Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 11 Geometry and Algebra Questions and Answers Notes Pdf to clear their doubts.
SSLC Maths Chapter 11 Geometry and Algebra Questions and Answers
Geometry and Algebra Class 10 Questions and Answers Kerala State Syllabus
SCERT Class 10 Maths Chapter 11 Geometry and Algebra Solutions
Class 10 Maths Chapter 11 Kerala Syllabus – Parallelograms
(Textbook Page No. 231-232)
Question 1.
Calculate the coordinates of the fourth vertex of the parallelogram in each picture below:

Answer:
(a)

The horizontal distance between A and B = |5 – 1|= 4
And also, the horizontal distance between C and D is 4.
The vertical distance between A and B = |6 – 4| = 2
And also, the vertical distance between C and D is 2.
Therefore, D(7 – 4, 10 – 2) = D(3, 8)
In easy way, D(x, y)
x = 1 + 7 – 5, y = 4 + 10 – 6
That is, x = 3, y = 8
Therefore coordinates of D are (3, 8).
(b)

The horizontal distance between C and D = |2 – (-3)| = 5.
And also, the horizontal distance between A and B is 5.
The vertical distance between C and D = |3 – 2| = 1
And also, the vertical distance between A and B is 1.
Therefore, A(1- 5, -2 – 1) = A(-4, -3)
In easy way, D(x, y)
x = -3 + 1 – 2, y = 2 + (-2) – 3
That is, x = -4, y = -3
Therefore coordinates of D are (-4, -3).
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Question 2.
The sides of the larger triangle in the picture are parallel to the sides of the smaller triangle.

Calculate the coordinates of the vertices of the large triangle.
Answer:

PQ is parallel to BR, and PB is parallel to QR.
Therefore, PQRB is a parallelogram.
The horizontal distance between P and Q is 5 – 3 = 2.
The vertical distance is 4 – 3 = 1
The horizontal distance between points B and R is 2.
The vertical distance between B and R is 1.
Therefore, B(4 – 2, 2 – 1) = B(2, 1)
So, PQCR is a parallelogram.
Therefore, C(4 + 5 – 3, 2 + 4 – 3) = C(6, 3)
Also, PRQA is a parallelogram,
Therefore, A(3 + 5 – 4, 3 + 4 – 2) = A(4, 5)
Question 3.
The adjacent sides of a parallelogram are the lines joining the origin to the points with coordinates (x1, y1) and (x2, y2). What are the coordinates of the fourth vertex?

Answer:
Let O be at (0, 0).
Let the fourth vertex be B(x, y).
The horizontal distance between O and A is x2 – 0 = x2
The vertical distance is y2 – 0 = y2
The horizontal and vertical distances between B and C are (x2, y2).
In B(x, y) is x = x1 + x2, y = y1 + y2
So B(x1 + x2, y1 + y2)
Question 4.
In the picture, the coordinates of the four vertices of a parallelogram are marked:

Prove that the coordinates are connected by the relations below:
x1 + x3 = x2 + x4
y1 + y3 = y2 + y4
Answer:
The horizontal distance between A and B is x2 – x1.
The horizontal distance between C and D is x3 – x4.
Since ABCD is a parallelogram, the horizontal distance between A and B is equal to the horizontal distance between C and D.
x2 – x1 = x3 – x4
⇒ x1 + x3 = x2 + x4
Similarly, y1 + y3 = y2 + y4
SCERT Class 10 Maths Chapter 11 Solutions – Midpoint
(Textbook Page No. 236)
Question 1.
A circle is drawn with the line joining (2, 3) and (6, 5) as the diameter. What are the coordinates of the centre of the circle?
Answer:
Let the centre of the circle be (x, y).
The centre of the circle is the midpoint of the diameter.
So, using the midpoint formula.
x = \(\frac{2+6}{2}=\frac{8}{2}\) = 4
y = \(\frac{3+5}{2}=\frac{8}{2}\) = 4
Therefore, the centre of the circle is (4, 4).
Question 2.
The coordinates of two opposite vertices of a parallelogram are (4, 5) and (1, 3). Calculate the coordinates of the point of intersection of the diagonals. What are the coordinates of the midpoint of the other diagonal?
Answer:
The point of intersection of the diagonals is the midpoint.
So the midpoint of the diagonals = \(\left(\frac{4+1}{2}, \frac{5+3}{2}\right)=\left(\frac{5}{2}, \frac{8}{2}\right)\)
Since in a parallelogram, the diagonals bisect each other.
Therefore the midpoint of the other diagonal = \(\left(\frac{5}{2}, \frac{8}{2}\right)\)
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Question 3.
A quadrilateral is drawn with vertices A(1, 3), B(8, 6), C(12, 13), D(5, 10). Prove that ABCD is a parallelogram.
Answer:
The midpoint of the diagonal AC = \(\left(\frac{1+12}{2}, \frac{3+13}{2}\right)=\left(\frac{13}{2}, 8\right)\)
And, the midpoint of the diagonal BD = \(\left(\frac{8+5}{2}, \frac{6+10}{2}\right)=\left(\frac{13}{2}, 8\right)\)
Here, the midpoints of both diagonals AC and BD are the same, that is (\(\frac {13}{2}\), 8).
Therefore, it is a parallelogram.
Question 4.
Prove that the triangle with vertices (3, 5), (9, 13), (10, 6) is isosceles. Calculate its area.
Answer:
Let A = (3, 5), B = (9, 13), C = (10, 6)

Two sides of the triangle are equal in length.
Here, the triangle ABC is an isosceles right triangle.
Area = \(\frac {1}{2}\) × AB × Height
= \(\frac {1}{2}\) × 5√2 × 5√2
= 25 sq units
Here we get an isosceles right-angled triangle.
Therefore, there is no need to find the midpoint of the triangle.
The two perpendicular sides are the base and its height, and the third side is its hypotenuse.
Question 5.
The centre of a circle is (1, 2) and a point on it is (3, 5). Find the coordinates of the other end of the diameter through this point.
Answer:
Centre = (1, 2)
Let AB be the diameter, and the coordinates of A(3, 5) and B(x, y)
Using the midpoint formula,
1 = \(\frac{3+x}{2}\)
⇒ 3 + x = 2
⇒ x = 2 – 3
⇒ x = -1
2 = \(\frac{5+y}{2}\)
⇒ 5 + y = 4
⇒ y = 4 – 5
⇒ y = -1
Therefore, the coordinates of the other end of the diameter are (-1, -1).
Question 6.
In the picture, the midpoints of the sides of a quadrilateral are joined to make a smaller quadrilateral within it:

(i) Calculate the coordinates of the other three vertices of the larger quadrilateral.
(ii) Calculate the coordinates of the fourth vertex of the smaller quadrilateral.
Answer:

(i) Consider the quadrilateral ABCD.
Let the coordinates of D be D(x, y).
Using the midpoint formula,
3 = \(\frac{2+x}{2}\)
⇒ 2 + x = 6
⇒ x = 6 – 2
⇒ x = 4
3 = \(\frac{1+y}{2}\)
⇒ 1 + y = 6
⇒ y = 6 – 1
⇒ y = 5
Therefore, D(4, 5)
Let the coordinates of C be (x, y).
Using the midpoint formula,
6 = \(\frac{4+x}{2}\)
⇒ 4 + x = 12
⇒ x = 12 – 4
⇒ x = 8
6 = \(\frac{5+y}{2}\)
⇒ 5 + y = 12
⇒ y = 12 – 5
⇒ y = 7
Therefore, C(8, 7)
Let the coordinates of B be B(x, y).
Using the midpoint formula,
9 = \(\frac{8+x}{2}\)
⇒ 8 + x = 18
⇒ x = 18 – 8
⇒ x = 10
Therefore, B(10, 3)
5 = \(\frac{7+y}{2}\)
⇒ 7 + y = 10
⇒ y = 10 – 7
⇒ y = 3
Therefore, B(10, 3)
(ii) Let the coordinates of Q be Q(x, y).
Using the midpoint formula,
x = \(\frac{2+10}{2}\) = 6
y = \(\frac{1+3}{2}\) = 2
Therefore, D(6, 2)
Question 7.
Calculate the coordinates of the circumcentre of the triangle with vertices (0, 0), (0, 4), (3, 0).
Answer:
These are the vertices of a right-angled triangle with a right angle at (0, 0).
The endpoints of the hypotenuse are (0, 4) and (3, 0).
For a right triangle, the circumcentre is the midpoint of the hypotenuse.
Midpoints = \(\left(\frac{0+3}{2}, \frac{4+0}{2}\right)=\left(\frac{3}{2}, 2\right)\)
Therefore the circumcentre is (\(\frac {3}{2}\), 2)
Class 10 Maths Kerala Syllabus Chapter 11 Solutions – Ratio
(Textbook Page No. 239-240)
Question 1.
The coordinates of point A are (3, 2) and the coordinates of point B are (8, 7). Calculate the coordinates of the points P and Q, which divide AB in the ratios given below:
(i) AP : PB = 2 : 3
(ii) AQ : QB = 3 : 2
Answer:
(i) In the displacement from A to B, the movement is towards the right and upward.
The coordinates of the dividing point are P(3 + a, 2 + b).
\(\frac{a}{8-3}=\frac{2}{5}\)
⇒ a = 2
\(\frac{b}{7-2}=\frac{2}{5}\)
⇒ b = 2
Therefore, P(3 + 2, 2 + 2) = (5, 4)
So the coordinates of P are P(5, 4)
(ii) \(\frac{a}{8-3}=\frac{3}{5}\)
⇒ a = 3
\(\frac{b}{7-2}=\frac{3}{5}\)
⇒ b = 3
Therefore, the coordinates of Q are Q(6, 5).
Question 2.
Calculate the coordinates of the points that divide the line joining (1, 6) and (5, 2) into three equal parts.
Answer:
In the displacement from A to B, the movement is towards the right and downward.
The points P and Q divide the line into three equal parts.
That is AP = PQ = QB
The coordinates of the dividing point P are P(1 + a, 6 – b).
\(\frac{a}{4}=\frac{1}{3}\)
⇒ 3a = 4
⇒ a = \(\frac {4}{3}\)
\(\frac{b}{4}=\frac{1}{3}\)
⇒ b = \(\frac {4}{3}\)
Therefore, \(P\left(1+\frac{4}{3}, 6-\frac{4}{3}\right)=\left(\frac{7}{3}, \frac{14}{3}\right)\)
So the coordinates of P are P\(P\left(\frac{7}{3}, \frac{14}{3}\right)\)
(ii) The coordinates of the dividing point Q are Q(1 + c, 6 – d)
\(\frac{c}{4}=\frac{2}{3}\)
⇒ 3c = 8
⇒ c = \(\frac {8}{3}\)
\(\frac{d}{4}=\frac{2}{3}\)
⇒ 3d = 8
⇒ d = \(\frac {8}{3}\)
Therefore, \(Q\left(1+\frac{8}{3}, 6-\frac{8}{3}\right)=Q\left(\frac{11}{3}, \frac{10}{3}\right)\)
So the coordinates of Q are Q\(\left(\frac{11}{3}, \frac{10}{3}\right)\)
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Question 3.
The vertices of a triangle are (-1, 5), (3, 7), (1, 1). Find the coordinates of its centroid.
Answer:
M is the midpoint of AB, CM is the median, and G is the centroid.
Centroid of a triangle: Draw a line from each corner of a triangle to the centre of the other side.
These are medians.
The centroid is the point where the three medians meet.
The centroid divides the median from the vertex in the ratio 2 : 1.
Let A = (-1, 5), B = (1, 1), and C = (3, 7)
CG : GM = 2 : 1
\(M\left(\frac{-1+1}{2}, \frac{5+1}{2}\right)\) = M(0, 3)
From C to M, the movement is towards the left and downwards.
The coordinates of G are G(3 – a, 7 – b)
\(\frac{a}{3}=\frac{2}{3}\)
⇒ a = 2
\(\frac{b}{4}=\frac{2}{3}\)
⇒ 3b = 8
⇒ b = \(\frac {8}{3}\)
Therefore, \(G\left(3-2,7-\frac{8}{3}\right)=G\left(1, \frac{13}{3}\right)\)
So the coordinates of G are G(1, \(\frac {13}{3}\))
SSLC Maths Chapter 11 Questions and Answers – Line Maths
(Textbook Page No. 245)
Question 1.
Find the coordinates of two other points on the line joining (-1, 4) and (1, 2).
Answer:
Let the given points be A(-1, 4) and B(1, 2).
The point A(-1, 4) moves to B(1, 2).
The change in x-coordinate is x2 – x1 = 1 – (-1) = 2
That is, it increases by 2.
The change in the x-coordinate is y2 – y1 = 2 – 4 = -2
That is, it decreases by 2.
So we can write two other points with these changes.
Consider the point B(1, 2), and add the changes in the x and y coordinates.
That is x = 1 + 2 = 3
y = 2 – 2 = 0
⇒ (3, 0)
Similarly, consider the points (3, 0)
x = 3 + 2 = 5
y = 0 – 2 = -2
⇒ (5, -2)
Therefore the points are C(3, 0) and D(5, -2).
Question 2.
Prove that the points (1, 2), (2, 4), (3, 6) lie on the same line. Find the coordinates of two other points on this line.
Answer:
Let A = (1, 2), B = (2, 4), C = (3, 6)
The decimal representing the change in the y-coordinates and the change in the x-coordinates for the points A = (1, 2) and B = (2, 4) is,
\(\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{2-1}\) = 2
The fraction representing the change in the y-coordinates and the change in the x-coordinates for the points B = (2, 4) and C = (3, 6) is,
\(\frac{y_2-y_1}{x_2-x_1}=\frac{6-4}{3-2}\) = 2
Since the changes are the same.
Therefore, the three points A, B, and C lie on the same line.
To find two other points on this line:
Here we can see that the coordinate x increases by 1 and y increases by 2.
By considering the point C = (3, 6).
We get two other points as D = (4, 8) and E = (5, 10).
Question 3.
y1, y2, y3… is an arithmetic sequence. Prove that the points with coordinates (1, y1), (2, y2), (3, y3),… all lie on the same line.
Answer:
Since y1, y2, y3 … is an arithmetic sequence.
Let the common difference be d.
That is y2 – y1 = y3 – y2 = d
For the points (1, y1) and (2, y2).
The decimal representation of the change in the y-coordinate and the x-coordinate is,
\(\frac{y_2-y_1}{2-1}=\frac{d}{1}\) = d
For the points (2, y2) and (3, y3)
The decimal representation of the change in the y-coordinate and the x-coordinate is,
\(\frac{y_3-y_2}{3-2}=\frac{d}{1}\) = d
Since this ratio is the same for all pairs of points (1, y1), (2, y2), (3, y3), ….
Therefore, the points lie on the same straight line.
Question 4.
x1, x2, x3,… and y1, y2, y3,… are arithmetic sequences. Prove that the points with coordinates (x1, y1), (x2, y2), (x3, y3),… all lie on the same line.
Answer:
If d1 is the common difference of the arithmetic sequences x1, x2, x3,….
xn – x1 = \(\frac{(n-1)}{d_1}\)
If d2 is the common difference of the arithmetic sequences y1, y2, y3,….
yn – y1 = \(\frac{(n-1)}{d_2}\)
⇒ \(\frac{y_{\mathrm{n}}-y_1}{x_{\mathrm{n}}-x_1}=\frac{d_2}{d_1}\)
No matter what natural number n, \(\frac{d_2}{d_1}\) does not change.
Therefore the points with coordinates (x1, y1), (x2, y2), (x3, y3),… lie on the same line.
Kerala Syllabus Class 10 Maths Chapter 11 Solutions – Slope of a Line
(Textbook Page No. 248)
Question 1.
Calculate the slope of the line joining each pair of points below:
(i) (2, 3), (4, 5)
(ii) (2, 3), (4, 1)
(iii) (1, 1), (-1, -1)
(iv) (0, 1), (1, 0)
Answer:
(i) Change in y-coordinate = 5 – 3 = 2
Change in x-coordinate = 4 – 2 = 2
Therefore, Slope = \(\frac {2}{2}\) = 1
(ii) Change in y-coordinate = 1 – 3 = -2
Change in x-coordinate = 4 – 2 = 2
Therefore, Slope = \(\frac {-2}{2}\) = -1
(iii) Change in y-coordinate = -1 – 1 = -2
Change in x-coordinate = -1 – 1 = -2
Therefore, Slope = \(\frac {-2}{-2}\) = 1
(iv) Change in y-coordinate = 0 – 1 = -1
Change in x-coordinate = 1 – 0 = 1
Therefore, Slope = \(\frac {-1}{1}\) = -1
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Question 2.
The slope of a line passing through (2, 5) is \(\frac {-2}{3}\). Find the coordinates of two other points on this line.
Answer:
From the point (2, 5):
When the change in the y-coordinate is -2, and the change in the x-coordinate is 3, the new point is (2 + 3, 5 – 2) = (5, 3).
When the change in the y-coordinate is 2, and the change in the x-coordinate is -3, the new point is (2 – 3, 5 + 2) = (-1, 7).
Question 3.
The slope of a line passing through (3, 1) is \(\frac {-1}{2}\). Check whether the points given below are on the line or not:
(i) (5, 2)
(ii) (1, 0)
(iii) (4, 3)
(iv) (2, -1)
Answer:
A line passes through (3, 1) and has a slope = \(\frac {-1}{2}\)
Use the point slope form of a linear equation y – y1 = m(x – x1)
That is, y – 1 = \(\frac {-1}{2}\)(x – 3)
⇒ y – 1 = \(-\frac{1}{2} x+\frac{3}{2}\)
⇒ y = \(-\frac{1}{2} x+\frac{5}{2}\)
(i) For the point (5, 2)
2 = \(-\frac{1}{2}(5)+\frac{5}{2}\)
⇒ 2 = \(-\frac{5}{2}+\frac{5}{2}\)
⇒ 2 = 0
This is false, so the point is not on the line.
(ii) For the point (1, 0)
0 = \(-\frac{1}{2}(1)+\frac{5}{2}\)
0 = \(-\frac{1}{2}(1)+\frac{5}{2}\)
0 = 2
This is false, so the point is not on the line.
(iii) For the point (4, 3)
3 = \(-\frac{1}{2}(4)+\frac{5}{2}\)
3 = \(-\frac{4}{2}+\frac{5}{2}\)
3 = \(\frac {1}{2}\)
This is false, so the point is not on the line.
(iv) For the point (2, -1)
-1 = \(-\frac{1}{2}(2)+\frac{5}{2}\)
-1 = \(-\frac{2}{2}+\frac{5}{2}\)
-1 = \(\frac {3}{2}\)
This is false, so the point is not on the line.
Question 4.
We have seen that if y1, y2, y3,… is an arithmetic sequence, then the points (1, y1), (2, y2), (3, y3),… are all on the same line. What is the relation between the slope of this line and the common difference of the arithmetic sequence?
Answer:
If y1, y2, y3,… form an arithmetic sequence with common difference d,
Then for any two consecutive points: (1, y1), (2, y2)
The slope of the line joining them is:
Slope = \(\frac{\text { the change in } \mathrm{y} \text { coordinate }}{\text { the change in } \mathrm{x} \text { coordinate }}\)
= \(\frac{y_2-y_1}{2-1}\)
= d
Therefore, the slope of the line is equal to the common difference of the arithmetic sequence.
Question 5.
We have seen that if x1, x2, x3,… and y1, y2, y3,… are arithmetic sequences, then the points with coordinates (x1, y1), (x2, y2), (x3, y3),… all lie on the same line. What is the relation between the slope of this line and the common differences of the arithmetic sequences?
Answer:
Let x1, x2, x3,.. and y1, y2, y3,… be arithmetic sequences with common differences dx and dy respectively.
The slope of the line joining the two consecutive terms (x1, y1) and (x2, y2) is:
Slope = \(\frac{\text { the change in } \mathrm{y} \text { coordinate }}{\text { the change in } \mathrm{x} \text { coordinate }}\) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{d_y}{d_x}\)
Therefore, the slope of the line is equal to the ratio of the common difference of the y-sequence to the common difference of the x-sequence.
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Question 6.
Find another point on the line through (1, 3) with slope \(\frac {1}{2}\) and another point on the line through the same point with slope -2. Prove that the lines are perpendicular. (Hint: Pythagoras’ Theorem)
Answer:
The first iine passes through the point P(1, 3) with slope \(\frac {1}{2}\).
Slope = \(\frac{\text { the change in } \mathrm{y} \text { coordinate }}{\text { the change in } \mathrm{x} \text { coordinate }}\)
If we move 2 units to the right from x = 1 we get
x2 = 1 + 2 = 3
The change in y coordinate will be 1 because the slope is \(\frac {1}{2}\).
Slope = \(\frac{y_2-y_1}{x_2-x_1}\)
\(\frac{1}{2}=\frac{y_2-3}{3-1}\)
y2 – 3 = 1
y2 = 4
Therefore, another point on the first line is A = (3, 4)
The first line passes through the point (1, 3) with slope -2.
Slope = \(\frac{\text { the change in } \mathrm{y} \text { coordinate }}{\text { the change in } \mathrm{x} \text { coordinate }}\)
If we move 1 unit to the right from x = 1, we get
x3 = 1 + 1 = 2
The change in y-coordinate will be 1 because the slope is \(\frac {1}{2}\).
Slope = \(\frac{y_3-y_1}{x_3-x_1}\)
-2 = \(\frac{y_3-3}{3-1}\)
y3 – 3 = -4
y3 = -1
Therefore, another point on the first line is B = (2, 1)
PA2 = (3 – 1)2 + (4 – 3)2
= 22 + 12
= 4 + 1
= 5
PB2 = (2 – 1)2 + (1 – 3)2
= 12 + (-2)2
= 1 + 4
= 5
AB2 = (3 – 2)2 + (4 – 1)2
= 12 + 32
= 1 + 9
= 10
Using Pythagoras theorem
PA2 + PB2 = AB2
5 + 5 = 10
10 = 10
Therefore, the two lines are perpendicular.
Class 10 Maths Chapter 11 Kerala Syllabus – Equation of a Line
(Textbook Page No. 251-252)
Question 1.
Find the equation of the line joining each pair of points below:
(i) (0, 0), (1, 1)
(ii) (0, 0), (1, -1)
(iii) (1, 0), (0, 1)
(iv) (-1, 0), (0, -1)
Answer:
(i) Slope of the line = \(\frac{1-0}{1-0}\) = 1
Let (x, y) be a point on the line.
The change in the y-coordinate is equal to the change in the x-coordinate multiplied by the slope of the line.
That means, y – 0 = 1(x – 0)
⇒ y = x
(ii) Slope of the line = \(\frac{-1-0}{1-0}\) = -1
Let (x, y) be a point on the line.
The change in the y-coordinate is equal to the change in the x-coordinate multiplied by the slope of the line.
That means, y – 0 = -1(x – 0)
⇒ y = -x
⇒ x + y = 0
(iii) Slope of the line = \(\frac{1-0}{0-1}\) = -1
Let (x, y) be a point on the line.
The change in the y-coordinate is equal to the change in the x-coordinate multiplied by the slope of the line.
That means, y – 0 = -1(x – 1)
⇒ y = -1(x – 1)
⇒ y = -x + 1
⇒ x + y – 1 = 0
(iv) Slope of the line = \(\frac{-1-0}{0-(-1)}\) = -1
Let (x, y) be a point on the line.
The change in the y-coordinate is equal to the change in the x-coordinate multiplied by the slope of the line.
That means, y – 0 = -1(x – (-1))
⇒ y = -1(x + 1)
⇒ y = -x – 1
⇒ x + y + 1 = 0
Question 2.
(i) Find the equation of the line joining (-1, 3) and (2, 5).
(ii) Prove that if (x, y) is a point on this line, then so is (x + 3, y + 2).
Answer:
(i) Slope of the line = \(\frac{5-3}{2-(-1)}=\frac{2}{3}\)
Let (x, y) be a point on the line.
Then, y – 3 = \(\frac {2}{3}\) × (x – (-1))
⇒ (y – 3) × 3 = 2 × (x + 1)
⇒ 3y – 9 = 2x + 2
⇒ 2x – 3y + 11 = 0
(ii) To check whether a point lies on a line, it is enough to see if the coordinates of that point satisfy the equation of the line.
To check whether (x + 3, y + 2) lies on a line,
2(x + 3) – 3(y + 2) + 11 = 2x + 6 – 3y – 6 + 11
⇒ 2x – 3y + 11 = 0
Therefore, the point (x + 3, y + 2) lies on the line.
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Question 3.
(i) Find the equation of the line joining (-1, 1) and (2, 7).
(ii) Prove that for any number x, the point (x, 2x + 3) is on this line.
Answer:
(i) Slope of the line = \(\frac{7-1}{2-(-1)}=\frac{6}{3}\) = 2
If (x, y) is a point on the line, then
y – 1 = 2(x – (-1))
⇒ y – 1 = 2(x + 1)
⇒ y – 1 = 2x + 2
⇒ y = 2x + 3
(ii) To prove (x, 2x + 3) lies on this line.
Let (x, y) be a point on the line.
The equation of the line is y = 2x + 3.
Therefore, (x, y) can be written as (x, 2x + 3).
That means this point is on this line.
Question 4.
Prove that for any point on the line joining the origin and the point (1, 2), the y-coordinate is twice the x-coordinate.
Answer:
The slope of the line passing through the points (0, 0), (1, 2) is \(\frac{2-0}{1-0}\) = 2
Let (x, y) be a point on the line.
y – 0 = 2(x – 0)
⇒ y = 2x
If we consider any point on the line, the y-coordinate is twice that of the x-coordinate.
Question 5.
Prove that for any point on the line joining (2, 0) and (0, 3), the sum of half the x-coordinate and one-third the y-coordinate is 1.
Answer:
Slope of the line = \(\frac{3-0}{0-2}=-\frac{3}{2}\)
Let (x, y) be a point on the line,
y – 0 = \(-\frac {3}{2}\)(x – 2)
⇒ y = \(-\frac {3}{2}\)(x-2)
⇒ 2y = -3(x – 2)
⇒ 3x + 2y = 6
If we divide 6 by both sides, we get
\(\frac{3 x+2 y}{6}=\frac{6}{6}\)
⇒ \(\frac{3 x}{6}+\frac{2 y}{6}=\frac{6}{6}\)
⇒ \(\frac{1}{2} x+\frac{1}{3} y=1\)
SCERT Class 10 Maths Chapter 11 Solutions – Equation of a Circle
(Textbook Page No. 253-254)
Question 1.
Find the equation of the circle with centre at the origin and radius 5. Write the coordinates of eight points on this circle.
Answer:
Let (x, y) be any point on the circle.
(x – 0)2 + (y – 0)2 = 52
⇒ x2 + y2 = 25
Question 2.
A circle is drawn with the line joining (2, 3) and (4, 7) as the diameter.
(i) What are the coordinates of the centre of this circle?
(ii) What is its radius?
(iii) Write the equation of the circle.
Answer:
(i) Centre of the circle = \(\left(\frac{2+4}{2}, \frac{3+7}{2}\right)\) = (3, 5)
(ii) Radius r = \(\sqrt{(3-2)^2+(5-3)^2}=\sqrt{5}\)
(iii) (x – 3)2 + (y – 5)2 = (√5)2
⇒ (x – 3)2 + (y – 5)2 = 5
⇒ x2 + y2 – 6x – 10x – 29 = 0
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Question 3.
What is the equation of the circle in the picture below?

Answer:
(0, 2) and (4, 0) are the ends of the diameter.
Centre = (2, 1)
Diameter = \(\sqrt{(4-0)^2+(0-2)^2}=\sqrt{20}=2 \sqrt{5}\)
Therefore, the radius of the circle r = √5
Equation of the circle
(x – 2)2 + (y – 1)2 = 5
⇒ x2 + y2 – 4x – 2y = 0
Question 4.
The equation of a circle is x2 + y2 – 2x – 4y – 11= 0. Find the coordinates of its centre and the radius.
Answer:
x2 + y2 – 2x – 4y = 11
⇒ (x2 – 2x + 1) + (y2 – 4y + 4) = 11 + 1 + 4
⇒ (x – 1)2 + (y – 2)2 = 42
Therefore, Centre of the circle = (1, 2)
Radius = 4
Geometry and Algebra Class 10 Notes Pdf
Class 10 Maths Chapter 11 Geometry and Algebra Notes Kerala Syllabus
Introduction
Geometry and Algebra are two fundamental branches of mathematics that work together to help us analyse a situation mathematically. Geometry, derived from Greek words meaning “earth measurement,” deals with shapes, sizes, and the properties of space. Algebra, on the other hand, uses symbols and letters to represent numbers and relationships, allowing us to solve problems and express patterns logically. While geometry gives mathematics a visual form, algebra provides the language to describe and analyse it. Straight lines and circles are the two important geometric shapes that we discuss in this unit.
→ A quadrilateral whose opposite sides are equal and parallel is called a parallelogram.
→ In a rectangle, both pairs of opposite sides are parallel to the axes.
→ The midpoint of the line joining (x1, y1) and (x2, y2) is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).
→ If AP is a certain part of AB, and PB is the remaining part of AB.
→ From point A to P, the horizontal distance is ‘a’ towards the right, and the vertical distance is ‘b’ upwards. The coordinates of P are P (x1 + a, y1 + b).
→ From A to B the horizontal distance is x2 – x1, and the vertical distance is y2 – y1.
→ The ratio AP : AB is derived from the similarity of the triangles. That is, \(\frac{a}{x_2-x_1}=\frac{a}{y_2-y_1}=\frac{A P}{A B}\)
→ For any two points on a line, not parallel to either axis, the change in the y-coordinate is the change in the x-coordinate multiplied by a fixed number.
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→ For any two points on a line not parallel to either axis, the change in y-coordinates divided by the change in x-coordinates gives the same number. This number is called the slope of the line.
→ If the slope is the negative of the coefficient of x divided by the coefficient of y.
→ The general form of the equation of a line is ax + by + c = 0 and its slope is \(-\frac {a}{b}\).
→ The relation between the coordinates of every point on the line is called the equation of the line.
→ The equation of the circle, r = \(\sqrt{(x-a)^2+(y-b)^2}\)
Parallelogram
A quadrilateral whose opposite sides are equal and parallel is called a parallelogram. In a rectangle, both pairs of opposite sides are parallel to the axes.
Suppose a rectangle is drawn such that one pair of its parallel sides is parallel to the x-axis and the other pair is parallel to the y-axis.
When moving from point A(3, 2) to point C(8, 6) in this square, both the x-coordinate and the y-coordinate change. The movement from A to C is towards the right and upward.

The change in the x-coordinate increases by 8 – 3 = 5
The change in the y-coordinate increases by 6 – 2 = 4
This change occurs by moving 3 units to the right from point A and then 4 units upward from there. It can be illustrated as shown below.

The movement to point D is 5 units to the left of point B and 4 units upward from point A. That is, we can write it as D(8 – 5, 2 + 4). The same kind of change can be seen in parallelograms as well. In a parallelogram, opposite sides are parallel. Therefore, the changes in the x-coordinates and y-coordinates are the same.
The parallelogram ABCD is drawn based on the coordinate axes. Since A is (1, 1) and B is (5, 3), the line is not parallel to either the x-axis or the y-axis. When moving from A to B, the x-coordinate increases by 4 and the y-coordinate increases by 2. The same change occurs when moving from D to C. However, when moving from C to D, the x-coordinate decreases, and the y-coordinate also decreases.
D(8 – 4,6 – 2) = D(4, 4)
If A(x1, y1), B(x2, y2), C(x3, y3), D(x4, y4) are the vertices of a parallelogram.

|x2 – x1| = |x3 – x4|
This can be written as follows.
x4 = x1 + x3 – x2
Similarly y4 = y1 + y3 – y2
Question 1.
In the figure P(1, 2), Q(5, 3), and R(3, 0) are the midpoints of the sides of triangle ABC.

(a) Suggest a suitable name for BPQR.
(b) Write the coordinates of the vertex B.
(c) Write the coordinates of A.
Answer:
(a) Parallelogram
(b) B(1 + 3 – 5, 2 + 0 – 3) = B(-1, -1)
(c) A(3, 5)
Question 2.
If O is the origin and OABC is a parallelogram, and B(7, 5), then

(a) Write the coordinates of A and C?
(b) What is the area of OABC?
Answer:
(a) A(4, 0), C(0 + 7 – 4, 0 + 5 – 0) = C(3, 5)
(b) Area = 4 × 5 = 20
Question 3.
ABCD is a parallelogram. If A(1, 1), B(3, 2), C(5, 7), then write the coordinates of D.
Answer:
D(1 + 5 – 3, 1 + 7 – 2) = D(3, 6)
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Question 4.
A(1, 2), B(4, y), C(x, 6) and D(4, 2) are the vertices of a parallelogram.
(a) Find x.
(b) Find y.
Answer:
(a) The shift of the x-coordinates of A and B is 3.
So the shift of the y-coordinates of C and D is also 3.
4 – 1 = x – 4
⇒ x = 7
(b) The shift of the y-coordinates of A and B is the same as the shift of the y-coordinates of C and D
y – 2 = 6 – 2
⇒ y = 6
Midpoint
The point that divides the line into two equal parts is called the midpoint of the line.
If the coordinates of the endpoints of the line are (x1, y1) and (x2, y2), we need to think about what the coordinates of the midpoint will be. Look at the figure.

Let the coordinates of the midpoint be P(x, y).
The triangles AMP and ACB are similar.
Therefore, \(\frac{A M}{A C}=\frac{P M}{B C}=\frac{A P}{A B}\)
Since P is the midpoint,
\(\frac{A P}{A B}=\frac{1}{2}\)
That is, \(\frac{x-x_1}{x_2-x_1}=\frac{1}{2}\)
from this we get x = \(\frac{x_1+x_2}{2}\)
Similarly, y = \(\frac{y_1+y_2}{2}\)
The midpoint of the line joining (x1, y1) and (x2, y2) is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Question 1.
Based on the coordinate axes, the coordinates of the diameter AB of the circle are A(2, 2) and B(4, 2).

(a) Write the coordinates of the centre of the circle.
(b) Write the coordinates of the endpoints of the diameter perpendicular to AB.
Answer:
(a) (\(\frac{2+4}{2}\), 2) = (3, 2)
(b) C(3, 3), D(3, 1)
Question 2.
The sides of the square are parallel to the coordinate axes. Given that A(1, 1) and C(7, 4).

(a) Write the coordinates of the other two vertices.
(b) Write the coordinates of the midpoints of the sides.
Answer:
(a) B(7, 1), D(1, 4)
(b) (\(\frac{1+7}{2}\), 1) = S(4, 1)
R(7, \(\frac {5}{2}\))
Q(4, 4)
P(1, \(\frac {5}{2}\))
Question 3.
One endpoint of a diameter of the circle is (2, 3). If the centre is (7, 4), find the coordinates of the other endpoint of the diameter.
Answer:
Let the other endpoint be (x, y).
Using the midpoint formula:
\(\frac{2+x}{2}\) = 7
⇒ 2 + x = 14
⇒ x = 12
\(\frac{3+y}{2}\) = 4
⇒ 3 + y = 8
⇒ y = 5
Hence, the other endpoints of the diameter are (12, 5).
Question 4.
AB is a diameter parallel to the x-axis, and CD is a diameter parallel to the y-axis. If the center of the circle is (4, 4) and the radius is 3, then

(a) Write the coordinates of A and B.
(b) Write the coordinates of C and D.
Answer:
(a) A(1, 4), B(7, 4)
(b) C(4, 7), D(4, 1)
Ratio
This section is about the condition when a point divides a line in a given ratio.

The endpoints of the line are A(x1, y1) and B(x2, y2).
The point P(x, y) lies between them and divides the line segment AB.
AP is a certain part of AB, and PB is the remaining part of AB.
From point A to P, the horizontal distance is ‘a’ towards the right, and the vertical distance is ‘b’ upwards.
Therefore, the coordinates of P can be written as P(x1 + a, y1 + b).
From A to B, the horizontal distance is x2 – x1, and the vertical distance is y2 – y1.
The ratio AP : AB is derived from the similarity of the triangles shown in the figure.
That is, \(\frac{a}{x_2-x_1}=\frac{a}{y_2-y_1}=\frac{A P}{A B}\)
From this, ‘a’ and ‘b’ can be found, and then the coordinates of P can be as explained in the example.
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Question 1.
Write the coordinates of the point that divides the line joining A(2, 4) and B(8, 7) in the ratio 1 : 2.
Answer:
Look at the coordinates of A and B.
To move from A to B, you must go horizontally to the right and vertically upward.
The point that divides the line from A to B is one-third of the distance from A.
If the horizontal and vertical distances from A to P (the dividing point) are a and b, then the coordinates of that point are P(2 + a, 4 + b).
From A to B:
the horizontal distance = 8 – 2 = 6
the vertical distance = 7 – 4 = 3
Since the point divides AB in the ratio 1 : 2.
\(\frac{a}{6}=\frac{1}{3}\)
⇒ a = 2
\(\frac{b}{3}=\frac{1}{3}\)
⇒ b = 1
Therefor P = (2 + 2, 4 + 1) = (4, 5)
So the coordinates of P are (4, 5).
Question 2.
Write the coordinates of the point which divides the line joining A(1, 1) and B(7, -4) in the ratio 1 : 3.
Answer:
Look at the coordinates of A and B.
To move from A to B, you must go horizontally to the right and vertically downward.
The point that divides the line from A to B is one-fourth of the distance from A.
If the horizontal and vertical distances from A to P (the dividing point) are a and b, then (since the movement is downwards, that is, ‘-b’) the coordinates of that point are P(1 + a, 1 – b).
From A to B:
the horizontal distance = 7 – 1 = 6
the vertical distance = 1 – (-4) = 5
Since the point divides AB in the ratio 1 : 2
\(\frac{a}{6}=\frac{1}{4}\)
⇒ a = \(\frac {3}{2}\)
\(\frac{b}{5}=\frac{1}{4}\)
⇒ b = \(\frac {5}{4}\)
Therefor P = \(\left(1+\frac{3}{2}, 1-\frac{5}{4}\right)=\left(\frac{5}{2},-\frac{1}{4}\right)\)
So the coordinates of P are \(\left(\frac{5}{2},-\frac{1}{4}\right)\)
Question 3.
Write the coordinates of the point which divides the line joining A(1, 1) and B(-3, -7) in the ratio 2 : 3.
Answer:
In the displacement from A to B, the movement is towards the left and downward.
The coordinates of the dividing point are P(1 – a, 1 – b)
\(\frac{a}{1-(-3)}=\frac{2}{5}\)
⇒ \(\frac{a}{4}=\frac{2}{5}\)
⇒ 5a = 8
⇒ a = \(\frac {8}{5}\)
\(\frac{b}{1-(-7)}=\frac{2}{5}\)
⇒ \(\frac{b}{8}=\frac{2}{5}\)
⇒ 5b = 16
⇒ b = \(\frac {16}{5}\)
Therefor P = \(\left(1-\frac{8}{5}, 1-\frac{16}{5}\right)=\left(\frac{-3}{5}, \frac{-11}{5}\right)\)
So the coordinates of P are \(\left(\frac{-3}{5}, \frac{-11}{5}\right)\)
Line Maths
Consider the line passing through the points A(2, 5) and B(6, 7).

To go from A to B, the movement is 4 units horizontally, that is 6 – 2 = 4, and 2 units vertically, that is 7 – 5 = 2.
Here, in the movement from A to B, the vertical distance is half of the horizontal distance. This property will hold throughout the entire line.
For any two points on this line, the vertical distance will always be half of the horizontal distance.

If (x1, y1) and (x2, y2) are any two points on this line. Then the horizontal distance is x2 – x1 and the vertical distance is y2 – y1.
This can be written as, y2 – y1 = \(\frac {1}{2}\) × (x2 – x1)
For any two points on a line, not parallel to either axis, the change in the y-coordinate is the change in the x-coordinate multiplied by a fixed number.
Using this idea, if the coordinates of three points are known, we can check whether all three points lie on the same line.
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Question 1.
Check whether these points lie on the same line: (1, 8), (-2, 10), (-5, 12).
Answer:
Let A(1, 8) and B(-2, 10).
The change in x-coordinate is x2 – x1 = -2 – 1 = -3
That is, it decreases by 3.
The change in y-coordinate is y2 – y1 = 10 – 8 = 2
That is, it increases by a factor of 2.
The decimal form of the change is \(\frac{2}{-3}=-\frac{2}{3}\)
Now, consider B(-2, 10) and C(-5, 12).
The change in x-coordinate is x2 – x1 = -5 – (-2) = -3.
That is, it decreases by 3.
The change in y-coordinate is y2 – y1 = 12 – 10 = 2
That is, it increases by a factor of 2.
The decimal form of the change is \(\frac{2}{-3}=-\frac{2}{3}\)
Since the change is the same for any two points among the three.
Therefore, all three points lie on the same straight line.
Slope of a Line
Lines based on the coordinate axes can be drawn as horizontal, vertical, or slanted lines.
Horizontal lines are parallel to the x-axis, and vertical lines are parallel to the y-axis.
A slanting line is not parallel to either the x-axis or the y-axis.

Consider the line passing through the points (2, 8) and (4, 9).
The change in the y-coordinates is 1, and the change in the x-coordinates is 2.
We can find the relationship between the change in the y-coordinates and the change in the x-coordinates.
For any two points on a line not parallel to either axis, the change in y-coordinates divided by the change in x-coordinates gives the same number. This number is referred to as the slope of the line.
For any two points on a line, the change in the y-coordinate is equal to the product of the slope and the change in the x-coordinate.
The slope of a line can be positive, negative, or zero.
If a line is parallel to the x-axis, its slope is zero.
If, for two points on a line, the changes in both the y-coordinates and the x-coordinates are positive, or if both are negative, then the slope is positive.
If the line slants to the right, and the angle between the line and the x-axis on the right side is less than 90°.

If, for two points on a line, the change in one coordinate, either x or y, is positive and the change in the other is negative, then the slope of the line is negative.
The line slants downward to the left, and the angle between the line and the x-axis on the right side is greater than 90°.

To find the slope of a line, divide the change in the y-coordinates by the change in the x-coordinates of any two points on the line.
Question 1.
The two points on a line are (1, 1) and (-5, -5).
(a) Find the slope of this line.
(b) Will there be other lines with the same slope? What is the special feature of such lines?
(c) If a line makes an angle of 45° with the x-axis on the right-hand side, what is its slope?
Answer:
(a) The change in y-coordinates = -5 – 1 = -6
The change in x-coordinate = -5 – 1 = -6
Therefore, slope = \(\frac {-6}{-6}\) = 1
(b) There will be many other lines having the same slope.
If we take two points on each of these lines and divide the changes in y-coordinates by the changes in the x-coordinates will get 1.
Hence, all these lines are parallel to each other.
(c) For a line making a 45° angle with the x-axis on the right-hand side, the horizontal and vertical distances between any two points will be equal.
Therefore, the slope = 1.
Question 2.
In each of the following lines, two points are marked.

Find the slope of each line.
Answer:
Consider the points (-1, 0) and (3, 2):
Change in y-coordinate = 2 – 0 = 2
Change in x-coordinate = 3 – (-1) = 4
Therefore, the slope of the line = \(\frac{2}{4}=\frac{1}{2}\)
Consider the points (2, 0) and (3, 2):
Change in y-coordinate = 2 – 0 = 2
Change in x-coordinate = 3 – 2 = 1
Therefore, the slope of the line = \(\frac {2}{1}\) = 2
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Question 3.
A line makes an angle of 45° with the x-axis on the right-hand side. The line cuts the x-axis at a point 4 units away from the origin.
(a) What are the coordinates of the point where the line cuts the x-axis?
(b) What is the slope of the line?
(c) Write the coordinates of any two other points on this line.
Answer:
(a) The point where the line cuts the x-axis is (4, 0)
(b) Since the line makes a 45° angle with the x-axis on the right-hand side, the horizontal and vertical distances between any two points are equal.
Therefore, the slope = 1
(c) One point on the line: (4 + 3, 0 + 3) = (7, 3)
Another point on the line: (4 + 10, 0 + 10) = (14, 10)
Question 4.
Prove that A(1, 1), B(7, 3), C(10, 5), and D(4, 3) are the vertices of a parallelogram.
Answer:
The slope of line AB = \(\frac{3-1}{7-1}=\frac{2}{6}=\frac{1}{3}\)
The slope of line CD = \(\frac{3-5}{4-10}=\frac{-2}{-6}=\frac{1}{3}\)
∴ AB is parallel to CD.
The slope of line BC = \(\frac{5-3}{10-7}=\frac{2}{3}=\frac{2}{3}\)
The slope of line AD = \(\frac{3-1}{4-1}=\frac{2}{3}=\frac{2}{3}\)
∴ AD is parallel to BC.
Since both pairs of opposite sides are parallel,
Therefore, ABCD is a parallelogram.
Equation of a Line
Now it’s clearly understood what the slope of a line means. A line has only one slope, and for any two points on that line, the change in their y-coordinates is always a fixed multiple of the change in their x-coordinates. This fixed number is referred to as the slope of the line. There can be many lines with the same slope; all such lines are parallel to each other. For every line, the x-and y-coordinates of its points have a common relationship. The algebraic expression of this relationship is called the equation of the line.
The points (1, 1), (2, 2), (3, 3)… lie on the same line. What is the relationship between the x-coordinate and the y-coordinate of all the points on this line?
It can be written as y = x or y – x = 0.
This relationship is the equation of the line.
The points (-1, 1), (0, 0), and (-5, -5) also lie on this same line.
Isn’t the figure of this line shown below?

The points (1, -1), (-1, 1), (3, -3), and (-3, 3) all lie on the same line.
The coordinates of the points on this line satisfy the relation y = -x, that is, x + y = 0.
This is the equation of the line.

The equation of a line can be written using its slope.
If the coordinates of a point and the slope are known, the equation of the line can be found.
Let’s find the equation of the line passing through the point (1, 3) with a slope of -2.
Let (x, y) be any point on the line.
The change in the y-coordinate between the points (1, 3) and (x, y) is equal to -2 times the change in the x-coordinate.
In equation form: y – 3 = -2(x – 1)
That is, y – 3 = -2x + 2
⇒ 2x + y – 5 = 0
This is the equation of the line.
If the equation of a line is written in this form,
The slope is the negative of the coefficient of x divided by the coefficient of y.
The general form of the equation of a line is ax + by + c = 0
Therefore its slope is \(-\frac {a}{b}\)
Question 1.
Find the slope of the line passing through the points (1, 0) and (0, 1). Write the equation of this line.
Answer:
Slope = \(\frac{1-0}{0-1}=\frac{1}{-1}\) = -1
Let (x, y) be any point on the line.
Then, y – 0 = -1(x – 1)
So, y = -x + 1
⇒ x + y – 1 = 0
This is the required equation of the line.
Question 2.
Find the slope of the line x + 2y – 3 = 0. Write the equation of another line having the same slope. What is the special property of these two lines?
Answer:
The slope m = \(-\frac {1}{2}\)
The slope of a line depends only on the coefficients of x and y.
If these coefficients or their ratio remain unchanged, the slope will also remain the same.
Hence, x + 2y – 1 = 0 is another line with the same slope.
These two lines are parallel to each other.
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Question 3.
Write the equation of the median passing through point C of the triangle shown in the figure.

Answer:
Midpoint of AB is \(M\left(\frac{7}{2}, \frac{3}{2}\right)\)
Slope of the median = \(\frac{3-\frac{3}{2}}{2-\frac{7}{2}}\) = -1
Let (x, y) be a point in the median.
The equation is y – 3 = -1(x – 2)
⇒ y – 3 = -x + 2
⇒ x + y – 5 = 0
Equation of a Circle
In the previous classes and lessons, we learned many things about the circle.
In those lessons, we studied the circle geometrically.
Here, we will understand the algebraic form of the circle.
Using the coordinates that define the circle, we can write the equation of the circle.
Let (x, y) be any point on the circle and (a, b) be the centre of the circle.
The distance between these points is called the radius of the circle.
Let r be the radius. Then we can write:
r = \(\sqrt{(x-a)^2+(y-b)^2}\)
This is the equation of the circle.
Question 1.
Write the equation of the circle with centre (1, 1) and radius 2.
Answer:
Let (x, y) be any point on the circle.
(x – 1)2 + (y – 1)2 = 22
⇒ x2 + 12 – 2x + y2 + 12 – 2y = 4
⇒ x2 + y2 – 2x – 2y + 2 – 4 = 0
⇒ x2 + y2 – 2x – 2y – 2 = 0
Question 2.
Write the equation of the circle whose centre is the origin and whose radius is √2.
Answer:
Let (x, y) be any point on the circle.
(x – 0)2 + (y – 0)2 = (√2)2
⇒ x2 + y2 = 2
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Question 3.
Write the equation of the circle with centre (1, 1) and radius √5. Calculate the coordinates of the points where the circle intersects the x-axis.
Answer:
Let (x, y) be any point on the circle.
(x – 1)2 + (y – 1)2 = (√5)2
⇒ x2 + 1 – 2x + y2 + 1 – 2y = 5
⇒ x2 + y2 – 2x – 2y – 3 = 0
The y-coordinates of the points where it intersects the x-axis.
x2 + 02 – 2x – 0 – 3 = 0
⇒ x2 + 2x – 3 = 0
To find the values of x,
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
Therefore, x = 3, -1.
The points that intersect the x-axis are (3, 0) and (-1, 0).
Question 4.
Find the radius and centre of the circle x2 + y2 – 4x + 2y – 4 = 0.
Answer:
x2 + y2 – 4x + 2y – 4 = 0
⇒ x2 + y2 – 4x + 2y = 4
⇒ (x2 – 4x + 4) + (y2 + 2y + 1) = 4 + 5
⇒ (x – 2)2 + (x + 1)2 = 32
Therefore, Centre of the circle = (2, -1)
Radius = 3.

























































































































































