Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions

I. Multiple-choice Questions:

Question 1.

The co-ordinates of origin are

a. 0,0.

b. 0, 1.

c. 1, 0.

d. 1, 1.

Answer:

a. 0,0.

Question 2.

The distance of the point (3, 4) x – co-ordinate is equal to

a. 3

b. 4

c. 1

d. 7

Answer:

b. 4

Question 3.

The distance of the point (3, 4) from y-axis is

a. 3

b. 4

c. 5

d. 1

Answer:

a. 3

Question 4.

The angle between x-axis and y-axis is

a. 0°

b. 45°

c. 60°

d. 90°

Answer:

d. 90°

Question 5.

The distance of the point (α, β) from origin is

a. α.β

b. α^{2}. β^{2}

c. \(\sqrt{\alpha^{2}-\beta^{2}}\)

d. \(\sqrt{\alpha^{2}+\beta^{2}}\)

Answer:

d. \(\sqrt{\alpha^{2}+\beta^{2}}\)

Question 6.

The distance between the point (x_{1},y_{1}) and (x_{2}, y_{2}) is

Answer:

Question 7.

Which of the following is distance formula?

Answer:

a. The distance between 2 points (x_{1},y_{1}) and (x_{2}, y_{2}) is

b. The mid point of the lines segment join the points (x, y) and (x_{2}, y_{2})

c. The co-ordinates of the centroid of the ∆^{le} ABC where A, B, C are (x_{1},y_{1})

d.

Question 8.

The distance b/w the point (0, 3) and (- 2, 0) is

a. \(\sqrt{14}\)

b. \(\sqrt{15}\)

c. \(\sqrt{13}\)

d. \(\sqrt{5}\)

Answer:

c. \(\sqrt{13}\)

Question 9.

The ∆^{le} with vertices (- 2, 1) (2, – 2) and (5, 2) is

a. Scalene

b. Equilateral

c. Isosceles

d. right angled isosceles

Answer:

d. right angled isosceles

Question 10.

The Co-ordinates of the point which divides the join of (x, y) and (x_{2}, y_{2}) in the ratio m_{1} : m_{2} internally are

Answer:

Question 11.

If the distance b/w the point (3, a) and (4,1) is \(\sqrt{10}\), then, find the values of a

a. 3,- 1

b. 4, – 2

c. 2, – 2

d. 5, – 3

Answer:

c. 2, – 2

Question 12.

The distance of the mid – point of the line -segment joining the points 6, 8. and 2, 4. from the point (1, 2) is

a. 3

b. 4

c. 5

d. 6

Answer:

c. 5

Question 13.

The coordinates of the mid – point of the line – segment joining (- 8, 13) and (x, – 7)

a. 16

b. 10

c. 4

d. 8

Answer:

a. 16

Question 14.

Find the value of k if the distance b/w (k, 3) and (2, 3) is 5

a. 5

b. 6

c. 7

d. 8

Answer:

c. 7

Question 15.

The distance b/w the point (Sin θ, Sin θ) & (cos θ, – cos θ) is

a. \(\sqrt{2}\)

b. 2

c. \(\sqrt{3}\)

d. 1

Answer:

a. \(\sqrt{2}\)

II. Short Answers Questions:

Question 1.

Write the coordinate of the origin.

Answer:

(0, 0)

Question 2.

What is the ordinate of all the point on x-axis

Answer:

0(zero)

Question 3.

Find the perpendicular distance of the point (- 3, 5) form the x – axis.

Answer:

5

Question 4.

Find the distance of B(3 + \(\sqrt{3}\), 3 – \(\sqrt{3}\)) form origin.

Answer:

d = \(\sqrt{24}\)

d = \(\sqrt{4 \times 6}\)

d = 2\(\sqrt{6}\)

Question 5.

Write the formula to find distance b/w two points

Answer:

Question 6.

Write the formula to find distance b/W point and origin.

Answer:

d = \(\sqrt{x^{2}+y^{2}}\)

Question 7.

Write the formula to find area of triangle.

Answer:

area of triangle

Question 8.

Find the value of a, so that the point (3, a) lie on the line 2x – 3y = 5.

Answer:

Since (3, a) lies on the line 2x – 3y = 5

Then 2(3) – 3(a) = 5

6 – 3a = 5

– 3a = 5 – 6

– 3a = – 1

a = 1/3

Question 9.

Write the section formula when it divide in ration p : q

Answer:

Co-ordinates

III. Long Answer Questions:

Question 1.

Find the distance between the following pairs of points.

i) (6, 5) and (4, 4)

Answer:

Question 2.

The distance between the point (3, 1) and (0, x) in 5 unit. Find x.

Answer:

(x_{1}, y_{1}) = (3, 1); (x_{2}, y_{2}) = (o, x)

d = 5

(∵ squaring on both the sides)

25 = (3)_{2} + (x – 1)_{2}

x_{2} + 1 – 2x + 9 = 25

x_{2} + 1 – 2x + 9 – 25 = 0

x_{2} + 1 – 2x + 9 – 25 = 0

x_{2} – 2x – 15 = 0

x_{2} – 5x + 3x – 15 = 0

x(x – 5) + 3(x – 5) = 0

(x – 5)(x + 3) = 0

x – 5 = 0, x + 3 = 0

x = 5, x = – 3

Question 3.

Find a point on y-axis which is equidistant from the points (5, 2) and (- 4, 3)

Answer:

A Point lie on y – axis = (0, y) the points are equidistant from the point on y – axis.

6y – 4y = 25 – 29

2y = – 4

y = \(\frac{-4}{2}\) ⇒ y = – 2

∴ point on y – axis is y = – 2

Question 4.

Find the radius of a circle whose centre is (- 5, 4) and which passes through the point (- 7,1)

Question 5.

Prove that each of the set of co¬ordinates are the vertices of parallelograms.

i. (- 5, – 3), (1, – 11), (7, – 6), (1, 2).

Answer:

∴ AD = BC & AB = CD

∴ Given vertices from a parallelogram.

Question 6.

In what ratio does the point (- 2, 3) divide the line segment joining the point (- 3, 5) and (4, – a)?

Answer:

(x_{1}, y_{1}) = (3, 5)

(x_{2}, y_{2}) = (4, – a)

p(x, y) = (- 2, 3)

4m – 3n = – 2(m + n)

4m – 3n = – 2m – 2n

4m + 2m = – 2n + 3n

6m = n

\(\frac{m}{n}=\frac{1}{6}\)

m : n = 1 : 6 (or)

\(\frac{-9 m+5 n}{m+n}\) = 3

– 9m + 5n = 3(m + n)

– 9m + 5n = 3m + 3n

– 9m – 3m = 3n – 5n

– 12m = – 2n

\(\frac{m}{n}=\frac{-2}{-12}=\frac{1}{6}\)

m : n = 1 : 6

Question 7.

In the point C(1, 1) divides the line segment joining A(- 2, 7) and B in the ration 3 : 2 find the co-ordinates of B.

Answer:

Point C(1, 1)

(x_{1}, y_{1}) = A(- 2, 7) m : n = 3 : 2

B(x_{2}, y_{2}) =?

∴ Co-ordinate of B = (3, – 3)

Question 8.

Find the ratio in which the point (- 1, k) divides the line joining the points (- 3, 10) and (6, – 8) and also find ‘k’.

Answer:

P(x, y) = (- 1, k); (x_{1}, y_{1}) = (- 3,10)

(x_{2}, y_{2}) = (6,- 8)

6m – 3n = – (m + n)

6m – 3n = – m – n

6m + m = 3n – n, \(\frac{-8 m+10 n}{m+n}\) = k

Question 9.

Three Consecutive vertices of a parallelogram are A(1, 2), B(2, 3) and (8, 5). Find the fourth vertex.

Answer:

‘O’ is the mid point of AC and by using mid point formulas AC is a line and ‘O’ is the point

A(x_{1}, y_{1}) =(1, 2);

c(x_{2}, y_{2}) = (8, 5)

Co-ordinates of

Co-ordinates of ‘O’ = \(\frac{9}{2}, \frac{7}{2}\)

BD is a line and ‘O’ is midpoint

B(2, 3) = (x_{1}, y_{1}); D(x, y) = (x_{2}, y_{2})

\(\left(\frac{9}{2} \frac{7}{2}\right)\) = (x, y)

By midpoint formula

Co-ordinates of

2 + x = 9 3 + y = 7

x = 9 – 2 y = 7 – 3

x = 7 y = 4

the fourth vertex is (7, 4)

Question 10.

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

Answer:

Given point are A(x, y), B(1, 2) and C(7, 0)

These points will be collinear if the area of the triangle formed by them is zero.

Now, ar(ABC)

⇒ 2x + 6y – 14 = 0

⇒ x + 3y = 7

which is the required relation between x & y

Question 11.

Find the ratio in which the line segment joining the points A(3, – 3) and B(- 2,7) is divides by x-axis. Also find the Co-ordinates of the point of division.

Answer:

Here, point Q is an x-axis its ordinate & 0

Let ratio be k : 1 and Co-ordinate of point Q be (x, 0)

Question 12.

If A(4, 2), B = (7, 6) and C(1, 4) are the vertices of ∆ ABC and AD is its median, P.T that median AD divides ∆ ABC into two triangle of equal areas.D is the median on BC

⇒ The Co-ordinates of midpoint D are given by

Co-ordinates of D are (4, 5)

Now, Area of triangle

ABD = \(\frac{1}{2}\)[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})

= \(\frac{1}{2}\)[4(6 – 5) + 7(52) + 4(2 – 6)

= \(\frac{1}{2}\)[4 + 21 – 16] = \(\frac{1}{2}\) sq.units

Area of

∆ ACD = \(\frac{1}{2}\)[4[4 – 5] + 1[5 – 2] + 4[2 – 4]]

= \(\frac{1}{2}\)[- 4 + 3 – 8] = \(\frac{-9}{2}=\frac{9}{2}\) sq.units

Question 13.

In what ratio does the y – axis divide the line segment joining the point p(- 4, 5) and Q(3, 7). Also, find the co-ordinates of the intersection.

Answer:

Suppose y – axis divides PQ in the ratio k : 1. Then the co-ordinates of the point of division are

Since, R lies on y-axis and X Co-ordinate of every point on y-axis is zero.

∴ \(\frac{3 \mathrm{K}-4}{\mathrm{K}+1}\) = 0

⇒ 3k – 4 = 0 ⇒ K = \(\frac{4}{3}\)

Hence, the required ratio is 4/3 : 1 i.e, 4 : 3

Putting K = 4/3 in the Co-ordinates of R, we find that its Co-ordinates are (0, – 13/7)

Question 14.

Show that ∆ ABC with vertices A(- 2, 0), B(2, 0) and C(0, 2) is similar to ∆ DEF with vertices D(- 4, 0), E(4, 0) and F(0, 4)

Answer:

Given vertices of ∆ ABC and ∆ DEF are

A(- 2, 0), B(2, 0), C(0, 2), D(- 4, 0), E(4, 0) and F(0, 4)

Here, we see that sides of ∆ DEF are twice the sides of a ∆ ABC.

Hence, both triangle are similar.