solve y log y dx – x dy = 0

 

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

 

 7. solve y log y dx – x dy = 0

This is a variable separable differential equation

separating the variables [dividing each term with y.log y . x

[dx/x ] -  [ dy / (y.log y ) ] = 0

integrating using integration formula

and rearranging [ dy / (y.log y ) ] =[ (1/y)dy / logy ] and noting

that integral of {(f ') / f }form is ln(f)  

log x - log{logy}+logC = 0

here log stands for ln (natural logarithm)

 

using properties of log

log[ Cx / logy] = 0

Cx / log y = e^0

Cx / logy =1

log y = Cx

or y = e^(Cx)

 

8. solve (x^5) {dy/dx} = − (y^5 )

This is a variable separable differential equation

separating the variables

dy/(y^5 ) = -dx /(x^5 )

(y^(-5) )dy =- (x^(-5) )dx

 

integrating using integration formula

[(y^(-4)]/(-4) = -[(x^(-4)]/(-4) +C/(-4)

so that

[(x^(-4)]+[(y^(-4)] =C

=================================================

 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 solution

   

5.solve  ( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

solution

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

solution

7. solve y log y dx – x dy = 0

solution

8. solve (x^5) {dy/dx} = − (y^5 )

solution

 

solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

 

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

 

5.solve  ( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

This is a differential equation of the type variable separable.

( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

dy ={ [ e^x – e^(–x)  ]  / [e^x + e^( –x)] } dx

now integrate both sides

keeping in mind that integral of {(f ') / f }form is ln(f)  

{ here ln refers to natural logarithms }

 

y=ln{[e^x + e^( –x)]} +C

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

This is a differential equation of the type variable separable.

dy / [  1 + (y^2) ] = [ 1+( x^2) ]dx

now integrating on both sides using integration formula

arc[tan y] =x +[(x^3)/3] +C

note that arc[tan y] stands for  inverse  tan  of (y)

=================================================

 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 solution

   

5.solve  ( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

solution

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

solution

 

 

solve (dy/dx)+y=1

 

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

3. solve (dy/dx)+y=1

(dy/dx)+y=1

(dy/dx)=1-y

 

separating the variables

dy / [1-y] =dx

integrating,

[remember to divide by (-1), the coefficient of y]

{ln[1-y] }/(-1)=x+ln C

 

{ here ln refers to natural logarithms }

-ln(1-y) =x+ln C

lnC +ln(1-y) = -x

 

using properties of logarithms

ln [C(1-y)]= (-x)

getting rid of logarathmic funtion

C(1-y)= e^(-x)

(1-y) = (1/C) e^(-x)

y=1-(1/C) e^(-x)

 

put (1/C) =(-A)

 

y = 1 + A e^(-x)

 

 4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 

divide each term by tan y tan x 

[ { (sec x)^2}/ tanx ]dx + [ { (sec y)^2}/ tan y ]dy = 0

integrating term by term 

keeping in mind that integral of {(f ') / f }form is ln(f)

 

{ here ln refers to natural logarithms }

 

ln(tan x) +ln(tan y) =lnC

using properties of logarithms

 

ln(tanx tany ) =lnC

tan x tan y = C

=================================================

 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

solve dy/dx = [1-cosx]/[1+cosx]

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

solve dy/dx = [1-cosx]/[1+cosx]

here this is of the type variable separable

so

dy/dx = [1-cosx]/[1+cosx]

 

dy = { [1-cosx]/[1+cosx] }dx

 

using trigonometric formula

1-cosx =2 [ sin(x/2)] ^2

1+cosx =2 [ cos(x/2)] ^2

 

 

dy = { [{2 [ sin(x/2)] ^2} ] / {2 [ sin(x/2)] ^2} }dx

 

dy =[tan(x/2)] ^2 dx

we  again have to use trigonometric formula to change the

[tan(x/2)] ^2 in terms of [sec(x/2)] ^2  -1 before integrating

dy ={[sec(x/2)] ^2  -1 }dx

 

now integrating on both sides using integration formula

 

y = { [tan(x/2)] /[1/2] } -x  + C

do not forget to divide by the coefficient of x while integrating

y=2 [tan(x/2)] - x +C

 

you can use the following for more explanation

 

2. solve (dy/dx) = sqrt[4-(y^2)]

separating the variables

dy/{sqrt[4-(y^2)]} =dx

now integrating on both sides using integration formula

 

and noting that 4=(2^2) so that a=2

we get

arc[sin(y/2)]=x+C

y/2 =sin(x+C)

y=2 sin(x+C)

 

 

=================================================

 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 solution

(ii) (– 5, 7), (– 1, 3)

solution

 

(iii) (a, b), (– a, – b)

solution

2. Find the distance between the points (0, 0) and (36, 15).

 solution 

3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

 

 

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

 exercise 7.1 co ordinate geometry chapter 7 cbse ncert 10th mathematics distance formula, collinear points using distance formula

3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

let

A= (1, 5)

B= (2, 3)

C= (– 2, – 11) 

 

A= (1, 5)=(x1,y1)

B= (2, 3)=(x2,y2)

 

using distance formula

AB= sqrt{(2-1)^2 +(3-5)^2} =sqrt(5)

B= (2,3)=(x1,y1)

C= (-2, -11)=(x2,y2)

 

using distance formula

BC= sqrt{(-2-2)^2 +(-11-3)^2} =sqrt(212)=2sqrt(53)

 

 

A= (1, 5)=(x1,y1)

C= (-2, -11)=(x2,y2)

 

AC=sqrt{(-2-1)^2 +(-11-5)^2}  =sqrt{265}

clearly AC is the biggest

but AB+BC is clearly not equal to AC

so the points are not collinear

 

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 solution

(ii) (– 5, 7), (– 1, 3)

solution

 

(iii) (a, b), (– a, – b)

solution

2. Find the distance between the points (0, 0) and (36, 15).

 solution 

3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

Find the distance between the following pairs of points (ii) (– 5, 7), (– 1, 3)

 

 exercise 7.1 co ordinate geometry chapter 7 cbse ncert 10th mathematics distance formula, collinear points using distance formula

Find the distance between the following pairs of points :((ii) (– 5, 7), (– 1, 3)

(-5,7) =(x1,y1)

(-1,3))=(x2,y2)

distance =sqrt{(-1-(-5) )^2 +(3-7)^2} =sqrt(32)=4sqrt(2) units

 

(iii) (a, b), (– a, – b)

 

(a,b) =(x1,y1)

(-a,-b))=(x2,y2)

distance =sqrt{(-a-a)^2 +(-b-b)^2}  =sqrt{4 ( a^2  + b^2 )}

=2sqrt{ ( a^2  + b^2 )}

 

2. Find the distance between the points (0, 0) and (36, 15).

(0,0) =(x1,y1)

(36,15))=(x2,y2)

 

distance ={(36-0)^2 +(15-0)^2}  = sqrt(1521)=39 units

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 solution

(ii) (– 5, 7), (– 1, 3)

solution

 

(iii) (a, b), (– a, – b)

solution

2. Find the distance between the points (0, 0) and (36, 15).

 solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order

 exercise 7.2 co ordinate geometry chapter 7 cbse ncert 10th mathematics section formula , midpoint formula  

 

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order

take 

A= (3, 0)

B=(4, 5)

C= (– 1, 4) 

D=(– 2, – 1)

 

diagonals are AC and BD

 

A= (3, 0) =(x1,y1)

C= (– 1, 4)=(x2,y2)

 

using distance formula

AC= sqrt{(-1-3)^2  + (4-0)^2} =sqrt(32)=4sqrt(2)

B= (4,5) =(x1,y1)

D= (– 2, -1)=(x2,y2)

 

using distance formula

BD=sqrt{ (-2-4)^2  +(-1-5)^2 }  =sqrt(72)=6sqrt(2)

area of a rhombus =(1/2)*AC*BD=24 sq.units

 

 exercise 7.1 co ordinate geometry chapter 7 cbse ncert 10th mathematics distance formula, collinear points using distance formula

Find the distance between the following pairs of points :(2, 3), (4, 1)

(2,3) =(x1,y1)

(4,1))=(x2,y2)

 

using distance formula

 

distance = sqrt{(4-2)^2 +(1-3)^2} =sqrt(8) =2sqrt(2) units.

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 

solution

 

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If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

 exercise 7.2 cordinate geometry chapter 7 cbse ncert 10th mathematics section formula , midpoint formula  

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

AP =(3/7) AB

means AP = [3/ (3+4)] AB

 

so that P divides AB internally in the ratio 3:4 [draw a rough figure]

 

using section formula for internal division

3 : 4 = m1 : m2

(–2, -2)=(x1,y1)

(2, –4)=(x2,y2)

 

P = [    { 3(2)+4(-2) }/{3+4}  ,{  3(-4) +4(-2) } / {3+4} ]

P = ( (-2)/7,  (-20)/7 )

 

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

let the required points be P, Q, R respectively

These points divide AB internally in the ratio 1:3 , 2:2 , 3:1 respectively

 {draw a rough figure}

 

To find P

1 : 3 = m1 : m2

(–2, 2)=(x1,y1)

(2, 8)=(x2,y2)

 

P = [ { 1(2)+3(-2) }  / {1+3} , {1(8)+3(2)}/{1+3}  ]

P= ( -1 , (7/2) )

To find Q

2 : 2  or  1:1 = m1 : m2

(–2, 2)=(x1,y1)

(2, 8)=(x2,y2)

 

Q = [ { 1(2)+1(-2) }  / {1+1} , {1(8)+1(2)}/{1+1}  ]

Q=( 0 , 5)

 

To find R

3 : 1 = m1 : m2

(–2, 2)=(x1,y1)

(2, 8)=(x2,y2)

 

R = [ { 3(2)+1(-2) }  / {3+1} , {3(8)+1(2)}/{3+1}  ]

R =( 1 , 13/2 )

 

 

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

 

 

 

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

  exercise 7.2 cordinate geometry chapter 7 cbse ncert 10th mathematics 

section formula , midpoint formula 

6.

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

let

A(1, 2),B (4, y),C (x, 6) and D(3, 5)

 

since the vertices are said to be taken in order

we take the diagonals as AC and BD

 

A(1, 2),C (x, 6)

using midpoint formula,

midpoint of AC =[{1+x}/2  , {2+6}/2 ] = [{1+x}/2  ,4 ]  

 

B (4, y), D(3, 5)

using midpoint formula,

midpoint of BD =[{4+3}/2  , {y+5}/2 ] = [7/2  ,{y+5}/2 ]  

We know that the diagonals of a parallelogram bisect each other

so the midpoints of AC and BD are the same

[{1+x}/2  ,4 ] =[7/2  ,{y+5}/2 ]  

using separate co ordinates

{1+x}/2 = 7/2

1+x=7

x=6

4={y+5}/2

8=y+5

y=3

 

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 

let A=(h.k)

 B = (1,4)

 using formula for midpoint of AB,

midpoint of AB is  [ {h+1}/2 , {k+4}/2  ]

This is the same as the midpoint (2, -3)

 

[ {h+1}/2 , {k+4}/2  ] = (2, -3)

 

separate the coordinates

 

{h+1}/2 = 2

h+1 = 4

h=3

 

(k+4)/2 =   -3

k+4 = (-6)

k=-6-4

k= (-10)

so that A is (3, -10)

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

 

6.If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. 

solution

 

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

 

Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6

  exercise 7.2 coordinate geometry chapter 7 cbse ncert 10th mathematics

 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 

let the ratio be k:1 internally

using section formula for internal division

k : 1 = m1 : m2

(–3, 10)=(x1,y1)

(6, –8)=(x2,y2)

 

point of division is [{k(6)+1(-3)} / {k+1} , {k(-8)+1(10)} / {k+1}]

 

This is the given point  (– 1, 6)

 

using the x- coordinate

 

{6k -3} /{k+1}=(-1)

 

so that k=2/7

 

and ratio is 2:7 internally..

 

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

 

let the ratio be k:1 internally

using section formula for internal division

k : 1 = m1 : m2

(1,-5)=(x1,y1)

(-4, 5)=(x2,y2)

 

point of division is P [{k(-4)+1(1)} / {k+1} , {k(5)+1(-5)} / {k+1}] -----------(1)

This is the point of division and it is given that this lies on the x-axis

so that the y-coordinate has to be zero

so

{k(5)+1(-5)} / {k+1} =0

 

5k-5=0

5k=5

k=1

ratio is 1:1 internally

 

using this value of k=1 in P in eqn(1)

point of division is 

P= [{k(-4)+1(1)} / {k+1} , {k(5)+1(-5)} / {k+1}]

P=[{1(-4)+1(1)} / {1+1} , {1(5)+1(-5)} / {1+1}]=( (-3)/2  , 0 )

 

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.