Geometry and Algebra Class 10 Extra Questions Kerala State Syllabus Maths Chapter 11

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 11 Geometry and Algebra Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 11 Geometry and Algebra Important Questions and Answers

Geometry and Algebra Class 10 Extra Questions Kerala Syllabus

Geometry and Algebra Class 10 Kerala Syllabus Extra Questions

Question 1.
The endpoints of the diameter of a circle are (1, 1) and (7, 9). What is the center of the circle?
(a) (4, 5)
(b) (1, 3)
(c) (3, 1)
(d) (0, 2)
Answer:
(a) (4, 5)

Question 2.
What is the slope of the line passing through (0, 3) and (7, 3)?
(a) 0
(b) 1
(c) -1
(d) 2
Answer:
(a) 0
Note the equality of the y-coordinates of the given points. The line is parallel to the x-axis.

Question 3.
Which of the following is the center of the circle (x – 1)² + y² = 1
(a) (1, 0)
(b) (1, 1)
(c) (1, 2)
(d) (1,-1)
Answer:
(a) (1, 0)
Compare with (x – a)² + (y – b)² = r²

Question 4.
Read the two statements given below.
Statement 1: The midpoint of the line joining (2, 3) and (4, 7) is (3, 5).
Statement 2: The midpoint of the line joining (x₁, y₁) and (x₂, y₂) is given by \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Choose the correct answer from those given below.
(a) Statement 1 is true, and Statement 2 is false.
(b) Statement 1 is false, and Statement 2 is true.
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both statements are true, but Statement 2 is not the correct reason for Statement 1.
Answer:
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.

Question 5.
Read the two statements given below.
Statement 1: The coordinates of the midpoint of a line joining (-3, 6) and (1, -2) are (-1, 2).
Statement 2: A midpoint divides a line segment joining two points in the ratio 1 : 1.
Choose the correct answer from those given below.
(a) Statement 1 is true, and Statement 2 is false.
(b) Statement 1 is false, and Statement 2 is true.
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both statements are true, but Statement 2 is not the correct reason for Statement 1.
Answer:
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.

Question 6.
The line passing through (k, 4) and (-3, -2) has slope \(\frac {1}{2}\).
(a) Find k.
(b) Write the coordinates of another point on this line.
Answer:
(a) Slope = \(\frac {1}{2}\)
Change in the y-coordinate = -2 – 4 = -6
Change in the x-coordinate = -3 – k
⇒ \(\frac{-6}{-3-k}=\frac{1}{2}\)
⇒ -12 = -3 – k
⇒ k = 9
(b) Another point is (-1, -1).

Question 7.
A line makes an angle of 45° with the positive direction of the x-axis
(a) What is the slope of this line?
(b) (4, 0) is a point on this line. Write the coordinates of one more point on this line?
Answer:
(a) According to the property of a 45° – 45° – 90° triangle,
The change in the y-coordinates and x-coordinates of the two points on this line will be equal.
Therefore, slope = 1

(b) Another point on the line = (4 + 1, 0 + 1) = (5, 1)

Question 8.
The vertices of a triangle are A(-3, 2), B(3, -4), and C(1, 5).
(a) Find the midpoint of the sides AB.
(b) Calculate the length of the median to the side AB.
(c) Find the centroid.
Answer:
(a) Midpoint of AB = \(\left(\frac{-3+3}{2}, \frac{2 \pm 4}{2}\right)\) = (0, -1)

(b) Median is the line joining a vertex to the midpoint of the opposite side.
Length of the median to the side AB = \(\sqrt{(1-0)^2+(5–1)^2}=\sqrt{37}\)

(c) The centroid divides the median in the ratio 1 : 2
Let the point be G(x, y) then, x = 1 – a, y = 5 – b
\(\frac{a}{1-0}=\frac{2}{3}\)
⇒ a = \(\frac {2}{3}\)
⇒ x = 1 – \(\frac {2}{3}\) = \(\frac {1}{3}\)
\(\frac{b}{5-(-1)}=\frac{2}{3}\)
⇒ b = \(\frac {12}{3}\) = 4
⇒ y = 5 – 4 = 1
Therefore the centroid (\(\frac {1}{3}\), 1)

Question 9.
A line passes through (1, 1) and (3, 4).
(a) What is the slope of the line?
(b) Write the equation of this line.
Answer:
(a) Slope = \(\frac{4-1}{3-1}=\frac{3}{2}\)

(b) Let (x, y) be a point on this line.
y – 1 = \(\frac {3}{2}\)(x – 1)
⇒ 2(y -1) = 3(x – 1)
⇒ 2y – 2 = 3x – 3
⇒ 3x – 2y = 1

Question 10.
Consider the line 2x + 3y = 6
(a) What are the points at which the line cuts the coordinate axes?
(b) Write the slope of this line.
Answer:
(a) When the line cuts the x-axis at y = 0
2x + 3 × 0 = 6
⇒ 2x = 6
⇒ x = 3
The point on the x-axis is (3, 0)
When the line cuts the y-axis, x = 0.
2 × 0 + 3y = 6
⇒ 3y = 6
⇒ y = 2
The point is (0, 2)

(b) Slope = \(\frac{0-2}{3-0}=\frac{-2}{3}\)

Question 11.
The line x + y = 2 intersects the x-axis at A and y-axis at B.
(a) Write the coordinates of A and B .
(b) Find the coordinates of the circumcentre of the triangle AOB?
Answer:
(a) At A, y = 0, x + 0 = 2 ⇒ x = 2.
The point A(2, 0)
At B, x = 0, 0 + y = 2
The point B(0, 2)

(b) Circumcentre is the midpoint of the hypotenuse AB.
Midpoint of A(2, 0) and B(0, 2) is \(\left(\frac{2+0}{2}, \frac{0+2}{2}\right)\) = (1, 1)
Therefore the circumcentre is (1, 1).

Question 12.
3x – y = 9 is the equation of a line.
(a) What is the slope of this line?
(b) Write the equation of another line parallel to this line.
Answer:
(a) Slope = \(\frac {-3}{-1}\) = 3
(b) 3x – y = 1

Question 13.
The lines x + y = 4, x + y = -4, x – y = 4, x – y = -4 encloses a quadrilateral.
(a) What are the vertices of this quadrilateral?
(b) Suggest a suitable name to his quadrilateral.
(c) What is the area of this quadrilateral?
Answer:
(a) A(4, 0), B(0, 4), C(-4, 0), D(0, -4)
(b) Square
(c) Side is 4√2
Area of the square = (4√2)² = 32

Question 14.
A line passes through the points (a, 0) and (0, b).
(a) Write the equation of the line.
(b) Rewrite the equation as \(\frac{x}{a}+\frac{y}{b}\) = 1
(c) If (p, q) is the mid point of the line joining (a, 0) and (0, b) then prove that \(\frac{x}{p}+\frac{y}{q}\) = 2
Answer:
(a) Let (x, y) be a point on the line.
\(\frac{y-0}{x-a}=\frac{b-0}{0-a}\)
⇒ \(\frac{y}{x-a}=\frac{-b}{a}\)
⇒ ay + bx = ab

(b) Dividing both sides by ab, we get
\(\frac{a y}{a b}+\frac{b x}{a b}=\frac{a b}{a b}\)
⇒ \(\frac{x}{a}+\frac{y}{b}\) = 1

(c) If (p, q) is the mid point of the line joining (a, 0) and (0, b) then
p = \(\frac{a+0}{2}\), q = \(\frac{0+b}{2}\)
⇒ a = 2p, b = 2q
Equation becomes \(\frac{x}{2 p}+\frac{y}{2 q}\) = 1
That is \(\frac{x}{p}+\frac{y}{q}\) = 2

Question 15.
Consider the points (6, 0) and (0, 6).
(a) What are the coordinates of the midpoint of the line joining these points.
(b) Find the slope of the line passing through this point.
(c) Write the equation of the line.
(d) At what point the line cut x-axis?
(e) At what point the line cut y-axis?
Answer:
(a) Slope = \(\frac{7-2}{5-1}=\frac{5}{4}\)
(b) Let (x, y) be a point on the line.
\(\frac{y-2}{x-1}=\frac{5}{4}\)
(c) Equation of the line is 4y – 5x = 3
(d) (\(\frac {-3}{5}\), 0)
(e) (0, \(\frac {3}{4}\))

Question 16.
If the endpoints of the diameter are A(-1, 0) and B(1, 0):
(a) Write the coordinates of the centre of the circle.
(b) What is the radius?
Answer:
(a) Centre of the circle is the midpoint of AB.
Centre = \(\left(\frac{-1+1}{2}, \frac{0+0}{2}\right)\) = (0, 0)

(b) The radius is half the length of the diameter.
AB = \(\sqrt{(1-(-1))^2+(0-0)^2}=\sqrt{(2)^2}\) = 2
Radius r = \(\frac{A B}{2}=\frac{2}{2}\) = 1

Question 17.
If the centre of the circle is (3, 4) and the radius is 4:
(a) Write the coordinates of a point on the circle.
(b) Write the equation of the circle.
Answer:
(a) The perpendicular distance from the centre to the x-axis is 4, which is the radius.
The x-axis is the tangent to the circle.
The point of contact is (3, 0), which is a point on the circle.
(b) (x – 3)² +(y – 4)² = 42
⇒ (x – 3)² + (y – 4)² = 16

Question 18.
A circle touches the y-axis at the point (0, 5). If the radius of the circle is 2, then:
(a) Write the coordinates of the two possible centres of the circle.
(b) Write the equation of one of these circles.
Answer:
(a) (2, 5), (-2, 5)
(b) (x – 2)² + (y – 5)² = 4

Question 19.
A circle touches the sides of a square whose vertices are (4, 0), (0, 4), (-4, 0), and (0, -4).
(a) Write the coordinates of the points where the circle touches the sides of the square.
(b) Write the equation of the circle.
Answer:
(a) (2, 2), (-2, 2), (2, -2), (-2, -2)
(b) x² + y² = (2√2)²
⇒ x² + y² = 8