KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry

In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 13 Coordinate Geometry for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 13 Coordinate Geometry pdf, free KSEEB solutions for Class 10 Maths Chapter 13 Coordinate Geometry book pdf download. Now you will get step by step solution to each question.

KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry (English Medium)

KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry Exercise 13.1:

Question I:

Locate the following points of a graph sheet.

1. P (4, -3)
2. R (-1, -1)
3. I (0, -5)
4. X(-5, -2)
5. Y (3, 2)
6. Z (4, 0)
7. E (0, 6)
8. F(-2, 5)

Solution :

Question II:

In which quadrants do these points lie?

i. (4, -6)
ii. (3, 1)
iii. (-10, -2)
iv. (-5, -2)
v. (-5, -1)
vi. (5, -7)
vii. (9, 9)
viii. (-2, 7)

Solution :

i. The point (4, -6) has positive x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (4, -6) lies in IV^th quadrant.
ii. The point (3, 1) has positive x-coordinate (abscissa) and positive y-coordinate (ordinate). The point (3, 1) lies in I^st quadrant.
iii. The point (-10, -2) has negative x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (-10, -2) lies in III^rd quadrant.
iv. The point (-5, -2) has negative x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (-5, -2) lies in III^rd quadrant.
v. The point (-5, -1) has negative x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (-5, -1) lies in III^rd quadrant.
vi. The point (5, -7) has positive x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (5, -7) lies in IV^th quadrant.
vii. The point (9, 9) has positive x-coordinate (abscissa) and positive y-coordinate (ordinate). The point (9, 9) lies in I^st quadrant.
viii. The point (-2, 7) has negative x-coordinate (abscissa) and positive y-coordinate (ordinate). The point (-2, 7) lies in II^nd quadrant.

Question III:

Plot the points on a Cartesian plane whose

1. Ordinate is 4, abscissa is 0
2. Ordinate is -3, abscissa is -1
3. Ordinate is 5, abscissa is 4
4. Ordinate is -1, abscissa is 7

Solution :

1. Ordinate is 4, abscissa is 0 = (0, 4)
2. Ordinate is -3, abscissa is -1 = (-1, -3)
3. Ordinate is 5, abscissa is 4 = (4, 5)
4. Ordinate is -1, abscissa is 7= (7, -1)

KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry Exercise 13.2:

Question 1:

Find the slope of the line whose inclination is

i. 90°
ii. 45°
iii. 30°
iv. 0°

Solution :

Question 2:

Find the angles of inclination of straight lines whose slopes are

Solution :

Question 3:

Find the slope of the line joining the points

Solution :

Question 4:

Find whether the lines drawn through the two pairs of points are parallel or perpendicular

i. (5,2), (0,5) and (0,0), (-5,3)
ii. (3,3), (4,6) and (4,1), (6,7)
iii. (4,7), (3,5) and (-1,7), (1,6)
iv. (-1,-2),(1,6) and (-1,1), (-2,-3)

Solution :

Question 5:

Find the slope of the line perpendicular to the line joining the points

i. (1, 7) and (-4, 3)
ii. (2, -3) and (1, 4)

Solution :

Question 6:

Find the slope of the line parallel to the line joining the points

i. (-4, 3) and (2, 5)
ii. (1, -5) and (7, 1)

Solution :

Question 7:

A line passing through the points (2,7) and (3,6) is parallel to the line joining (9, a) and (11, 3). Find the value of a.

Solution :

Question 8:

A line passing through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0). Find the value of m.

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry Exercise 13.3:

Question 1:

Find the equation of the line whose angle of inclination and y- intercept are given.

i. θ = 60° , y – intercept is – 2
ii. θ = 45° , y – intercept is 3

Solution :

Question 2:

Find the equation of the line whose slops and y-intercept are given.

i. Slope = 2, y- intercept = -4

iii. Slope = -2, y-intercept = 3

Solution :

Question 3:

Find the slope and y-intercept of the lines

i. 2x + 3y = 4
ii. 3x = y
iii. x – y + 5 = 0
iv. 3x – 4y = 5

Solution :

Question 4:

Is the line x = 2y parallel to 2x – 4y + 7 = 0 [Hint : Parallel lines have same slopes]

Solution :

Question 5:

Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50=0 are perpendicular to each other. [Hint : For perpendicular lines, m₁m₂=-1

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry Exercise 13.4:

Question 1:

Find the distance between the following pairs of pointsi. (8, 3) and (8, -7)
ii. (1, -3) and (-4, 7)
iii. (-4, 5) and (-12, 3)
iv. (6, 5) and (4, 4)
v. (2, 0) and (0, 3)
vi. (2, 8) and (6, 8)
vii.(a, b) and (c, b)
viii. (cosθ, -sinθ) and (sinθ, -cosθ)

Solution :

Question 2:

Find the distance between the origin and the point

i. (-6, 8)ii
ii. (5, 12)
iii. (-8, 15)

Solution :

Question 3(i):

The distance between the points (3,1) and (0,x) is 5 units. Find x.

Solution :

Question 3(ii):

A point P (2,-1) is equidistant from the points (a, 7) and (-3, a). Find ‘a’.

Solution :

Question 3(iii):

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

Solution :

Question 4:

Find the perimeter of the triangle whose vertices have the following coordinates

i. (-2, 1), (4, 6),(6, -3)
ii. (3, 10), (5, 2), (14, 12)

Solution :

Question 5:

Prove that the points A(1, -3), B(-3, 0) and C(4, 1) are the vertices of a right isosceles triangle.

Solution :

Question 6:

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

Solution :

Question 7:

Prove that each of the set of coordinates are the vertices of parallelogram.

1. (-5, -3), (1, -11), (7, -6), (1, 2)
2. (4, 0), (-2, -3), (3, 2), (-3, -1)

Solution :

Question 8:

The coordinates of vertices of triangles are given. Identify the types of triangles.

i. (2,1) (10,1) (6,9)
ii. (1,6) (3,2) (10,8)
iii. (3,5) (-1,1) (6,2)
iv. (3,-3) (3,5) (11,-3)

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 13 Coordinate Geometry Exercise 13.5:

Question 1:

In what ratio does the point (-2,3) divide the line segment joining the points (-3,5) and (4,-9) ?

Solution :

Question 2:

In the point C(1,1) divides the line segment joining A(-2,7) and B in the ratio 3:2, Find the coordinates of B.

Solution :

Question 3:

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

Solution :

Question 4:

Find the coordinates of the midpoint of the line joining the points (-3,10) and (6,-8)

Solution :

Question 5:

Three consecutive vertices of a parallelogram are
A (1, 2), B (2, 3) and C (8, 5). Find the fourth vertex. (Hint : diagonals of a parallelogram bisect each other)

Solution :

All Chapter KSEEB Solutions For Class 10 Maths

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