Coordinate Geometry Exercise 14.2 – Questions:
- 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.
- Find the equation of the line whose slope and y-intercept are given.
(i) Slope = 2, y-intercept = -4
(ii) Slope = –2/3, y-intercept = 1/2
(iii) Slope = -2, y-intercept = 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
- Is the line x = 2y parallel to 2x – 4y + 7 = 0[Hint: Parallel lines have some slopes]
- Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other.[Hint: For perpendicular lines, m1.m2 = -1].
Coordinate Geometry Exercise 14.2 – Solutions:
- Find the equation of the line whose angle of inclination and y-intercept are given.
(i) θ = 60˚, y-intercept is -2.
Solution:
m = tan θ = tan 60˚ = √3
y – intercept = c = -2
Equation: y = mx + c
y = √3x – 2
(ii) θ = 45˚, y-intercept is 3.
Solution:
m = tan θ = tan 45˚ = 1
y – intercept = c = 3
Equation: y = mx + c
y = x + 3
- Find the equation of the line whose slope and y-intercept are given.
(i) Slope = 2, y-intercept = -4
Solution:
m = slope = 2
y – intercept = c = -4
Equation: y = mx + c
y = 2x – 4
(ii) Slope = –2/3, y-intercept = 1/2
Solution:
m = slope = –2/3
y – intercept = c = 1/2
Equation: y = mx + c
y = –2/3 x + 1/2
(iii) Slope = -2, y-intercept = 3
Solution:
m = slope = -2
y – intercept = c = 3
Equation: y = mx + c
y = -2x + 3
- Find the slope and y-intercept of the lines
(i) 2x + 3y = 4
Solution:
2x + 3y = 4
3y = 4 – 2x
y = – 2/3x + 4/3
Standard equation is y = mx + c
Therefore, m = –2/3 and c = 4/3
(ii) 3x = y
Solution:
Given, 3x = y
y = 3x + 0
Standard equation is y = mx + c
Therefore, m = 3 and c = 0
(iii) x – y + 5 = 0
Solution:
x – y + 5 = 0
-y = – x – 5
y = x + 5
Standard equation is y = mx + c
Therefore, m = 1 and c = 5
(iv) 3x – 4y = 5
Solution:
3x – 4y = 5
-4y = 5 – 3x
4y = 3x – 5
y = 3/4 x – 5/4
Standard equation is y = mx + c
Therefore, m = 3/4 and c = – 5/4
- Is the line x = 2y parallel to 2x – 4y + 7 = 0[Hint: Parallel lines have same slopes]
Solution:
line 1:
x = 2y
y = 1/2 x + 0
Standard equation is y = mx + c
Therefore, m1 = 1/2 and c1 = 0
line 2:
2x – 4y + 7 = 0
-4y = 7 – 2x
4y = 2x – 7
y = 2/4 x – 7/4
Standard equation is y = mx + c
Therefore, m2 = 2/4 = 1/2 and c2 = –7/4
Since slope of the line x = 2y and the line 2x – 4y + 7 = 0 are same i.e., m1 = m2 = 1/2.
Therefore, the line x = 2y parallel to 2x – 4y + 7 = 0.
- Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other.[Hint: For perpendicular lines, m1.m2 = -1].
Solution:
line 1:
3x + 4y + 7 = 0
4y = – 3x – 7
y = – 3/4 x – 7/4
Standard equation is y = mx + c
Therefore, m1 = – 3/4 and c1 = –7/4
line 2:
28x – 21y + 50 = 0
-21y = – 28x – 50
21y = 50 + 28x
y = 28/21 x + 50/21
Standard equation is y = mx + c
Therefore, m2 = 28/21 and c2 = 50/21
If slope of the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular then we must get m1.m2 = -1
i.e.,
– 3/4 . 28/21 = -1
Therefore, the line 3x + 4y + 7 = 0 is perpendicular to 28x – 21y + 50 = 0.
