Coordinate Geometry Exercise 14.1 – Class 10

Coordinate Geometry Exercise 14.1 – Questions:

  1. Find the slope of the curve whose inclination is

(i) 90˚

(ii) 45˚

(iii) 30˚

(iv) 0˚

  1. Find the angles of inclination of straight lines whose slopes are

(i)√3

(ii) 1

(iii) 1/√3

  1. Find the slope of the line joining the points

(i) (-4, 1) and (-5, 2)

(ii) 4, -8) and (5, -2)

(iii) (0, 0) and (√3, 3)

(iv) (-5, 0) and (0, -7)

(v) (2a, 3b) and (a, -b)

  1. 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)

  1. 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)

  1. 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)

  1. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a) and (11, 3). Find a.
  2. A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0). Find the value of m.

 


Coordinate Geometry Exercise 14.1 – Solutions:

  1. Find the slope of the curve whose inclination is

(i) 90˚

Solution:

θ = 90˚

Slope = m = tan θ

m = tan 90˚

m = undefined

 

 (ii) 45˚

Solution:

θ = 45˚

Slope = m = tan θ

m = tan 45˚

m = 1

 

 (iii) 30˚

Solution:

θ = 30˚

Slope = m = tan θ

m = tan 30˚

m = 1/√3

 

 (iv) 0˚

Solution:

θ = 0˚

Slope = m = tan θ

m = tan 0˚

m = 0

 

  1. Find the angles of inclination of straight lines whose slopes are

(i)√3

Solution:

Slope = tan θ

√3 = tanθ

We know that, tan 60˚ = √3

θ = 60˚

 

 (ii) 1

Solution:

Slope = tan θ

1 = tan θ

We know that, tan 45˚ = 1

θ = 45˚

 

 (iii) 1/√3

Solution:

Slope = tan θ

1/√3 = tanθ

We know that, tan 30˚ = 1/√3

θ = 30˚

 

  1. Find the slope of the line joining the points

(i) (-4, 1) and (-5, 2)

Solution:

Let (x1 ,y1) = (-4, 1) and (x2 ,y2) (-5, 2)

Slope = y2-y1/x2-x1 = 2-1/-5-(-4) = 1/-5+4 = -1

 

 (ii) 4, -8) and (5, -2)

Solution:

Let (x1 ,y1) = (4, -8) and (x2 ,y2) = (5, -2)

Slope = y2-y1/x2-x1 = -2-(-8)/5-4 = -2+8/1 = 6

 

 (iii) (0, 0) and (√3, 3)

Solution:

Let (x1 ,y1) = (0, 0) and (x2 ,y2) = (√3, 3)

Slope = y2-y1/x2-x1 = 3-0/√3-0 = 3/√3 = √3

 

 (iv) (-5, 0) and (0, -7)

Solution:

Let (x1 ,y1) = (-5, 0) and (x2 ,y2) = (0, -7)

Slope = y2-y1/x2-x1 = -7-0/0-(-5) = -7/5

 

 (v) (2a, 3b) and (a, -b)

Solution:

Let (x1 ,y1) = (2a, 3b) and (x2 ,y2) = (a, -b)

Slope = y2-y1/x2-x1 = -b-3b/a-2a = -4b/-a = 4b/a

 

  1. 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)

Solution:

For line 1:

Let (x1 ,y1) = (5, 2) and (x2 ,y2)= (0, 5)

Slope = m1 = y2-y1/x2-x1 = 5-2/0-5 = 3/-5

For line 2:

Let (x1 ,y1) = (0, 0) and (x2 ,y2)= (-5, 3)

Slope = m2 = y2-y1/x2-x1 = 3-0/-5-0 = 3/-5

We have m1 = m2

Therefore, (5, 2), (0, 5) and (0, 0), (-5, 3) lines are parallel.

 

 (ii) (3, 3), (4, 6) and (4, 1), (6, 7)

Solution:

For line 1:

Let (x1 ,y1) = (3, 3) and (x2 ,y2)= (4, 6)

Slope = m1 = y2-y1/x2-x1 = 6-3/4-3 = 3/1 = 3

For line 2:

Let (x1 ,y1) = (4, 1) and (x2 ,y2)= (6, 7)

Slope = m2 = y2-y1/x2-x1 = 7-1/6-4 = 6/2 = 3

We have m1 = m2

Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) parallel.

 

 (iii) (4, 7), (3, 5) and (-1, 7), (1, 6)

Solution:

For line 1:

Let (x1 ,y1) = (4, 7) and (x2 ,y2)= (3, 5)

Slope = m1 = y2-y1/x2-x1 = 5-7/3-4 = -2/-1 = 2

For line 2:

Let (x1 ,y1) = (-1, 7) and (x2 ,y2)= (1, 6)

Slope = m2 = y2-y1/x2-x1 = 6-7/1-(-1) = -1/2

We have m1 ≠ m2

m1.m2 = 2 x (1/-2) = – 1

Therefore, (4, 7), (3, 5) and (-1, 7), (1, 6) are perpendicular.

 

 (iv) (-1, -2), (1, 6) and (-1, 1), (-2, -3)

Solution:

For line 1:

Let (x1 ,y1) = (-1, -2) and (x2 ,y2)= (1, 6)

Slope = m1 = y2-y1/x2-x1 = 6-(-2)/1-(-1) = 8/2 = 4

For line 2:

Let (x1 ,y1) = (-1, 1) and (x2 ,y2)= (-2, -3)

Slope = m2 = y2-y1/x2-x1 = -3-1/-2-(-1) = -4/-1 = 4

We have m1 = m2

Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) are parallel.

 

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

(i)(1, 7) and (-4, 3)

Solution:

Let (x1 ,y1) = (1, 7) and (x2 ,y2)= (-4, 3)

Slope = m1 = y2-y1/x2-x1 = 3-7/-4-1 = -4/-5 = 4/5

The slope of the line perpendicular to the line joining points (1, 7) and (-4, 3) is  -5/4.[since m1.m2 = -1]

 

 (ii) (2, -3) and (1, 4)

Solution:

Let (x1 ,y1) = (2, 3) and (x2 ,y2)= (1, 4)

Slope = m1 = y2-y1/x2-x1 = 4-3/1-2 = 1/-1 = -1

The slope of the line perpendicular to the line joining points (2, -3) and (1, 4) is 1. [Since m1.m2 = -1]

 

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

(i) (-4, 3) and (2, 5)

Solution:

Let (x1 ,y1) = (-4, 3) and (x2 ,y2)= (2, 5)

Slope = m1 = y2-y1/x2-x1 = 5-3/2-(-4) = 2/6 = 1/3

The slope of the line parallel to the line joining points (-4, 3) and (2, 5) is 1/3.

 

 (ii) (1, -5) and (7, 1)

Solution:

Let (x1 ,y1) = (1, -5) and (x2 ,y2)= (7, 1)

Slope = m1 = y2-y1/x2-x1 = 1-(-5)/7-1 = 6/6 = 1

The slope of the line parallel to the line joining points (1, -5) and (7, 1) is 1.

 

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

Solution:

Line 1:

Let (x1 ,y1) = (2, 7) and (x2 ,y2)= (3, 6)

Slope = m1 = y2-y1/x2-x1 = 6-7/3-2 = -1/1 = -1

Line 2:

Let (x1 ,y1) = (9, a) and (x2 ,y2)= (11, 3)

Slope = m2 = y2-y1/x2-x1 = 3-a/11-9 = 3-a/2

Since m1 = m2 as line (2, 7) and (3, 6) is parallel to (9, a) and (11, 3).

We have, -1 = 3-a/2

-2 = 3 – a

-2 – 3 = – a

– 5 = -a

a = 5

 

  1. A line joining 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:

Line 1:

Let (x1 ,y1) = (1, 0) and (x2 ,y2)= (4, 3)

Slope = m1 = y2-y1/x2-x1 = 3-0/4-1 = 3/3 = 1

Line 2:

Let (x1 ,y1) = (-2, -1) and (x2 ,y2)= (m, 0)

Slope = m2 = y2-y1/x2-x1 = 0-(-1)/m-(-2) = 1/m+2

Since m1.m2 = -1 as A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0).

We have, m1.m2 = -1

1.(1/m+2) = -1

1/m+2 = -1

1 = -(m+2)

1 = -m – 2

-m = 1 + 2

m = -3

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