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 $\displaystyle \text{Given: } m = \tan 150^{\circ} = - \tan 30^{\circ} = - \frac{1}{\sqrt{3}}$ $\displaystyle c = \text{y-intercept} = 2$ $\displaystyle \text{Substituting in } y = mx+c \text{ we get } y = - \frac{1}{\sqrt{3}} x + 2$ $\displaystyle \Rightarrow x + \sqrt{3}y = 2 \sqrt{3}$ $\displaystyle \text{Hence the equation of the line is } x + \sqrt{3}y = 2 \sqrt{3}$ $\displaystyle \\$

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. $\displaystyle \text{i) Given } m = 2 , \hspace{0.5cm} c=3$ $\displaystyle \text{Substituting in } y = mx+c$ 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 $\displaystyle y = -2x - 6$ $\displaystyle \\$

Question 3: Find the equations of the bisectors of the angles, between the coordinate axes. $\displaystyle \text{For line 1, } m_1 = \tan 45^{\circ} = 1, c = \text{ y-intercept } = 0$ $\displaystyle \text{Substituting in } y = mx+c$ 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 $\displaystyle x \pm y = 0$ $\displaystyle \\$

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. $\displaystyle \text{Here } m = \tan^{-1} (3) = 3, \hspace{0.5cm} c = \text{y-intercept } = -4$ $\displaystyle \text{Substituting in } y = mx+c \text{ we get the equation as } y = 3x - 4$

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 $\displaystyle y = 3x - 4$ $\displaystyle \\$

Question 5: Find the equation of a line that has y-intercept $\displaystyle -4$ and is parallel to the line joining $\displaystyle (2, -5) \text{ and } (1,2)$. $\displaystyle \text{Here slope } (m) = \frac{2-(-5)}{1-2} = \frac{7}{-1} = -7$ $\displaystyle c = \text{y-intercept } = -4$ $\displaystyle \text{Substituting in } y = mx+c \text{ we get the equation as } y = -7x - 4 \Rightarrow 7x +y+4=0$

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 $\displaystyle 7x +y+4=0$ $\displaystyle \\$

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. $\displaystyle \text{Slope of line joining } (4,2) \text{ and } (3, 5) = \frac{5-2}{3-4} = \frac{3}{-1} = -3$ $\displaystyle \text{The slope of the required line, which is } \perp \text{ to the above line } = \frac{-1}{-3} = \frac{1}{3}$ $\displaystyle c = \text{y-intercept } = 3$ $\displaystyle \text{Substituting in } y = mx+c \text{ we get the equation as } y = \frac{1}{3} x + 3 \Rightarrow x-3y+ 9 = 0$

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

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. $\displaystyle \text{Slope of line joining } (4,3) \text{ and } (-1, 1) = \frac{1-3}{-1-4} = \frac{-2}{-5} = \frac{2}{5}$ $\displaystyle \text{The slope of the required line, which is } \perp \text{ to the above line } = \frac{-1}{2/5} = \frac{-5}{2}$ $\displaystyle c = \text{y-intercept } = -3$ $\displaystyle \text{Substituting in } y = mx+c$ 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 $\displaystyle 5x+2y+6 = 0$ $\displaystyle \\$

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 $\displaystyle \text{Here } m = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ $\displaystyle c = \text{y-intercept } = -3$ $\displaystyle \text{Substituting in } y = mx+c$ 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 $\displaystyle x - \sqrt{3}y + 2 \sqrt{3}= 0$