Class 10 Optional Mathematics Note

Co-ordinate Geometry

 

Equation of straight line in slope intercepts form:

 

Let P(x, y) be the any point line AB which makes a positive angle θ with x-axis.

From figure

∠ ACD= ∠ABE = θ

Slope of line AB = tanθ = m

PD = y- c

CD = OE =x

Tanθ =$\frac{{{\rm{PD}}}}{{{\rm{CD}}}}$

 

m =$\frac{{{\rm{y}} - {\rm{\: c}}}}{{\rm{x}}}$

or y= mx+ cis the requiredequation of AB

 

 

Angle betweenlines y = m1x+ c2andy =m2 x+ c1

Tanθ = $ \pm \frac{{{{\rm{m}}_1} - {{\rm{m}}_2 }}}{{1 + {{\rm{m}}_1}.{{\rm{m}}_1}}}$

Condition for parallelism

If two lines are parallel to each other then m1 = m2.

Condition for perpendicularity

If two lines are perpendicular to each other then m1.m 2 = -1

The equation that is perpendicular to the equation ax + by + c = 0 is bx – ay + k = 0.

The equation that is parallel to the equation ax + by + c = 0 is ax + by + K = 0.

Example 1

Find the equation of the straight line passing through the points of intersection of lines x-y+ 2=0 and x+2y -6= 0 and parallel to line 2x+y= 4

Given lines are

x - y +2=0………..1

x + 2y - 6 = 0………………….2

Solving 1 and 2

x = $\frac{2}{3}$and y = $\frac{8}{3}$

∴ Point of intersection (x, y) = (2/3, 8/3 )

The equation of given lineis 2x + y = 4 forslope of this line y = -2x + 4

 Comparing this equation with y = mx + cm = -2

The equation of the line passing through the point (2/3, 8/3) and slope m = -2

Is (y – $\frac{8}{3}{\rm{\: }}){\rm{\: }} = {\rm{\: }} - 2{\rm{\: }}\left( {{\rm{x}} - \frac{2}{3}{\rm{\: }}} \right)$

${\rm{or}},{\rm{\: }}\left( {3{\rm{y}} - 8{\rm{\: }}} \right) = {\rm{\: }} - 2{\rm{\: }}\left( {3{\rm{x}} - 2{\rm{\: }}} \right)$

 Or, 3y- 8= -6x+ 4

Or, 6x + 3y= 12

 Or , 2x+ y = 4is the required equation for the straight line.

 

Equation of pair of straight lines

Theequations of a1x + b1y = 0 and a2x + b2y =0 represent the straight lines passing through the origin. Multiplying we get

(a1x + b1y) (a2x + b2y) =0

Or, a1a2x2 + (a2b1 + b2a1) xy + b1 b2y2 =0

or ax2 + 2hxy + by2 =0 represents the second degree homogeneous equation in x and y

The angle between the two lines of a equation ax2 + 2hxy + by2 = 0 is tan? = $ \pm \frac{{2\sqrt {{{\rm{h}}^2} - {\rm{ab}}} }}{{{\rm{a}} + {\rm{b}}}}.$

If the lines of the equation ax2 + 2hxy + by2 = 0 are perpendicular then they should fulfill the condition a + b = 0 and for parallelism they should fulfill the condition h2 = ab

 

 

Circle

The equation of a circle with centre (0,0) and radius r is x2 + y2 = r2.

The equation of a circle with centre (h,k) and radius r is (x-h)2 + (y-k)2 = r2.

The equation of circle with end – points (x1,y1) and(x2, y2) is:

(x – x1)(x – x2) + (y – y1)(y – y2) = 0

 

 

If the two circle touch externallyr1+r2= C1C2

 

 

Example2

Find the equation center, radius and equation of the circlepassing through the following points

 ( 3,-1 )((6,2 )( 3,5 )

let o ( h,k ) be the centre of the circle passing through the point (3,-1 ) , (6,2 ) (3,5 )

A ( 3,-1 )B (6,2 )C ( 3,5 )centre of the circle O

 OA = $\sqrt {{{\left( {{\rm{h}} - 3} \right)}^2} + {\rm{\: }}{{\left( {{\rm{k}} + 1{\rm{\: }}} \right)}^2}} $

OA =$\sqrt {{\rm{\: }}{{\rm{h}}^2} - 6{\rm{h}} + 9{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} + 2{\rm{k}} + 1{\rm{\: }}} $

${\rm{O}}{{\rm{A}}^2} = {\rm{\: }}{{\rm{h}}^2} - 6{\rm{h}} + 9{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} + 2{\rm{k}} + 1$

${\rm{\: O}}{{\rm{A}}^2} = {\rm{\: }}{{\rm{h}}^2} - 6{\rm{h}} + 9{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} + 2{\rm{k}} + 1$

${\rm{\: O}}{{\rm{A}}^2} = {\rm{\: }}{{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 6{\rm{h}} + 2{\rm{k}} + 10$

 

 

OB = $\sqrt {{{\left( {{\rm{h}} - 6} \right)}^2} + {\rm{\: }}{{\left( {{\rm{k}} - 2{\rm{\: }}} \right)}^2}} $

OB =$\sqrt {{\rm{\: }}{{\rm{h}}^2} - 12{\rm{h}} + 36{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} - 4{\rm{k}} + 4{\rm{\: }}} $

${\rm{O}}{{\rm{B}}^2} = {\rm{\: }}{{\rm{h}}^2} - 12{\rm{h}} + 36{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} - 4{\rm{k}} + 4$

${\rm{\: O}}{{\rm{B}}^2} = {\rm{\: }}{{\rm{h}}^2} - 12{\rm{h}} + 36 + {\rm{\: }}{{\rm{k}}^2} - 4{\rm{k}} + 4$

${\rm{\: O}}{{\rm{B}}^2} = {\rm{\: }}{{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 12{\rm{h\: }} - 4{\rm{k}} + 40{\rm{\: }}$

 

 

OC = $\sqrt {{{\left( {{\rm{h}} - 3} \right)}^2} + {\rm{\: }}{{\left( {{\rm{k}} - 5{\rm{\: }}} \right)}^2}} $

OC=$\sqrt {{\rm{\: }}{{\rm{h}}^2} - 6{\rm{h}} + 9{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} - 10{\rm{k}} + 25{\rm{\: }}} $

${\rm{O}}{{\rm{C}}^2} = {\rm{\: }}{{\rm{h}}^2} - 6{\rm{h}} + 9{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} - 10{\rm{k}} + 25$

${\rm{\: O}}{{\rm{C}}^2} = {\rm{\: }}{{\rm{h}}^2} - 6{\rm{h}} + 9{\rm{\: }} + {\rm{\: }}{{\rm{k}}^2} - 10{\rm{k}} + 25$

${\rm{\: O}}{{\rm{C}}^2} = {\rm{\: }}{{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 6{\rm{h\: }} - 10{\rm{k}} + 34$

 

 As OA = OB = OC

${\rm{O}}{{\rm{A}}^2} = {\rm{O}}{{\rm{B}}^2} = {\rm{O}}{{\rm{C}}^2}$

 

Taking two ratio

${\rm{O}}{{\rm{A}}^2} = {\rm{O}}{{\rm{B}}^2}$

 

${\rm{\: }}{{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 6{\rm{h}} + 2{\rm{k}} + 10 = $${\rm{\: }}{{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 12{\rm{h\: }} - 4{\rm{k}} + 40$

 

6h+6k-30 = 0………………..1

 

Taking other two ratio

${\rm{O}}{{\rm{B}}^2} = {\rm{O}}{{\rm{C}}^2}$

${\rm{\: }}{{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 12{\rm{h\: }} - 4{\rm{k}} + 40{\rm{\: }} = $${{\rm{h}}^2} + {\rm{\: }}{{\rm{k}}^2} - 6{\rm{h\: }} - 10{\rm{k}} + 34$

12h -6k -6 = 0 ……………….2

Solving 1 and 2 we get,

(h, k) = (2 , 3 )

For radius

                            ${\left( {3 - 2} \right)^2}{\rm{\: }} + {\left( { - 1 + 3} \right)^2} = {{\rm{r}}^2}$

                                       ${\left( 1 \right)^2}{\rm{\: }} + {\left( 2 \right)^2} = {{\rm{r}}^2}$

∴ r2 = 5 unit

Equation of circle ${\rm{\: is\: }}{\left( {{\rm{x}} - 2} \right)^2}{\rm{\: }} + {\left( {{\rm{y}} - 3} \right)^2} = 10$

 

 

Example 3

Prove that the following points are concyclic .

(1,0) ,(2,-7),(8,1 ), (9, -6)

The equation of the circle is given by${\left( {{\rm{x}} - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {{\rm{y}} - {\rm{k}}} \right)^2} = {{\rm{r}}^2}$

If it passes through point (1,0) and 2,-7 )

 Then

${\left( {1 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {0 - {\rm{k}}} \right)^2} = {{\rm{r}}^2}$ …………….1

${\left( {2 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {7 + {\rm{k}}} \right)^2} = {{\rm{r}}^2}$……………2

Equating 1 and 2

 

${\left( {1 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {0 - {\rm{k}}} \right)^2} = {\left( {2 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {7 + {\rm{k}}} \right)^2}$

1-2h = 4-4h +49 +14k

 

h-7k -26 = 0…………………3

Alsoif it pases through the point (8,1 )

 

${\left( {8 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {1 - {\rm{k}}} \right)^2} = {{\rm{r}}^2}$……………4

 From 2 and 3

${\left( {8 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {1 - {\rm{k}}} \right)^2} = {\left( {1 - {\rm{h}}} \right)^2} + {\rm{\: }}{\left( {0 - {\rm{k}}} \right)^2}{\rm{\: }}$

$64 - 16{\rm{h}} + 1 - 2{\rm{k}} = 1 - 2{\rm{h}}$

14 h +2k -64 = 0

7 h + k -32 =0

 

Solving 3 and 4

We get (h, k ) = (5 ,-3)

 

Putting this value in 1 we get

 

${\left( {1 - 5} \right)^2} + {\rm{\: }}{\left( {0 + 3} \right)^2} = {{\rm{r}}^2}$

R= 5 unit

 

The equation of the circle is given by${\left( {{\rm{x}} - 5} \right)^2} + {\rm{\: }}{\left( {{\rm{y}} + 3} \right)^2} = 25$

 

If it pae htrought point(9 -6 )

 Then ,

                                     ${\left( {9 - 5} \right)^2} + {\rm{\: }}{\left( { - 6 + 3} \right)^2} = 25$

 

25 =25

 Which I true

∴ Given 4 point lies on the circle and they are concyclic

 


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