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

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

(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 > x_1.

∴ d(P, Q) = x₂ — x₁.

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

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

The points Q lie above point P, ∴ y₂ > y_1.

∴ 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

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

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, y_l), 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

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

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.

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   \((\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})\)

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.