}\) = \(\sqrt{(4)^2+(1)^2}=\sqrt{16+1}\) = \(\sqrt{17}\)
(c) Find length of seg BC :
Co—ordinates of point B(2, 2) and C(—2, 2)
By distance formula
BC = \(\sqrt{(x_2x_1)^2+(y_2y_1)^2}\)
∴ BC = \(\sqrt{(22)^2+(22)^2}\) = \(\sqrt{(4)^2+(0)^2}=\sqrt{16}\) = 4
To be Remember :
Coordinates of origin are (0, 0). Hence if coordinates of point A are (x, y), then d(OA) = \(\sqrt{x^2+y^2}\) If points A(x_{1}, y_{1}), B(x_{2}, y_{2}) lie in the XY—plane. then d(A, B) = \(\sqrt{(x_2x_1)^2+(y_2y_1)^2}\) that is, AB^{2} = ( x_{2} — x_{1} )^{2} + ( y_{2} — y_{1} )^{ 2} = ( x_{1} — x_{2} )^{2} + ( y_{1} — y_{2} )^{ 2} 
Recall :
Property of intercepts made by three parallel lines : In the figure line l  line m  line n, line p and line q are transversals, then \(\frac{AB}{BC}=\frac{DE}{EF}\) 
Division of a line segment :
In the figure, AP = 6 and PB = 10. ∴ \(\frac{AP}{PB}=\frac{6}{10}=\frac{3}{5}\)
∴ We can say that, point P divides the line segment AB in the ratio 3 : 5.
Section formula :
In the ﬁgure, point P on the seg RS in XYplane, divides seg RS in the ratio m : n.
Assume R(x_{1}, y_{l}), S(x_{2}, y_{2}) and P(x, y). Draw seg RT,
seg PQ and seg SM perpendicular to X~axis.
∴ T(x_{1}, 0); Q(x, 0) and M(x_{2}, 0).
∴ TQ = x — x_{1} and QM = x_{2} — x …. (1)
seg RT seg PQ  seg SM.
∴ by the property of intercepts made by three parallel lines,
\(\frac{RP}{PS}=\frac{TQ}{QM}=\frac{m}{n}\)
Now TQ = x — x_{1} and QM = x_{2} — x …..[From(1)]
∴ \(\frac{xx_1}{x_2x}=\frac{m}{n}\)
∴ n(x—x_{1}) = m(x_{2}  x)
∴ nx — nx_{1} = mx_{2}—mx
∴ mx + nx = mx_{2} + nx_{1}
∴ x(m + n) = mx_{2} + nx_{1}
∴ x = \(\frac{mx_2+nx_1}{m+n}\)
Similarly, drawing perpendiculars from points R. P and S to Yaxis,
we get, y = \(\frac{my_2+ny_1}{m+n}\)
∴ coordinates of the point, which divides the line segment joining the points
R(x_{1}, y_{1}) and S(x_{2}, y_{2}) in the ratio m : n are given by
\((\frac{mx_2+nx_1}{m+n},\,\frac{my_2+ny_1}{m+n})\)
Example :
If A(3, 5), B(7, 9) and point Q divides seg AB in the ratio 2 : 3 then find coordinates of point Q.
Solution :
In the given example let (x_{1}, y_{1})= (3, 5) and (x_{2}, y_{2}) = (7, 9) . m : n = 2 : 3
According to section formula,
x = \(\frac{mx_2+nx_1}{m+n}=\frac{2×7+3×3}{2+3}=\frac{23}{5}\)
y = \(\frac{my_2+ny_1}{m+n}=\frac{2×9+3×5}{2+3}=\frac{33}{5}\)
∴ Coordinates of Q are \((\frac{23}{5},\frac{33}{5})\)
Coordinates of the midpoint of a segment :
If A(x_{1}, y_{1})and B(x_{2}, y_{2}) are two points and P (x, y) is the midpoint of seg AB, then m = n.
∴ values of x and y can be written as
x = \(\frac{mx_2+nx_1}{m+n}\)
= \(\frac{mx_2+mx_1}{m+m}\) ….(m = n) = \(\frac{m(x_2+x_1)}{2m}\) = \(\frac{x_2+x_1}{2}\) 
y = \(\frac{my_2+ny_1}{m+n}\)
= \(\frac{my_2+my_1}{m+m}\) ….(m = n) = \(\frac{m(y_2+y_1)}{2m}\) = \(\frac{y_2+y_1}{2}\) 
∴ coordinates of midpoint P are \((\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})\)
This is called as midpoint formula.
Example :
Find the coordinates of point P if P is the midpoint of a line segment AB with A(4, 2) and B(6, 2).
Solution :
In the given example, suppose
(4, 2) = (x_{1}, y_{1}) ; (6, 2) = (x_{2}, y_{2}) and coordinates of P are (x, y)
∴ according to midpoint theorem,
x = \(\frac{x_2+x_1}{2}=\frac{6+(4)}{2}=\frac{2}{2}\) = 1
y = \(\frac{y_2+y_1}{2}=\frac{2+2)}{2}=\frac{4}{2}\) = 2
∴ coordinates of midpoint P are (1, 2) .
Centroid formula :
 We know that, medians of a triangle are concurrent. The point of concurrence (centroid) divides the median in the ratio 2 : 1.
Suppose, in Δ ABC, A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are the vertices. Seg AM is a median and G(x, y) is the centroid.
By definition of the median, M is the midpoint of seg BC.
Coordinates of M, by midpoint formula is \((\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})\)
G(x, y) is the centroid of Δ ABC
∴ AG : GM = 2 : 1
By section formula,
x = \(\frac{m(\frac{x_2+x_3}{2})+nx_1}{m+n}\)
= \(\frac{2(\frac{x_2+x_3}{2})+1(x_1)}{2+1}\)
= \(\frac{x_2+x_3+x_1}{3}=\frac{x_1+x_2+x_3}{3}\)
y = \(\frac{m(\frac{y_2+y_3}{2})+ny_1}{m+n}\)
= \(\frac{2(\frac{y_2+y_3}{2})+1(y_1)}{2+1}\)
= \(\frac{y_2+y_3+y_1}{3}=\frac{y_1+y_2+y_3}{3}\)
Thus, if (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are the vertices of a triangle then the coordinates of the centroid are
\((\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})\)
This is called the centroid formula.
Remember :
(i) Section formula : The coordinates of a point which divides the line segment joined by two distinct points (x_{1}, y_{l}) and (x_{2}, y_{2}) in the ratio m : n are \((\frac{mx_2+nx_1}{m+n},\,\frac{my_2+ny_1}{m+n})\) (ii) Midpoint formula : The coordinates of midpoint of a line segment joining two distinct points (x_{1}, y_{l}) and (x_{2}, y_{2}) are \((\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})\) (iii) Centroid formula : If (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are the vertices of a triangle then the coordinates of the centroid are \((\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})\) 
Example :
Find the coordinates of point P if P divides the line segment joining points A(4, 6) and B(5, 10) in the ratio 3 : 1 externally.
Solution :
\(\frac{AP}{PB}=\frac{3}{1}\) that is AP is larger than PB and ABP.
\(\frac{AP}{PB}=\frac{3}{1}\) that is AP = 3k, BP = k, then AB = 2k
∴ \(\frac{AB}{BP}=\frac{2}{1}\)
Now point B divides seg AP in the ratio 2 : 1.
Here m = 2, n = 1
Coordinates x = 5, y = 10, x_{1} = 4, y_{1} = 6,
Find out coordinates of Point P x_{2} and y_{2}
By section formula,
x = \(\frac{mx_2+nx_1}{m+n}\)
5 = \(\frac{2x_2+1(4)}{2+1}=\frac{2x_24}{3}\) ∴ 5 × 3 = 2x_{2} 4 ∴ 2x_{2} = 15+4 = 19 ∴ x_{2} = \(\frac{19}{2}\) 
y = \(\frac{my_2+ny_1}{m+n}\)
10 = \(\frac{2y_2+1(6)}{2+1}=\frac{2y_2+6}{3}\) ∴ 10 × 3 = 2y_{2} + 6 ∴ 2y_{2} = 30  6 = 24 ∴ y_{2} = \(\frac{24}{2}\) = 12 
∴ The coordinates of Point P are \((\frac{19}{2},12)\)
Slope of a line :
In the ﬁgure P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) are any two points in XYplane.
the ratio \(\frac{y_2y_1}{x_2x_1}\) is constant.
It is true for any two points on line l.
The ratio \(\frac{y_2y_1}{x_2x_1}\) is called the slope of the line l.
Usually slope is denoted by letter m.
∴ m = \(\frac{y_2y_1}{x_2x_1}\)
Slope of line – using ratio in trigonometry :
The tangent ratio of an angle made by the line with the positive direction of Xaxis is called the slope of that line.
In the ﬁgure, line l makes an angle θ with the positive direction of Xaxis.
slope of line l = m = tan θ.
Remember :
Slope of Xaxis and any line parallel to Xaxis is 0.
Slope of Yaxis and any line parallel to Yaxis cannot be determined.
Parallel lines have equal slopes.
When any two lines have same slope, these lines make equal angles with the
positive direction of X axis. ∴ These two lines are parallel.
Example :
In the figure both line l and line t make angle θ with the positive direction of X axis.
∴ line l  line t .......... (corresponding angle test)
Consider, point A(3, 0) and point B(0, 3) on line l
Find the slope of line l.
Slope of line l = \(\frac{y_2y_1}{x_2x_1}=\frac{30}{0(3)}=\frac{3}{3}=1\)
slope, m = tan θ = 1, ∴ θ = 45^{0}
In the similar way, considering suitable points on the line t we can find the slope of
line t. From this, we can verify that parallel lines have equal slopes.
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