Mathematics Part ii Solutions Solutions for Class 10 Math Chapter 4 - Geometry And Algebra

Answer:

Let the centre of the circle be denoted by point O and the point through which the circle passes be denoted by A.

Given: Coordinates of point O and A are (3, 4) and (2, 5) respectively.

Radius of the circle = Distance between the points O and A

We know that the distance between two points is given by:

D =

Now, OA =

Thus, the radius of the circle is units.

Answer:

Let the centre of the circle be denoted by point O and the given point be denoted by A.

Given: Coordinates of point O and A are (−2, 1) and (4, 1) respectively.

Radius of the circle = 3 units

We know that distance between two points is given by:

D =

Now, OA =

So, the distance between the points O and A is 6 units, but the radius of the circle is 3 units.

Thus, the point (4, 1) lie outside the circle with centre at (−2, 1).

Answer:

Let the given points be A (2, 1), B (3, 4) and C (−3, 6). 

We know that distance between two points is given by:

D =

Now, AB =

⇒ AB² = 10 ……… (1)

BC =

⇒ BC² = 40 ……… (2)

CA =

CA² = 50 ……… (3)

From (1), (2) and (3), we have:

50 = 40 + 10

i.e. CA² = BC² + AB²

Therefore, the points A, B and C follow the Pythagoras theorem. 

Thus, the points A, B and C form a right angled triangle, right-angled at B.

Answer:

Let the given point be A (1, 3). 

On x-axis:

Let the coordinates of the required points be (h, k). But since the points lie on the x-axis, the y-coordinate should be zero.

So, the coordinates of the required points are (h, 0).

We know that distance between two points is given by:

D =

The required point is at a distance of 5 units from the point A 

Thus, there are two points on the x -axis which are at a distance of 5 units from the point (1, 3) and their coordinates are (−3, 0) and (5, 0).

On y-axis:

Let the coordinates of the required points be (h, k). But since the points lie on the y-axis, the x-coordinate should be zero.

So, the coordinates of the required points are (0, k).

The required point is at a distance of 5 units from the point A 

Thus, there are two points on y-axis which are at a distance of 5 units from the point (1, 3) and their coordinates are

Answer:

For knowing the coordinates of points B, C and D, we need to know the length of the line segments BA, CA and DA respectively.

We know that distance between two points is given by:

D =

Now, OA =

ΔBAO is a right angled triangle, right angled at A.

Here, ∠BOA = 30°

∴ Tan 30° =

Also, ΔCAO is a right angled triangle, right angled at A.

Here, ∠COA = 30° + 15° = 45°

∴ Tan 45° =

ΔDAO is a right angled triangle, right angled at A.

Here, ∠DOA = 30° + 15° + 15° = 60°

∴ Tan 60° =

Points A, B, C and D all lie on a line parallel to y-axis.

Since, the distance between point A and y-axis is 9 units, the distance between points B, C and D and y-axis is also 9 units.

Thus, the coordinates of points B, C and D are (9, ), (9, 9) and (9, ) respectively.

Now, AB =

BC = AC − AB =

CD = AD − AC =

Thus, the lengths of AB, BC and CD in order of their magnitudes are:

Answer:

Coordinates of the centre of the circle = (2, 3)

Radius of the circle = 5 units

Let the points of intersection of the circle on the x- axis be A (h, 0) and B (k, 0).

We know that distance between two points is given by: 

D =

Also, the distance of the required points from the centre of the circle is 5 units. 

Therefore, the coordinates of the required points are A (6, 0) and B (−2, 0). 

Distance between the points AB

_{Thus, the length of chord AB is 8 units.}

Answer:

Let the coordinates of the triangle be A (1, 2), B (2, 3) and C (3, 1).

If a circle is circumscribing the triangle ABC, then it should pass through the points A, B and C and the centre of this circle, say O should be equidistant from the points A, B and C.

⇒ OA = OB = OC

Let the coordinates of point O be (h, k).

We know that distance between two points is given by:

D =

⇒ OA =  

⇒ OB =  

⇒ OC =  

OA = OB

⇒=

Also, OB = OC

⇒=

Solving (1) and (2), we get:

Thus, the coordinates of the centre of the circumcircle are

Radius of the circle = OA

_{Thus, the radius of the circumcircle is}

Answer:

Let the given points be A (2, 3), B (3, −1) and C (5, 6).

For a line joining the points A and B to pass through the point C, the slope of the line BC must be equal to the slope of line AC.

Slope of line BC =

Slope of line AC =

⇒ Slope of line BC ≠ Slope of line AC

Therefore, the line joining the points (2, 3) and (3, −1) does not pass through the point (5, 6). 

Now, consider the point D (5, −9).

For a line joining the points A and B to pass through the point D, the slope of the line BD must be equal to slope of line AD.

Slope of line BD =

Slope of line AD =

⇒ Slope of line BD = Slope of line AD

Therefore, the line joining the points (2, 3) and (3, −1) passes through the point (5, −9). 

Answer:

Let the given points be A (1, 4), B (4, 1) and C

For the points A, B and C to lie on the same line, the line joining the points A and B should pass through the point C. So, the slope of the line BC should be equal to the slope of the line AC.

Slope of line BC =

Slope of line AC =

⇒ Slope of line BC = Slope of line AC

Therefore, the points (1, 4), (4, 1) and lie on the same line.

Answer:

Let the points be A (2, 3), B (7, 5), C (9, 8) and D (4, 6).

For the given four points to form a parallelogram, the opposite sides must be parallel, i.e., AB should be parallel to DC and BC should be parallel to AD.

Slope of line AB =

Slope of line DC =

Slope of line BC =

Slope of line AD =

⇒ Slope of line AB = Slope of line DC

Slope of line BC = Slope of line AD

Thus, the given vertices form a parallelogram.

Answer:

Let the given points be A (2, 1), B (1, 2), C (3, 5) and D (4, 7).

Slope of line AB =

Slope of line CD =

⇒ Slope of line AB ≠ Slope of line CD

Therefore, the given lines are not parallel.

Let the point of intersection of both the lines be E (x₁, y₁).

Therefore, the slope of the line BE should be equal to the slope of the line AB.

Slope of line BE =

⇒ y₁ − 2 = −x₁ + 1

⇒ y₁ + x₁ = 3 ….. (1)

Similarly, the slope of the line DE should be equal to the slope of the line CD.

Slope of line DE =

⇒ y₁−7 = 2x₁ − 8

⇒ y₁− 2x₁ = −1 ….. (2)

Solving (1) and (2), we get:

x₁ = , y₁ =

Therefore, the point of intersection of both the lines is

Answer:

Let the given point be A (1, 3). 

Let the point B (x, y) lie on the line passing through point A.

Given that slope of the line passing through point A is  

⇒

⇒ 2y − 6 = x − 1

⇒ x − 2y + 5 = 0

This is the equation of the line passing through point A and having slope

Any point on this line can be found out by assuming x to have any value, say 2 or 3 and putting it in the above equation.

2 − 2y + 5 = 0

⇒ −2y + 7 = 0

⇒ y =

Also, 3 − 2y + 5 = 0

⇒ −2y + 8 = 0

⇒ y = 4

Thus, the coordinates of two more points on this line are and (3, 4).

Answer:

Let the given point be A (1, 3). 

Let the point B (x, y) lie on the line passing through point A.

Given that the slope of the first line passing through point A is  

⇒

⇒ 2y − 6 = x − 1

⇒ x − 2y + 5 = 0

This is the equation of the line passing through point A and having slope

Any point on this line can be found out by assuming x to have any value, say 2 and putting it in the above equation.

2 − 2y + 5 = 0

⇒ −2y + 7 = 0

⇒ y =

Thus, the coordinates of one more point on the first line are

Also, given that the slope of the second line passing through point A is −2.

⇒

⇒ y − 3 = −2x + 2

⇒ 2x + y − 5 = 0

This is the equation of the line passing through point A and having slope −2.

Any point on this line can be found out by assuming x to have any value, say 2 and putting it in the above equation.

4 + y − 5 = 0

⇒ y − 1 = 0

⇒ y = 1

Thus, the coordinates of one more point on the second line are (2, 1).

Let m₁ = , m₂ = −2

For two lines to be perpendicular, the product of their slopes should be equal to −1.

m₁ m₂ = × (−2) = −1

Therefore, the given lines are perpendicular to each other.

Answer:

The equation of a line passing through points (x₁, y₁) and (x₂, y₂) is given by: 

The equation of the line passing through the points (0, 0) and (4, 2) is:

⇒ 4y = 2x 

⇒ 2y = x

This is the equation of the line passing through the points (0, 0) and (4, 2).

Observing the equation, we can say that the x-coordinate of all the points on the line joining the given points is double the y-coordinate.

Answer:

The equation of a line passing through points (x₁, y₁) and (x₂, y₂) is given by: 

The equation of the line passing through points (1, 3) and (2, 7) is:

⇒ y − 3 = 4x − 4 

⇒ y − 4x = −1

If (x, y) is a point on this line, then it must satisfy the above equation.

Similarly, if (x + 1, y + 4) is a point on this line, then it must also satisfy the above equation.

Putting the coordinates (x + 1, y + 4) in the above equation:

y + 4 − 4(x + 1) = −1

⇒ y + 4 − 4x − 4 = −1

⇒ y − 4x = −1

This equation is identical to the original equation.

So, the point (x + 1, y + 4) also lie on the line y − 4x = −1 as the point (x, y).

Answer:

If a line cuts the x-axis, then the y-coordinate of that point should be zero.

For the equation, 2x + 4y − 1= 0, putting y = 0, we get:

2x + 4(0) − 1= 0

⇒ 2x − 1= 0

⇒ x =

Thus, the point at which the line 2x + 4y − 1= 0 cuts the x-axis is

Similarly, if a line cuts the y-axis, then the x-coordinate of that point should be zero.

For the equation, 2x + 4y − 1= 0, putting x = 0, we get:

2(0) + 4y − 1= 0

⇒ 4y − 1= 0

⇒ y =

Thus, the point at which the line 2x + 4y − 1= 0 cuts the y-axis is

Answer:

We know that two lines are parallel if they have equal slope.

Given: Equation of lines 3x + 2y + 5 = 0 and 3x + 2y − 1 = 0

Taking the coordinates of the two points on the first line as (x₁, y₁) and (x₂, y₂):

3x₁ + 2y₁ + 5 = 0

3x₂ + 2y₂ + 5 = 0

(3x₁ + 2y₁ + 5) − (3x₂ + 2y₂ + 5) = 0

⇒ 3(x₁ − x₂) + 2(y₁ − y₂) = 0

⇒

Thus, the slope of the first line is

Taking the coordinates of the two points on the second line as (x₁, y₁) and (x₂, y₂):

3x₁ + 2y₁ − 1 = 0

3x₂ + 2y₂ − 1 = 0

(3x₁ + 2y₁ − 1) − (3x₂ + 2y₂ − 1) = 0

⇒ 3(x₁ − x₂) + 2(y₁ − y₂) = 0

⇒

Thus, the slope of the second line is

Thus, the given lines are parallel to each other.

If a line cuts the x-axis, then the y-coordinate of that point must be zero.

For the equation, 3x + 2y + 5 = 0, putting y = 0, we get:

3x + 2(0) + 5 = 0

⇒ 3x + 5= 0

⇒ x =

Therefore, the point at which the line 3x + 2y + 5 = 0 cuts the x-axis is

Similarly, for the equation 3x + 2y − 1= 0, putting y = 0, we get:

3x + 2(0) − 1 = 0

⇒ 3x − 1 = 0

⇒ x =

Therefore, the point at which the line 3x + 2y − 1 = 0 cuts the x-axis is

Similarly, if a line cuts the y-axis then the x-coordinate of that point must be zero.

For the equation, 3x + 2y + 5 = 0, putting x = 0, we get:

3(0) + 2y + 5 = 0

⇒ 2y + 5 = 0

⇒ y =

Therefore, the point at which the line 3x + 2y + 5 = 0 cuts the y-axis is

For the equation 3x + 2y − 1 = 0, putting x = 0, we get:

3(0) + 2y − 1 = 0

⇒ 2y − 1= 0

⇒ y =

Therefore, the point at which the line 3x + 2y − 1 = 0 cuts the y-axis is

Answer:

Intersection point of two lines is given by the coordinates obtained on solving the two linear equations.

Given equations of lines are 3x + 2y + 5 = 0 and 2x − 3y − 1 = 0.

On solving, we get x = −1, y = −1 

Therefore, the intersection point of these lines is (−1, −1).

Let the point (x₁, y₁) lie on the line, 3x + 2y + 5 = 0.

⇒ 3x₁ + 2y₁ + 5 = 0 

Any point on this line can be found out by assuming x to have any value, say 1 and putting it in the above equation.

⇒ 3(1) + 2y₁ + 5 = 0

⇒ 2y₁ = −8

⇒ y₁ = −4

Thus, the point (1, −4) lie on the line 3x + 2y + 5 = 0.

Similarly, let the point (x₂, y₂) lie on the line 2x − 3y − 1 = 0.

⇒ 2x₁ − 3y₁ − 1 = 0 

Any point on this line can be found out by assuming x to have any value, say 2 and putting it in the above equation.

⇒ 2(2) − 3y₁ − 1 = 0

⇒ −3y₁ = −3

⇒ y₁ = 1

Thus, the point (2, 1) lie on the line 2x − 3y − 1 = 0.

Now, let the slope of the line 3x + 2y + 5 = 0 be p and the slope of 2x − 3y − 1 = 0 be q.

We have just now found the points lying on these lines.

Points (−1, −1) and (1, −4) lie on the line 3x + 2y + 5 = 0.

∴ p =

Similarly, points (−1, −1) and (2, 1) lie on the line 2x − 3y − 1 = 0.

∴ q =

For two lines to be perpendicular, the product of their slopes should be equal to −1, i.e., pq = −1

pq = × = −1

Thus, the given lines are perpendicular.