KSEEB SSLC Solutions for Class 10 Maths – Coordinate Geometry (English Medium)
Exercise 13.1:
Question I:
Locate the following points of a graph sheet.
- P (4, -3)
- R (-1, -1)
- I (0, -5)
- X(-5, -2)
- Y (3, 2)
- Z (4, 0)
- E (0, 6)
- 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 IVth quadrant.
ii. The point (3, 1) has positive x-coordinate (abscissa) and positive y-coordinate (ordinate). The point (3, 1) lies in Ist quadrant.
iii. The point (-10, -2) has negative x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (-10, -2) lies in IIIrd quadrant.
iv. The point (-5, -2) has negative x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (-5, -2) lies in IIIrd quadrant.
v. The point (-5, -1) has negative x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (-5, -1) lies in IIIrd quadrant.
vi. The point (5, -7) has positive x-coordinate (abscissa) and negative y-coordinate (ordinate). The point (5, -7) lies in IVth quadrant.
vii. The point (9, 9) has positive x-coordinate (abscissa) and positive y-coordinate (ordinate). The point (9, 9) lies in Ist quadrant.
viii. The point (-2, 7) has negative x-coordinate (abscissa) and positive y-coordinate (ordinate). The point (-2, 7) lies in IInd quadrant.
Question III:
Plot the points on a Cartesian plane whose
- Ordinate is 4, abscissa is 0
- Ordinate is -3, abscissa is -1
- Ordinate is 5, abscissa is 4
- Ordinate is -1, abscissa is 7
Solution :
- Ordinate is 4, abscissa is 0 = (0, 4)
- Ordinate is -3, abscissa is -1 = (-1, -3)
- Ordinate is 5, abscissa is 4 = (4, 5)
- Ordinate is -1, abscissa is 7= (7, -1)
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 :
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 :
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. (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.
- (-5, -3), (1, -11), (7, -6), (1, 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 :
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 :