Coordinate Geometry Exercise 14.1 – Questions:
- Find the slope of the curve whose inclination is
(i) 90˚
(ii) 45˚
(iii) 30˚
(iv) 0˚
- Find the angles of inclination of straight lines whose slopes are
(i)√3
(ii) 1
(iii) 1/√3
- 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)
- 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)
- 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)
- 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)
- A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a) and (11, 3). Find a.
- 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:
- 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
- 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˚
- 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
- 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.
- 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]
- 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.
- 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
- 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