**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, m**_{1}.m_{2}= -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}/_{3}x + ^{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, m_{1} = ^{1}/_{2} and c_{1} = 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, m_{2} = ^{2}/_{4} = ^{1}/_{2} and c_{2} = –^{7}/_{4}

Since slope of the line x = 2y and the line 2x – 4y + 7 = 0 are same i.e., m_{1} = m_{2} = ^{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, m**_{1}.m_{2}= -1].

Solution:

line 1:

3x + 4y + 7 = 0

**4**y = – 3x – 7

y = – ^{3}/_{4} x – ^{7}/_{4}

Standard equation is y = mx + c

Therefore, m_{1} = – ^{3}/_{4} and c_{1} = –^{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, m_{2} = ^{28}/_{21} and c_{2} = ^{50}/_{21}

If slope of the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular then we must get m_{1}.m_{2} = -1

i.e.,

– ^{3}/_{4} . ^{28}/_{21} = -1

Therefore, the line 3x + 4y + 7 = 0 is perpendicular to 28x – 21y + 50 = 0.