\displaystyle \textbf{Question 1: }\text{Find the slope of the line whose inclination is:}
\displaystyle \text{i) }0^\circ\qquad \text{ii) }30^\circ\qquad \text{iii) }72^\circ30'\qquad \text{iv) }46^\circ
\displaystyle \text{Answer:}
\displaystyle \text{i) }\theta=0^\circ
\displaystyle \text{Slope}=\tan\theta=\tan0^\circ=0
\displaystyle \text{ii) }\theta=30^\circ
\displaystyle \text{Slope}=\tan\theta=\tan30^\circ=\frac{1}{\sqrt{3}}\approx0.577
\displaystyle \text{iii) }\theta=72^\circ30'
\displaystyle \text{Slope}=\tan\theta=\tan72^\circ30'=\tan72.5^\circ\approx3.172
\displaystyle \text{iv) }\theta=46^\circ
\displaystyle \text{Slope}=\tan\theta=\tan46^\circ\approx1.036
\\

\displaystyle \textbf{Question 2: }\text{Find the inclination of the line whose slope is:}
\displaystyle \text{i) }0\qquad \text{ii) }\sqrt{3}\qquad \text{iii) }0.7646\qquad \text{iv) }1.0875
\displaystyle \text{Answer:}
\displaystyle \text{i) }\tan\theta=0
\displaystyle \therefore \theta=0^\circ
\displaystyle \text{ii) }\tan\theta=\sqrt{3}
\displaystyle \therefore \theta=60^\circ
\displaystyle \text{iii) }\tan\theta=0.7646
\displaystyle \therefore \theta=37^\circ24'
\displaystyle \text{iv) }\tan\theta=1.0875
\displaystyle \therefore \theta=47^\circ24'
\\

\displaystyle \textbf{Question 3: }\text{Find the slope of the line passing through the following pairs of points:}
\displaystyle \text{i) }(-2,-3)\text{ and }(1,2)\qquad \text{ii) }(-4,0)\text{ and origin}\qquad \text{iii) }(a,-b)\text{ and }(b,-a)
\displaystyle \text{Answer:}
\displaystyle \text{i) Let }A(-2,-3)=(x_1,y_1)\text{ and }B(1,2)=(x_2,y_2).
\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-3)}{1-(-2)}=\frac{5}{3}
\displaystyle \text{ii) Let }A(-4,0)=(x_1,y_1)\text{ and }B(0,0)=(x_2,y_2).
\displaystyle \text{Slope}=\frac{0-0}{0-(-4)}=0
\displaystyle \text{iii) Let }A(a,-b)=(x_1,y_1)\text{ and }B(b,-a)=(x_2,y_2).
\displaystyle \text{Slope}=\frac{-a-(-b)}{b-a}=\frac{b-a}{b-a}=1
\\

\displaystyle \textbf{Question 4: }\text{Find the slope of the line parallel to }AB\text{ if:}
\displaystyle \text{i) }A=(-2,4)\text{ and }B=(0,6)\qquad \text{ii) }A=(0,-3)\text{ and }B=(-2,5)
\displaystyle \text{Answer:}
\displaystyle \text{i) Slope of }AB=\frac{6-4}{0-(-2)}=\frac{2}{2}=1
\displaystyle \therefore \text{The slope of the line parallel to }AB\text{ is }1.
\displaystyle \text{ii) Slope of }AB=\frac{5-(-3)}{-2-0}=\frac{8}{-2}=-4
\displaystyle \therefore \text{The slope of the line parallel to }AB\text{ is }-4.
\\

\displaystyle \textbf{Question 5: }\text{Find the slope of the line perpendicular to }AB\text{ if:}
\displaystyle \text{i) }A=(0,-5)\text{ and }B=(-2,4)\qquad \text{ii) }A=(3,-2)\text{ and }B=(-1,2)
\displaystyle \text{Answer:}
\displaystyle \text{i) Slope of }AB=\frac{4-(-5)}{-2-0}=\frac{9}{-2}=-\frac{9}{2}
\displaystyle \text{Slope of the perpendicular line}=-\frac{1}{-\frac{9}{2}}=\frac{2}{9}
\displaystyle \therefore \text{The required slope is }\frac{2}{9}.
\displaystyle \text{ii) Slope of }AB=\frac{2-(-2)}{-1-3}=\frac{4}{-4}=-1
\displaystyle \text{Slope of the perpendicular line}=-\frac{1}{-1}=1
\displaystyle \therefore \text{The required slope is }1.
\\

\displaystyle \textbf{Question 6: }\text{The line passing through }(0,2)\text{ and }(-3,-1)\text{ is parallel} \\ \text{to the line passing through }(-1,5)\text{ and }(4,a).\text{ Find }a.
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line through }(0,2)\text{ and }(-3,-1)=\frac{-1-2}{-3-0}=1
\displaystyle \text{Slope of the line through }(-1,5)\text{ and }(4,a)=\frac{a-5}{4-(-1)}=\frac{a-5}{5}
\displaystyle \text{Since the lines are parallel, their slopes are equal.}
\displaystyle \frac{a-5}{5}=1
\displaystyle a-5=5
\displaystyle a=10
\displaystyle \therefore \text{The required value of }a\text{ is }10.
\\

\displaystyle \textbf{Question 7: }\text{The line passing through }(-4,-2)\text{ and }(2,-3)\text{ is perpendicular} \\ \text{to the line passing through }(a,5)\text{ and }(2,-1).\text{ Find }a.
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line through }(-4,-2)\text{ and }(2,-3)=\frac{-3-(-2)}{2-(-4)}=-\frac{1}{6}
\displaystyle \text{Slope of the line through }(a,5)\text{ and }(2,-1)=\frac{-1-5}{2-a}=\frac{-6}{2-a}
\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1.
\displaystyle -\frac{1}{6}\times\frac{-6}{2-a}=-1
\displaystyle \frac{1}{2-a}=-1
\displaystyle 1=a-2
\displaystyle a=3
\displaystyle \therefore \text{The required value of }a\text{ is }3.
\\

\displaystyle \textbf{Question 8: }\text{Without using the distance formula, show that }A(4,-2),\;B(-4,4)\text{ and } \\ C(10,6)\text{ are the vertices of a right-angled triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }AB=\frac{4-(-2)}{-4-4}=\frac{6}{-8}=-\frac{3}{4}
\displaystyle \text{Slope of }AC=\frac{6-(-2)}{10-4}=\frac{8}{6}=\frac{4}{3}
\displaystyle \text{Slope of }BC=\frac{4-6}{-4-10}=\frac{-2}{-14}=\frac{1}{7}
\displaystyle m_{AB}\times m_{AC}=-\frac{3}{4}\times\frac{4}{3}=-1
\displaystyle \therefore AB\perp AC.
\displaystyle \therefore \triangle ABC\text{ is a right-angled triangle (right angle at }A\text{).}
\\

\displaystyle \textbf{Question 9: }\text{Without using distance formula, show that }A(4,5),\;B(1,2),\;C(4,3) \\ \text{ and }D(7,6)\text{ are the vertices of a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }AB=\frac{2-5}{1-4}=\frac{-3}{-3}=1
\displaystyle \text{Slope of }BC=\frac{3-2}{4-1}=\frac{1}{3}
\displaystyle \text{Slope of }CD=\frac{6-3}{7-4}=\frac{3}{3}=1
\displaystyle \text{Slope of }DA=\frac{5-6}{4-7}=\frac{-1}{-3}=\frac{1}{3}
\displaystyle \therefore \text{Slope of }AB=\text{Slope of }CD\text{ and slope of }BC=\text{Slope of }DA.
\displaystyle \therefore AB\parallel CD\text{ and }BC\parallel DA.
\displaystyle \therefore ABCD\text{ is a parallelogram.}
\\

\displaystyle \textbf{Question 10: }(-2,4),\;(4,8),\;(10,7)\text{ and }(11,-5)\text{ are the vertices of a quadrilateral.} \\ \text{Show that the quadrilateral obtained by joining the midpoints of its sides is a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,4),\;B(4,8),\;C(10,7)\text{ and }D(11,-5)\text{ be the vertices.}
\displaystyle \text{Let }P,Q,R\text{ and }S\text{ be the midpoints of }AB,BC,CD\text{ and }DA\text{ respectively.}
\displaystyle P=\left(\frac{-2+4}{2},\frac{4+8}{2}\right)=(1,6)
\displaystyle Q=\left(\frac{4+10}{2},\frac{8+7}{2}\right)=\left(7,\frac{15}{2}\right)
\displaystyle R=\left(\frac{10+11}{2},\frac{7+(-5)}{2}\right)=\left(\frac{21}{2},1\right)
\displaystyle S=\left(\frac{11+(-2)}{2},\frac{-5+4}{2}\right)=\left(\frac{9}{2},-\frac{1}{2}\right)
\displaystyle \text{Slope of }PQ=\frac{\frac{15}{2}-6}{7-1}=\frac{\frac{3}{2}}{6}=\frac{1}{4}
\displaystyle \text{Slope of }QR=\frac{1-\frac{15}{2}}{\frac{21}{2}-7}=\frac{-\frac{13}{2}}{\frac{7}{2}}=-\frac{13}{7}
\displaystyle \text{Slope of }RS=\frac{-\frac{1}{2}-1}{\frac{9}{2}-\frac{21}{2}}=\frac{-\frac{3}{2}}{-6}=\frac{1}{4}
\displaystyle \text{Slope of }SP=\frac{6-\left(-\frac{1}{2}\right)}{1-\frac{9}{2}}=\frac{\frac{13}{2}}{-\frac{7}{2}}=-\frac{13}{7}
\displaystyle \therefore \text{Slope of }PQ=\text{Slope of }RS\text{ and slope of }QR=\text{Slope of }SP.
\displaystyle \therefore PQ\parallel RS\text{ and }QR\parallel SP.
\displaystyle \therefore PQRS\text{ is a parallelogram.}
\\

\displaystyle \textbf{Question 11: }\text{Show that the points }P(a,b+c),\;Q(b,c+a)\text{ and }R(c,a+b) \\ \text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }PQ=\frac{(c+a)-(b+c)}{b-a}=\frac{a-b}{b-a}=-1
\displaystyle \text{Slope of }QR=\frac{(a+b)-(c+a)}{c-b}=\frac{b-c}{c-b}=-1
\displaystyle \text{Since slope of }PQ=\text{slope of }QR,
\displaystyle \therefore P,\;Q\text{ and }R\text{ are collinear.}
\\

\displaystyle \textbf{Question 12: }\text{Find }x,\text{ if the slope of the line joining }(x,2)\text{ and }(8,-11) \\ \text{ is }-\frac{3}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{Given slope}=-\frac{3}{4}
\displaystyle \frac{-11-2}{8-x}=-\frac{3}{4}
\displaystyle \frac{-13}{8-x}=-\frac{3}{4}
\displaystyle -52=-24+3x
\displaystyle 3x=-28
\displaystyle x=-\frac{28}{3}
\displaystyle \therefore \text{The required value of }x\text{ is }-\frac{28}{3}.
\\

\displaystyle \textbf{Question 13: }\text{The side }AB\text{ of an equilateral triangle }ABC\text{ is parallel to the } \\ x\text{-axis. Find the slopes of all the sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }AB=\tan0^\circ=0
\displaystyle \text{Slope of }AC=\tan60^\circ=\sqrt{3}\approx1.732
\displaystyle \text{Slope of }BC=\tan120^\circ=-\sqrt{3}\approx-1.732
\\

\displaystyle \textbf{Question 14: }\text{The side }AB\text{ of a square }ABCD\text{ is parallel to the }x\text{-axis.} \\ \text{Find the slopes of all its sides. Also, find: i) slope of }AC,\text{ ii) slope of }BD.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB\parallel x\text{-axis,}
\displaystyle \text{Slope of }AB=\tan0^\circ=0
\displaystyle \text{Slope of }DC=\text{slope of }AB=0
\displaystyle \text{Slope of }BC=\tan90^\circ\text{, which is undefined.}
\displaystyle \text{Slope of }AD=\text{slope of }BC\text{, which is undefined.}
\displaystyle \text{Slope of }AC=\tan45^\circ=1
\displaystyle \text{Slope of }BD=\tan135^\circ=-1
\\

\displaystyle \textbf{Question 15: }A(5,4),\;B(-3,-3)\text{ and }C(1,-8)\text{ are the vertices of }\triangle ABC. \\ \text{ Find: i) slope of altitude on }AB,\text{ ii) slope of median }AD,\text{ iii) slope of line parallel to }AC.
\displaystyle \text{Answer:}
\displaystyle \text{i) Slope of }AB=\frac{-3-4}{-3-5}=\frac{-7}{-8}=\frac{7}{8}
\displaystyle \text{Slope of altitude on }AB=-\frac{1}{\frac{7}{8}}=-\frac{8}{7}
\displaystyle \therefore \text{The slope of the altitude on }AB\text{ is }-\frac{8}{7}.
\displaystyle \text{ii) Let }D\text{ be the midpoint of }BC.
\displaystyle D=\left(\frac{-3+1}{2},\frac{-3+(-8)}{2}\right)=\left(-1,-\frac{11}{2}\right)
\displaystyle \text{Slope of median }AD=\frac{-\frac{11}{2}-4}{-1-5}=\frac{-\frac{19}{2}}{-6}=\frac{19}{12}
\displaystyle \therefore \text{The slope of median }AD\text{ is }\frac{19}{12}.
\displaystyle \text{iii) Slope of }AC=\frac{-8-4}{1-5}=\frac{-12}{-4}=3
\displaystyle \therefore \text{The slope of the line parallel to }AC\text{ is }3.
\\

\displaystyle \textbf{Question 16: }\text{The slope of the side }BC\text{ of a rectangle }ABCD\text{ is }\frac{2}{3}. \\ \text{ Find: i) the slope of }AB,\text{ ii) the slope of }AD.
\displaystyle \text{Answer:}
\displaystyle \text{Since }BC\parallel AD,\text{ slope of }AD=\frac{2}{3}.
\displaystyle \text{Since }AB\perp BC,\text{ slope of }AB=-\frac{1}{\frac{2}{3}}=-\frac{3}{2}.
\displaystyle \therefore \text{i) Slope of }AB=-\frac{3}{2}\qquad \text{ii) Slope of }AD=\frac{2}{3}.
\\

\displaystyle \textbf{Question 17: }\text{Find the slope and the inclination of the line }AB\text{ if:}
\displaystyle \text{i) }A=(-3,-2)\text{ and }B=(1,2)\qquad \text{ii) }A=(0,-\sqrt{3})\text{ and }B=(3,0)\qquad \text{iii) }A=(-1,2\sqrt{3})\text{ and }B=(-2,\sqrt{3})
\displaystyle \text{Answer:}
\displaystyle \text{i) Slope}=\frac{2-(-2)}{1-(-3)}=\frac{4}{4}=1
\displaystyle \tan\theta=1
\displaystyle \therefore \theta=45^\circ
\displaystyle \text{ii) Slope}=\frac{0-(-\sqrt{3})}{3-0}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}
\displaystyle \tan\theta=\frac{1}{\sqrt{3}}
\displaystyle \therefore \theta=30^\circ
\displaystyle \text{iii) Slope}=\frac{\sqrt{3}-2\sqrt{3}}{-2-(-1)}=\frac{-\sqrt{3}}{-1}=\sqrt{3}
\displaystyle \tan\theta=\sqrt{3}
\displaystyle \therefore \theta=60^\circ
\\

\displaystyle \textbf{Question 18: }\text{The points }(-3,2),\;(2,-1)\text{ and }(a,4)\text{ are collinear. Find }a.
\displaystyle \text{Answer:}
\displaystyle \text{Since the points are collinear,}
\displaystyle \text{Slope of }AB=\text{Slope of }BC
\displaystyle \frac{-1-2}{2-(-3)}=\frac{4-(-1)}{a-2}
\displaystyle \frac{-3}{5}=\frac{5}{a-2}
\displaystyle -3(a-2)=25
\displaystyle -3a+6=25
\displaystyle -3a=19
\displaystyle a=-\frac{19}{3}
\displaystyle \therefore \text{The required value of }a\text{ is }-\frac{19}{3}.
\\

\displaystyle \textbf{Question 19: }\text{The points }(k,3),\;(2,-4)\text{ and }(-k+1,-2)\text{ are collinear. Find }k.
\displaystyle \text{Answer:}
\displaystyle \text{Since the points are collinear,}
\displaystyle \text{Slope of }AB=\text{Slope of }BC
\displaystyle \frac{-4-3}{2-k}=\frac{-2-(-4)}{(-k+1)-2}
\displaystyle \frac{-7}{2-k}=\frac{2}{-k-1}
\displaystyle 7(k+1)=2(2-k)
\displaystyle 7k+7=4-2k
\displaystyle 9k=-3
\displaystyle k=-\frac{1}{3}
\displaystyle \therefore \text{The required value of }k\text{ is }-\frac{1}{3}.
\\

\displaystyle \textbf{Question 20: }\text{Plot the points }A(1,1),\;B(4,7)\text{ and }C(4,10).\text{ Which segment} \\ \text{appears steeper, }AB\text{ or }AC?\text{ Justify by calculating their slopes.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }AB=\frac{7-1}{4-1}=\frac{6}{3}=2
\displaystyle \tan\theta=2\Rightarrow \theta\approx63^\circ30'
\displaystyle \text{Slope of }AC=\frac{10-1}{4-1}=\frac{9}{3}=3
\displaystyle \tan\theta=3\Rightarrow \theta\approx71^\circ36'
\displaystyle \text{Since slope of }AC>\text{slope of }AB,\text{ segment }AC\text{ is steeper than }AB.
\\

\displaystyle \textbf{Question 21: }\text{Find the value(s) of }k\text{ so that }PQ\parallel RS.
\displaystyle \text{i) }P(2,4),\;Q(3,6),\;R(8,1),\;S(10,k)
\displaystyle \text{ii) }P(3,-1),\;Q(7,11),\;R(-1,-1),\;S(1,k)
\displaystyle \text{iii) }P(5,-1),\;Q(6,11),\;R(6,-4k),\;S(7,k^2)
\displaystyle \text{Answer:}
\displaystyle \text{i) Since }PQ\parallel RS,\text{ their slopes are equal.}
\displaystyle \frac{6-4}{3-2}=\frac{k-1}{10-8}
\displaystyle 2=\frac{k-1}{2}
\displaystyle k=5
\displaystyle \text{ii) Since }PQ\parallel RS,\text{ their slopes are equal.}
\displaystyle \frac{11-(-1)}{7-3}=\frac{k-(-1)}{1-(-1)}
\displaystyle \frac{12}{4}=\frac{k+1}{2}
\displaystyle 3=\frac{k+1}{2}
\displaystyle k=5
\displaystyle \text{iii) Since }PQ\parallel RS,\text{ their slopes are equal.}
\displaystyle \frac{11-(-1)}{6-5}=\frac{k^2-(-4k)}{7-6}
\displaystyle 12=k^2+4k
\displaystyle k^2+4k-12=0
\displaystyle (k+6)(k-2)=0
\displaystyle k=-6\text{ or }2
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