Notes-Class-10-Mathematics-2-Chapter-5-Coordinate Geometry-Maharashtra Board

Coordinate Geometry

Class-10-Mathematics-2-Chapter-5-Maharashtra Board

Notes

Topics to be learn 

  • Distance formula
  • Division of a line segment
  • Section formula
  • Slope of a line

Introduction :

How to find the distance between any two points on a number line :

If x1 and x2 are the co—ordinates of points A and B and x2 > x1 then length of

seg AB = d(A,B) = x2 — x1

Example :

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.

[collapse]

(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(x1, 0) and Q(x2, 0) lie on X—axis.

Q lies to the right side of P, ∴ x2 > x1.

∴ d(P, Q) = x2 — x1.

(ii) To find distance between any two points on Y—axis.

In the figure, points P(0, yl) and Q(0, y2) lie on Y—axis.

The points Q lie above point P, ∴ y2 > y1.

∴ d(P,Q) = y2 — y1

(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 = x2 — x1, AB = x2 — x1

(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 = y2 — y1

∴ d(P,Q) = y2 — y1

Example :

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 ]

AC2 = [ ] + [ ] = [ ] 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]

∴ AC2 = [ 12 ] + [ 42 ] = [ 17 ] ∴ AC = [ \(\sqrt{17}\) ]

[collapse]
Distance formula :

In the below figure, A(x1, y1) and B(x2, y2) 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 x2.

seg AD is parallel to X—axis.

∴ the y co—ordinate of point D is y1.

∴ AD =d(A, D) = x2 — x1 ; BD= d(B, D) = y2 — y1

In right angled triangle Δ ABD,

AB2 = AD2 + BD2 = ( x2 — x1 )2 + ( y2 — y1 ) 2

∴ AB =  \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

This is known as distance formula.

Note that, \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) = \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

Example :

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 = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

∴ AB = \(\sqrt{(2-2)^2+(2-3)^2}\)  = \(\sqrt{(0)^2+(-1)^2}=\sqrt{1}\)  = 1

(b) Find length of seg AC :

Co—ordinates of point A(2, 3) and C(—2, 2)

By distance formula

AC = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

∴ AC = \(\sqrt{(-2-2)^2+(2-3)^2}\)  = \(\sqrt{(-4)^2+(-1)^2}=\sqrt{16+1}\)  = \(\sqrt{17}\)

(c) Find length of seg BC :

Co—ordinates of point B(2, 2) and C(—2, 2)

By distance formula

BC = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

∴ BC = \(\sqrt{(-2-2)^2+(2-2)^2}\)  = \(\sqrt{(-4)^2+(0)^2}=\sqrt{16}\) = 4

[collapse]
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(x1, y1), B(x2, y2) lie in the XY—plane.

then d(A, B) = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

that is, AB2 = ( x2 — x1 )2 + ( y2 — y1 ) 2 = ( x1 — x2 )2 + ( y1 — y2 ) 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 XY-plane, divides seg RS in the ratio m : n.

Assume R(x1, yl), S(x2, y2) and P(x, y). Draw seg RT,

seg PQ and seg SM perpendicular to X~axis.

 ∴ T(x1, 0); Q(x, 0) and M(x2, 0).

∴ TQ = x — x1 and QM = x2 — 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 — x1 and QM = x2 — x     …..[From(1)]

∴ \(\frac{x-x_1}{x_2-x}=\frac{m}{n}\)

∴ n(x—x1) = m(x2 - x)

∴ nx — nx1 = mx2—mx

∴ mx + nx = mx2 + nx1

∴ x(m + n) = mx2 + nx1

∴ x = \(\frac{mx_2+nx_1}{m+n}\)

Similarly, drawing perpendiculars from points R. P and S to Y-axis,

we get, y = \(\frac{my_2+ny_1}{m+n}\)

∴ coordinates of the point, which divides the line segment joining the points

R(x1, y1) and S(x2, y2) in the ratio m : n are given by

\((\frac{mx_2+nx_1}{m+n},\,\frac{my_2+ny_1}{m+n})\)

Example :

Example :

If A(3, 5), B(7, 9) and point Q divides seg AB in the ratio 2 : 3 then find co-ordinates of point Q.

Solution :

In the given example let (x1, y1)= (3, 5) and (x2, y2) = (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 = \(\frac{my_2+ny_1}{m+n}=\frac{2×9+3×5}{2+3}=\frac{33}{5}\)

∴ Co-ordinates of Q are \((\frac{23}{5},\frac{33}{5})\)

[collapse]

Co-ordinates of the midpoint of a segment :

If A(x1, y1)and B(x2, y2) 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)

= \(\frac{m(y_2+y_1)}{2m}\) 

= \(\frac{y_2+y_1}{2}\) 

∴ co-ordinates of midpoint P are \((\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})\) 

This is called as midpoint formula.

Example :

Example :

Find the co-ordinates 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) = (x1, y1) ; (6, 2) = (x2, y2) and co-ordinates 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

∴  co-ordinates of midpoint P are (1, 2) .

[collapse]

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(x1, y1), B(x2, y2)  and C(x3, y3)  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 (x1, y1), (x2, y2)  and (x3, y3)  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 (x1, yl) and (x2, y2) 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 (x1, yl) and (x2, y2)  are  

\((\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})\) 

(iii) Centroid formula  :

If (x1, y1), (x2, y2)  and (x3, y3)  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 :

Example :

Find the co-ordinates 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 A-B-P.

\(\frac{AP}{PB}=\frac{3}{1}\) 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, x1 = -4, y1 = 6,

Find out coordinates of Point P x2 and y2

By section formula,

x = \(\frac{mx_2+nx_1}{m+n}\)

5 = \(\frac{2x_2+1(-4)}{2+1}=\frac{2x_2-4}{3}\)

∴ 5 × 3 = 2x2 -4

2x2 = 15+4 = 19

x2 = \(\frac{19}{2}\)

y = \(\frac{my_2+ny_1}{m+n}\)

10 = \(\frac{2y_2+1(6)}{2+1}=\frac{2y_2+6}{3}\)

∴ 10 × 3 = 2y2 + 6

∴ 2y2 = 30 - 6 = 24

∴ y2 = \(\frac{24}{2}\) = 12

∴ The coordinates of Point P are \((\frac{19}{2},12)\)

[collapse]

Slope of a line :

In the figure P(x1, y1) and Q(x2, y2) are any two points in XY-plane.

the ratio \(\frac{y_2-y_1}{x_2-x_1}\)  is constant.

It is true for any two points on line l.

The ratio \(\frac{y_2-y_1}{x_2-x_1}\)  is called the slope of the line l.

Usually slope is denoted by letter m.

∴ m = \(\frac{y_2-y_1}{x_2-x_1}\) 

Slope of line – using ratio in trigonometry :

The tangent ratio of an angle made by the line with the positive direction of X-axis is called the slope of that line.

In the figure, line l makes an angle θ with the positive direction of X-axis.

 slope of line l = m = tan θ.

Remember :

Slope of X-axis and any line parallel to X-axis is 0.

Slope of Y-axis and any line parallel to Y-axis 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 :

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 = \(\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{0-(-3)}=\frac{3}{3}=1\) 

slope, m = tan θ = 1, ∴ θ = 450

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.

[collapse]

PDF-Notes,Solution,Text Book

Click on link to get PDF from store

PDF : Class 10th-Mathematics-2-Chapter-5-Coordinate Geometry-Text Book

PDF : Class 10th-Mathematics-2-Chapter-5-Coordinate Geometry-Notes

PDF : Class 10th-Mathematics-2-Chapter-5-Coordinate Geometry-Solution

Useful Links

Main Page : – Maharashtra Board Class 10th-Mathematics  – All chapters notes, solutions, videos, test, pdf.

Previous Chapter : Chapter-4-Geometric Construction – Online Notes

Next Chapter : Chapter-6-Co-ordinate GeometryOnline Notes

Leave a Reply

Write your suggestions, questions in comment box

Your email address will not be published. Required fields are marked *

We reply to valid query.