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

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

Answer:

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

\displaystyle c = \text{y-intercept} = 2

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

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

\displaystyle \text{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 \displaystyle 2 and y-intercept \displaystyle 3

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

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

Answer:

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

\displaystyle \text{Substituting in } y = mx+c we get \displaystyle y = 2x + 3

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

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

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

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

\displaystyle \text{iii) Given } m = -2 , and passes through \displaystyle (-3,0)

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

\displaystyle \text{Substituting in } y = mx+c we get the equation of the line is \displaystyle y = -2x - 6

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

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

Answer:

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

\displaystyle \text{Substituting in } y = mx+c we get the equation of line 1 is \displaystyle y = x

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

\displaystyle \text{Substituting in } y = mx+c we get the equation of line 2 is \displaystyle y = -x

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

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

Answer:

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

\displaystyle \text{Substituting in } y = mx+c \text{ we get the equation as } y = 3x - 4

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

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

Answer:

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

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

\displaystyle \text{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 \displaystyle 5x+2y+6 = 0

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

Answer:

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

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

\displaystyle \text{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 \displaystyle x - \sqrt{3}y + 2 \sqrt{3}= 0