Note: The equation of a line having slope of m and \text{y-intercept} as c is given by y= mx + c

Question 1: Find the equation of a line making an angle of  150^{\circ} with the x-axis and cutting off an intercept 2 from y-axis

Answer:

Given: \displaystyle m = \tan 150^{\circ} = - \tan 30^{\circ} = - \frac{1}{\sqrt{3}}

c = \text{y-intercept} = 2

Substituting in y = mx+c we get \displaystyle y =  - \frac{1}{\sqrt{3}}  x + 2

\Rightarrow x + \sqrt{3}y = 2 \sqrt{3}

Hence the equation of the line is x + \sqrt{3}y = 2 \sqrt{3}

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Question 2: Find the equation of a straight line:

(i) with slope 2 and y-intercept 3

(ii) with slope -1/3 and y-intercept - 4

(iii) with slope -2 and intersecting the x-axis at a distance of 3 units to the left of origin.

Answer:

i)        Given m = 2 , \hspace{0.5cm} c=3

Substituting in y = mx+c we get  y = 2x + 3

Hence the equation of a straight line with slope 2 and y-intercept 3 is y = 2x + 3

ii)     Given \displaystyle m = - \frac{1}{3} , \hspace{0.5cm} c=-4

Substituting in y = mx+c we get  \displaystyle y = - \frac{1}{3} x -4

Hence the equation of a straight line with slope \displaystyle - \frac{1}{3} and y-intercept - 4 is \displaystyle y = - \frac{1}{3}  x -4

iii)     Given m = -2 , and passes through (-3,0)

Substituting in y = mx+c we get  0 = -2(-3) + c \Rightarrow c = -6

Substituting in y = mx+c we get  the equation of the line is y = -2x - 6

Hence the equation of a straight line with slope -2 and passes through (-3,0) is y = -2x - 6

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Question 3: Find the equations of the bisectors of the angles, between the coordinate axes.

Answer:

For line 1, m_1 = \tan 45^{\circ} = 1, c = \text{ y-intercept } = 0

Substituting in y = mx+c we get  the equation of line 1 is y = x

For line 2, m_2 = \tan 135^{\circ} = -1, c = \text{ y-intercept } = 0

Substituting in y = mx+c we get  the equation of line 2 is y = -x

Hence the equations of the bisectors of the angles, between the coordinate axes x \pm y = 0

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Question 4: Find the equation of a line which makes an angle of  \tan^{-1} (3) with the x-axis and cuts off an intercept of 4 units on negative direction of y-axis.

Answer:

Here m = \tan^{-1} (3) = 3, \hspace{0.5cm} c = \text{y-intercept } = -4

Substituting in y = mx+c we get the equation as y = 3x - 4

Hence the equation of a line which makes an angle of  \tan^{-1} (3) with the x-axis and cuts off an intercept of 4 units on negative direction of y-axis is y = 3x - 4

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Question 5: Find the equation of a line that has y-intercept -4 and is parallel to the line joining (2, -5) and (1,2) .

Answer:

Here slope \displaystyle (m) =  \frac{2-(-5)}{1-2}  =  \frac{7}{-1}  = -7

c = \text{y-intercept } = -4

Substituting in y = mx+c we get the equation as y = -7x - 4 \Rightarrow 7x +y+4=0

Hence equation of a line that has y-intercept -4 and is parallel to the line joining (2, -5) and (1,2) is 7x +y+4=0

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Question 6: Find the equation of a line which is perpendicular to the line joining (4,2) and (3, 5) and cuts off an intercept of length 3 on y-axis.

Answer:

Slope of line joining (4,2) and \displaystyle (3, 5) =  \frac{5-2}{3-4} = \frac{3}{-1}  = -3

The slope of the required line, which is \perp to the above line \displaystyle =  \frac{-1}{-3}  =  \frac{1}{3}

c = \text{y-intercept } = 3

Substituting in y = mx+c we get the equation as \displaystyle y =  \frac{1}{3}  x + 3 \Rightarrow  x-3y+ 9 = 0

Hence the equation of the required equation is x-3y+ 9 = 0

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Question 7: Find the equation of the perpendicular to the line segment joining (4, 3) and (-1, 1) if it cuts off an intercept -3 from y-axis.

Answer:

Slope of line joining (4,3) and \displaystyle (-1, 1) = \frac{1-3}{-1-4} = \frac{-2}{-5} = \frac{2}{5}

The slope of the required line, which is \perp to the above line \displaystyle =  \frac{-1}{2/5}  =  \frac{-5}{2}

c = \text{y-intercept } = -3

Substituting in y = mx+c we get the equation as

\displaystyle y =  \frac{-5}{2}  x - 3 \Rightarrow  5x+2y+6 = 0

Hence the equation of the required equation is 5x+2y+6 = 0

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Question 8: Find the equation of the straight line intersecting y-axis at a distance of 2 units above the origin and making an angle of 30^{\circ} with the positive direction of the x-axis

Answer:

Here \displaystyle m = \tan 30^{\circ} =  \frac{1}{\sqrt{3}}

c = \text{y-intercept } = -3

Substituting in y = mx+c we get the equation as

\displaystyle y =  \frac{1}{\sqrt{3}}  x + 2 \Rightarrow x - \sqrt{3}y + 2 \sqrt{3}= 0

Hence the equation of the required equation is x - \sqrt{3}y + 2 \sqrt{3}= 0