KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions

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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. α². β²
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 (x₁,y₁) and (x₂, y₂) is

Answer:

Question 7.
Which of the following is distance formula?
Answer:
a. The distance between 2 points (x₁,y₁) and (x₂, y₂) is

b. The mid point of the lines segment join the points (x, y) and (x₂, y₂)

c. The co-ordinates of the centroid of the ∆^le ABC where A, B, C are (x₁,y₁)

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 (x₂, y₂) in the ratio m₁ : m₂ 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:
(x₁, y₁) = (3, 1); (x₂, y₂) = (o, x)
d = 5

(∵ squaring on both the sides)
25 = (3)₂ + (x – 1)₂
x₂ + 1 – 2x + 9 = 25
x₂ + 1 – 2x + 9 – 25 = 0
x₂ + 1 – 2x + 9 – 25 = 0
x₂ – 2x – 15 = 0
x₂ – 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:
(x₁, y₁) = (3, 5)
(x₂, y₂) = (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)
(x₁, y₁) = A(- 2, 7) m : n = 3 : 2
B(x₂, y₂) =?


∴ 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); (x₁, y₁) = (- 3,10)
(x₂, y₂) = (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(x₁, y₁) =(1, 2);
c(x₂, y₂) = (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) = (x₁, y₁); D(x, y) = (x₂, y₂)
\(\left(\frac{9}{2} \frac{7}{2}\right)\) = (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 = \(\frac{1}{2}\)[x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
= \(\frac{1}{2}\)[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.