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

\displaystyle \textbf{Question 1: }\text{Find the equation of a line making an angle of }150^\circ\text{ with the}
\displaystyle x\text{-axis and cutting off an intercept }2\text{ on the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope }m=\tan150^\circ=-\tan30^\circ=-\frac{1}{\sqrt{3}}.
\displaystyle \text{The }y\text{-intercept }c=2.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=-\frac{1}{\sqrt{3}}x+2.
\displaystyle \Rightarrow x+\sqrt{3}y=2\sqrt{3}.
\displaystyle \therefore \text{The required equation of the line is }x+\sqrt{3}y=2\sqrt{3}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of a straight line:}
\displaystyle \text{i) with slope }2\text{ and }y\text{-intercept }3
\displaystyle \text{ii) with slope }-\frac{1}{3}\text{ and }y\text{-intercept }-4
\displaystyle \text{iii) with slope }-2\text{ and intersecting the }x\text{-axis at a distance of }3\text{ units}
\displaystyle \text{to the left of the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{i) Given }m=2,\ c=3.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=2x+3.
\displaystyle \therefore \text{The required equation is }y=2x+3.
\displaystyle \text{ii) Given }m=-\frac{1}{3},\ c=-4.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=-\frac{1}{3}x-4.
\displaystyle \therefore \text{The required equation is }y=-\frac{1}{3}x-4.
\displaystyle \text{iii) Given }m=-2\text{ and the line passes through }(-3,0).
\displaystyle \text{Using }y=mx+c,
\displaystyle 0=-2(-3)+c
\displaystyle \Rightarrow c=-6.
\displaystyle \therefore \text{The required equation is }y=-2x-6.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equations of the bisectors of the angles between the}
\displaystyle \text{coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{For the first bisector, }m=\tan45^\circ=1,\ c=0.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=x.
\displaystyle \text{For the second bisector, }m=\tan135^\circ=-1,\ c=0.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=-x.
\displaystyle \therefore \text{The equations of the bisectors are }y=x\text{ and }y=-x,
\displaystyle \text{or equivalently }x-y=0\text{ and }x+y=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equation of a line which makes an angle of }\tan^{-1}(3)
\displaystyle \text{with the }x\text{-axis and cuts off an intercept of }4\text{ units on the negative}
\displaystyle \text{direction of the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \theta=\tan^{-1}(3).
\displaystyle \therefore m=\tan\theta=\tan(\tan^{-1}3)=3.
\displaystyle \text{The }y\text{-intercept is }c=-4.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=3x-4.
\displaystyle \therefore \text{The required equation is }y=3x-4.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of a line that has }y\text{-intercept }-4\text{ and is}
\displaystyle \text{parallel to the line joining }(2,-5)\text{ and }(1,2).
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(2,-5)\text{ and }(1,2)\text{ is}
\displaystyle m=\frac{2-(-5)}{1-2}=\frac{7}{-1}=-7.
\displaystyle \text{The required line is parallel to this line, so its slope is }-7.
\displaystyle \text{Also, }c=-4.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=-7x-4.
\displaystyle \therefore \text{The required equation is }7x+y+4=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equation of a line which is perpendicular to the line joining}
\displaystyle (4,2)\text{ and }(3,5)\text{ and cuts off an intercept of length }3\text{ on the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(4,2)\text{ and }(3,5)\text{ is}
\displaystyle m_1=\frac{5-2}{3-4}=\frac{3}{-1}=-3.
\displaystyle \text{If }m\text{ is the slope of the required line, then}
\displaystyle m_1m=-1.
\displaystyle \therefore (-3)m=-1
\displaystyle \Rightarrow m=\frac{1}{3}.
\displaystyle \text{Also, }c=3.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=\frac{1}{3}x+3.
\displaystyle \therefore x-3y+9=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the equation of the perpendicular to the line segment joining}
\displaystyle (4,3)\text{ and }(-1,1)\text{ if it cuts off an intercept }-3\text{ on the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(4,3)\text{ and }(-1,1)\text{ is}
\displaystyle m_1=\frac{1-3}{-1-4}=\frac{-2}{-5}=\frac{2}{5}.
\displaystyle \text{If }m\text{ is the slope of the required line, then}
\displaystyle m_1m=-1.
\displaystyle \therefore \frac{2}{5}m=-1
\displaystyle \Rightarrow m=-\frac{5}{2}.
\displaystyle \text{Also, }c=-3.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=-\frac{5}{2}x-3.
\displaystyle \therefore 5x+2y+6=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equation of the straight line intersecting the }y\text{-axis at a}
\displaystyle \text{distance of }2\text{ units above the origin and making an angle of }30^\circ\text{ with the}
\displaystyle \text{positive direction of the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle m=\tan30^\circ=\frac{1}{\sqrt3}.
\displaystyle c=2.
\displaystyle \text{Substituting in }y=mx+c,
\displaystyle y=\frac{1}{\sqrt3}x+2.
\displaystyle \therefore x-\sqrt3y+2\sqrt3=0.
\displaystyle \\

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