\displaystyle \textbf{Question 1: }\text{Find the slopes of the lines which make the following angles with the}
\displaystyle \text{positive direction of the }x\text{-axis: } \text{i) }\frac{-\pi}{4}\qquad\text{ii) }\frac{2\pi}{3}\qquad\text{iii) }\frac{3\pi}{4}\qquad\text{iv) }\frac{\pi}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{i) }\theta=\frac{-\pi}{4}
\displaystyle \therefore m=\tan\theta=\tan\left(\frac{-\pi}{4}\right)=-1
\displaystyle \text{ii) }\theta=\frac{2\pi}{3}
\displaystyle \therefore m=\tan\theta=\tan\left(\frac{2\pi}{3}\right)=-\sqrt{3}
\displaystyle \text{iii) }\theta=\frac{3\pi}{4}
\displaystyle \therefore m=\tan\theta=\tan\left(\frac{3\pi}{4}\right)=-1
\displaystyle \text{iv) }\theta=\frac{\pi}{3}
\displaystyle \therefore m=\tan\theta=\tan\left(\frac{\pi}{3}\right)=\sqrt{3}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the slope of a line passing through the following points:}
\displaystyle \text{i) }(-3,2)\text{ and }(1,4)\qquad\text{ii) }(at_1^2,2at_1)\text{ and }(at_2^2,2at_2)\qquad\text{iii) }(3,-5)\text{ and }(1,2).
\displaystyle \text{Answer:}
\displaystyle \text{i) }\text{The slope of the line passing through }(-3,2)\text{ and }(1,4)\text{ is}
\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{1-(-3)}=\frac{2}{4}=\frac{1}{2}
\displaystyle \text{ii) }\text{The slope of the line passing through }(at_1^2,2at_1)\text{ and }(at_2^2,2at_2)\text{ is}
\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{2at_2-2at_1}{at_2^2-at_1^2}
\displaystyle =\frac{2a(t_2-t_1)}{a(t_2-t_1)(t_2+t_1)}=\frac{2}{t_1+t_2}
\displaystyle \text{iii) }\text{The slope of the line passing through }(3,-5)\text{ and }(1,2)\text{ is}
\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-5)}{1-3}=\frac{7}{-2}=-\frac{7}{2}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{State whether the two lines in each of the following are parallel,}
\displaystyle \text{perpendicular or neither:}
\displaystyle \text{i) Through }(5,6)\text{ and }(2,3);\ \text{Through }(9,-2)\text{ and }(6,-5)
\displaystyle \text{ii) Through }(9,5)\text{ and }(-1,1);\ \text{Through }(3,-5)\text{ and }(8,-3)
\displaystyle \text{iii) Through }(5,3)\text{ and }(1,1);\ \text{Through }(-2,5)\text{ and }(2,-5)
\displaystyle \text{iv) Through }(3,15)\text{ and }(16,6);\ \text{Through }(-5,3)\text{ and }(8,2)
\displaystyle \text{Answer:}
\displaystyle \text{i) Let }m_1\text{ be the slope of the line joining }(5,6)\text{ and }(2,3).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{3-6}{2-5}=\frac{-3}{-3}=1
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(9,-2)\text{ and }(6,-5).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-(-2)}{6-9}=\frac{-3}{-3}=1
\displaystyle \text{Since }m_1=m_2,\text{ the two lines are parallel.}
\displaystyle \text{ii) Let }m_1\text{ be the slope of the line joining }(9,5)\text{ and }(-1,1).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{5-1}{9-(-1)}=\frac{4}{10}=\frac{2}{5}
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(3,-5)\text{ and }(8,-3).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-(-5)}{8-3}=\frac{2}{5}
\displaystyle \text{Since }m_1=m_2,\text{ the two lines are parallel.}
\displaystyle \text{iii) Let }m_1\text{ be the slope of the line joining }(5,3)\text{ and }(1,1).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{1-3}{1-5}=\frac{-2}{-4}=\frac{1}{2}
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(-2,5)\text{ and }(2,-5).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-5}{2-(-2)}=\frac{-10}{4}=-\frac{5}{2}
\displaystyle m_1m_2=\frac{1}{2}\times\left(-\frac{5}{2}\right)=-\frac{5}{4}\neq-1
\displaystyle \therefore \text{The two lines are neither parallel nor perpendicular.}
\displaystyle \text{iv) Let }m_1\text{ be the slope of the line joining }(3,15)\text{ and }(16,6).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-15}{16-3}=-\frac{9}{13}
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(-5,3)\text{ and }(8,2).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{2-3}{8-(-5)}=-\frac{1}{13}
\displaystyle \text{Since }m_1\neq m_2\text{ and }m_1m_2=\frac{9}{169}\neq-1,\text{ the two lines are neither parallel nor perpendicular.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the slope of a line } \text{i) which bisects the first quadrant angle}
\displaystyle \text{ii) which makes an angle of }30^\circ\text{ with the positive direction of the }y\text{-axis}
\displaystyle \text{measured anticlockwise.}
\displaystyle \text{Answer:}
\displaystyle \text{i) The angle between the coordinate axes is }\frac{\pi}{2}.
\displaystyle \text{The line bisects the first quadrant angle.}
\displaystyle \therefore \text{Its inclination with the positive }x\text{-axis is }\frac{\pi}{4}.
\displaystyle \therefore m=\tan\frac{\pi}{4}=1.
\displaystyle \text{ii) The line makes an angle of }30^\circ\text{ with the positive }y\text{-axis measured anticlockwise.}
\displaystyle \therefore \text{Its inclination with the positive }x\text{-axis is }90^\circ+30^\circ=120^\circ.
\displaystyle \therefore m=\tan120^\circ=-\tan60^\circ=-\sqrt{3}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Using the method of slope, show that the following points are collinear:}
\displaystyle \text{i) }A(4,8),\ B(5,12),\ C(9,28)\qquad\text{ii) }A(16,-18),\ B(3,-6),\ C(-10,6)
\displaystyle \text{Answer:}
\displaystyle \text{i) Given }A(4,8),\ B(5,12)\text{ and }C(9,28).
\displaystyle \text{Slope of }AB=\frac{12-8}{5-4}=\frac{4}{1}=4.
\displaystyle \text{Slope of }BC=\frac{28-12}{9-5}=\frac{16}{4}=4.
\displaystyle \text{Slope of }CA=\frac{8-28}{4-9}=\frac{-20}{-5}=4.
\displaystyle \text{Since the slopes are equal, the three points are collinear.}
\displaystyle \text{ii) Given }A(16,-18),\ B(3,-6)\text{ and }C(-10,6).
\displaystyle \text{Slope of }AB=\frac{-6-(-18)}{3-16}=\frac{12}{-13}=-\frac{12}{13}.
\displaystyle \text{Slope of }BC=\frac{6-(-6)}{-10-3}=\frac{12}{-13}=-\frac{12}{13}.
\displaystyle \text{Slope of }CA=\frac{-18-6}{16-(-10)}=\frac{-24}{26}=-\frac{12}{13}.
\displaystyle \text{Since the slopes are equal, the three points are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{What is the value of }y\text{ so that the line through }(3,y)\text{ and }(2,7)
\displaystyle \text{is parallel to the line through }(-1,4)\text{ and }(0,6)?
\displaystyle \text{Answer:}
\displaystyle \text{Let }m_1\text{ be the slope of the line joining }(-1,4)\text{ and }(0,6).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-4}{0-(-1)}=\frac{2}{1}=2
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(3,y)\text{ and }(2,7).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{7-y}{2-3}=\frac{7-y}{-1}=y-7
\displaystyle \text{Since the two lines are parallel, }m_1=m_2.
\displaystyle \therefore 2=y-7
\displaystyle \therefore y=9
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{What can be said regarding a line if its slope is }
\displaystyle \text{i) zero \qquad ii) positive \qquad iii) negative?}
\displaystyle \text{Answer:}
\displaystyle \text{i) If }m=\tan\theta=0,\text{ then }\theta=0^\circ.
\displaystyle \therefore \text{The line is parallel to the }x\text{-axis.}
\displaystyle \text{ii) If }m\text{ is positive, then }\tan\theta\text{ is positive.}
\displaystyle \therefore 0^\circ<\theta<90^\circ.
\displaystyle \therefore \text{The line makes an acute angle with the positive }x\text{-axis.}
\displaystyle \text{iii) If }m\text{ is negative, then }\tan\theta\text{ is negative.}
\displaystyle \therefore 90^\circ<\theta<180^\circ.
\displaystyle \therefore \text{The line makes an obtuse angle with the positive }x\text{-axis.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Show that the line joining }(2,-3)\text{ and }(-5,1)\text{ is parallel}
\displaystyle \text{to the line joining }(7,-1)\text{ and }(0,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let }m_1\text{ be the slope of the line joining }(2,-3)\text{ and }(-5,1).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{1-(-3)}{-5-2}=\frac{4}{-7}=-\frac{4}{7}
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(7,-1)\text{ and }(0,3).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{3-(-1)}{0-7}=\frac{4}{-7}=-\frac{4}{7}
\displaystyle \text{Since }m_1=m_2,\text{ the two lines are parallel.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Show that the line joining }(2,-5)\text{ and }(-2,5)\text{ is perpendicular}
\displaystyle \text{to the line joining }(5,3)\text{ and }(1,1).
\displaystyle \text{Answer:}
\displaystyle \text{Let }m_1\text{ be the slope of the line joining }(2,-5)\text{ and }(-2,5).
\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{5-(-5)}{-2-2}=\frac{10}{-4}=-\frac{5}{2}
\displaystyle \text{Let }m_2\text{ be the slope of the line joining }(5,3)\text{ and }(1,1).
\displaystyle m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{1-3}{1-5}=\frac{-2}{-4}=\frac{1}{2}
\displaystyle m_1m_2=-\frac{5}{2}\times\frac{1}{2}=-\frac{5}{4}\neq-1
\displaystyle \therefore \text{The two lines are not perpendicular.}
\displaystyle \text{Hence, the given statement is incorrect.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Without using Pythagoras theorem, show that the points }A(0,4),
\displaystyle B(1,2)\text{ and }C(3,3)\text{ are the vertices of a right-angled triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(0,4),\ B(1,2)\text{ and }C(3,3).
\displaystyle \text{Slope of }AB=\frac{2-4}{1-0}=-2.
\displaystyle \text{Slope of }BC=\frac{3-2}{3-1}=\frac{1}{2}.
\displaystyle \text{Slope of }CA=\frac{4-3}{0-3}=-\frac{1}{3}.
\displaystyle \text{Slope of }AB\times\text{Slope of }BC=-2\times\frac{1}{2}=-1.
\displaystyle \therefore AB\perp BC.
\displaystyle \therefore \triangle ABC\text{ is a right-angled triangle with }\angle B=90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If the points }A(h,0),\ P(a,b)\text{ and }B(0,k)\text{ lie on a line,}
\displaystyle \text{show that }\frac{a}{h}+\frac{b}{k}=1.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(h,0),\ P(a,b)\text{ and }B(0,k)\text{ are collinear.}
\displaystyle \therefore \text{Slope of }PB=\text{Slope of }BA.
\displaystyle \Rightarrow \frac{k-b}{0-a}=\frac{0-k}{h-0}
\displaystyle \Rightarrow \frac{k-b}{-a}=\frac{-k}{h}
\displaystyle \Rightarrow h(k-b)=ak
\displaystyle \Rightarrow kh-bh=ak
\displaystyle \Rightarrow ak+bh=kh
\displaystyle \Rightarrow \frac{ak}{hk}+\frac{bh}{hk}=\frac{kh}{hk}
\displaystyle \therefore \frac{a}{h}+\frac{b}{k}=1.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The slope of a line is double the slope of another line. If the tangent}
\displaystyle \text{of the angle between them is }\frac{1}{3},\text{ find the slopes of the two lines.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }m_1\text{ and }m_2\text{ be the slopes of the two lines.}
\displaystyle \text{Given }m_2=2m_1.
\displaystyle \text{Let }\theta\text{ be the angle between the two lines.}
\displaystyle \therefore \tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|
\displaystyle \Rightarrow \frac{1}{3}=\left|\frac{m_2-m_1}{1+m_1m_2}\right|
\displaystyle \Rightarrow \frac{1}{3}=\left|\frac{2m_1-m_1}{1+2m_1^2}\right|
\displaystyle \Rightarrow \frac{1}{3}=\left|\frac{m_1}{1+2m_1^2}\right|
\displaystyle \Rightarrow \frac{m_1}{1+2m_1^2}=\pm\frac{1}{3}
\displaystyle \text{Case 1: Positive sign}
\displaystyle \frac{m_1}{1+2m_1^2}=\frac{1}{3}
\displaystyle \Rightarrow 3m_1=1+2m_1^2
\displaystyle \Rightarrow 2m_1^2-3m_1+1=0
\displaystyle \Rightarrow (2m_1-1)(m_1-1)=0
\displaystyle \Rightarrow m_1=\frac{1}{2}\text{ or }1
\displaystyle \therefore m_2=1\text{ or }2
\displaystyle \text{Case 2: Negative sign}
\displaystyle \frac{m_1}{1+2m_1^2}=-\frac{1}{3}
\displaystyle \Rightarrow 3m_1=-1-2m_1^2
\displaystyle \Rightarrow 2m_1^2+3m_1+1=0
\displaystyle \Rightarrow (2m_1+1)(m_1+1)=0
\displaystyle \Rightarrow m_1=-\frac{1}{2}\text{ or }-1
\displaystyle \therefore m_2=-1\text{ or }-2
\displaystyle \therefore \text{The possible pairs of slopes are }\left(\frac{1}{2},1\right),\ (1,2),\ \left(-\frac{1}{2},-1\right)\text{ and }(-1,-2).
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Consider the following population versus year graph. Find the slope of}
\displaystyle \text{the line }AB\text{ and using it, find the population in the year }2010.
\displaystyle \text{Answer:}
\displaystyle \text{From the graph: }A(1985,92),\ B(1995,97)\text{ and }C(2010,P).
\displaystyle \text{Slope of }AB=\frac{97-92}{1995-1985}=\frac{5}{10}=\frac{1}{2}.
\displaystyle \text{Since }A,\ B\text{ and }C\text{ lie on the same straight line,}
\displaystyle \text{Slope of }BC=\text{Slope of }AB.
\displaystyle \therefore \frac{P-97}{2010-1995}=\frac{1}{2}
\displaystyle \Rightarrow \frac{P-97}{15}=\frac{1}{2}
\displaystyle \Rightarrow 2(P-97)=15
\displaystyle \Rightarrow 2P-194=15
\displaystyle \Rightarrow 2P=209
\displaystyle \Rightarrow P=\frac{209}{2}=104.5
\displaystyle \therefore \text{The population in the year }2010\text{ is }104.5\text{ crore.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Without using the distance formula, show that the points }(-2,-1),
\displaystyle (4,0),\ (3,3)\text{ and }(-3,2)\text{ are the vertices of a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,-1),\ B(4,0),\ C(3,3)\text{ and }D(-3,2)\text{ be the vertices of the quadrilateral.}
\displaystyle \text{Slope of }AB\ (m_1)=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}.
\displaystyle \text{Slope of }BC\ (m_2)=\frac{3-0}{3-4}=\frac{3}{-1}=-3.
\displaystyle \text{Slope of }CD\ (m_3)=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}.
\displaystyle \text{Slope of }DA\ (m_4)=\frac{-1-2}{-2-(-3)}=\frac{-3}{1}=-3.
\displaystyle \text{Since }m_1=m_3,\ AB\parallel CD.
\displaystyle \text{Also, }m_2=m_4,\ BC\parallel AD.
\displaystyle \therefore ABCD\text{ is a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the angle between the }x\text{-axis and the line joining the}
\displaystyle \text{points }(3,-1)\text{ and }(4,-2).
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(3,-1)\text{ and }(4,-2)\text{ is}
\displaystyle m=\frac{-2-(-1)}{4-3}=\frac{-1}{1}=-1.
\displaystyle \text{If }\theta\text{ is the inclination of the line with the positive }x\text{-axis, then}
\displaystyle m=\tan\theta.
\displaystyle \therefore \tan\theta=-1.
\displaystyle \text{Since }0^\circ<\theta<180^\circ,\ \theta=135^\circ.
\displaystyle \therefore \text{The angle between the positive }x\text{-axis and the given line is }135^\circ.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The line through the points }(-2,6)\text{ and }(4,8)\text{ is}
\displaystyle \text{perpendicular to the line through the points }(8,12)\text{ and }(x,24).
\displaystyle \text{Find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,6),\ B(4,8),\ P(8,12)\text{ and }Q(x,24).
\displaystyle \text{Slope of }AB\ (m_1)=\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}.
\displaystyle \text{Slope of }PQ\ (m_2)=\frac{24-12}{x-8}=\frac{12}{x-8}.
\displaystyle \text{Since }AB\perp PQ,\ m_1m_2=-1.
\displaystyle \therefore \frac{1}{3}\times\frac{12}{x-8}=-1
\displaystyle \Rightarrow \frac{4}{x-8}=-1
\displaystyle \Rightarrow 4=-(x-8)
\displaystyle \Rightarrow 4=-x+8
\displaystyle \Rightarrow x=4.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the value of }x\text{ for which the points }(x,-1),\ (2,1)\text{ and }
\displaystyle (4,5)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(x,-1),\ B(2,1)\text{ and }C(4,5).
\displaystyle \text{Slope of }AB\ (m_1)=\frac{1-(-1)}{2-x}=\frac{2}{2-x}.
\displaystyle \text{Slope of }BC\ (m_2)=\frac{5-1}{4-2}=\frac{4}{2}=2.
\displaystyle \text{Since }A,\ B\text{ and }C\text{ are collinear, }m_1=m_2.
\displaystyle \therefore \frac{2}{2-x}=2
\displaystyle \Rightarrow 2=2(2-x)
\displaystyle \Rightarrow 2=4-2x
\displaystyle \Rightarrow 2x=2
\displaystyle \therefore x=1.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the angle between the }x\text{-axis and the line joining the}
\displaystyle \text{points }(-2,-2)\text{ and }(2,2).
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(-2,-2)\text{ and }(2,2)\text{ is}
\displaystyle m=\frac{2-(-2)}{2-(-2)}=\frac{4}{4}=1.
\displaystyle \text{If }\theta\text{ is the inclination of the line with the positive }x\text{-axis, then}
\displaystyle m=\tan\theta.
\displaystyle \therefore \tan\theta=1.
\displaystyle \therefore \theta=45^\circ.
\displaystyle \therefore \text{The angle between the positive }x\text{-axis and the given line is }45^\circ.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{By using the concept of slope, show that the points }(-2,-1),\ (4,0),
\displaystyle (3,3)\text{ and }(-3,2)\text{ are the vertices of a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,-1),\ B(4,0),\ C(3,3)\text{ and }D(-3,2)\text{ be the vertices of the quadrilateral.}
\displaystyle \text{Slope of }AB\ (m_1)=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}.
\displaystyle \text{Slope of }BC\ (m_2)=\frac{3-0}{3-4}=\frac{3}{-1}=-3.
\displaystyle \text{Slope of }CD\ (m_3)=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}.
\displaystyle \text{Slope of }DA\ (m_4)=\frac{-1-2}{-2-(-3)}=\frac{-3}{1}=-3.
\displaystyle \text{Since }m_1=m_3,\ AB\parallel CD.
\displaystyle \text{Also, }m_2=m_4,\ BC\parallel AD.
\displaystyle \therefore ABCD\text{ is a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A quadrilateral has vertices }(4,1),\ (1,7),\ (-6,0)\text{ and }(-1,-9).
\displaystyle \text{Show that the mid-points of its sides form a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(4,1),\ B(1,7),\ C(-6,0)\text{ and }D(-1,-9).
\displaystyle \text{Let }E\text{ be the mid-point of }AB.
\displaystyle E=\left(\frac{4+1}{2},\frac{1+7}{2}\right)=\left(\frac{5}{2},4\right).
\displaystyle \text{Let }F\text{ be the mid-point of }BC.
\displaystyle F=\left(\frac{1+(-6)}{2},\frac{7+0}{2}\right)=\left(-\frac{5}{2},\frac{7}{2}\right).
\displaystyle \text{Let }G\text{ be the mid-point of }CD.
\displaystyle G=\left(\frac{-6+(-1)}{2},\frac{0+(-9)}{2}\right)=\left(-\frac{7}{2},-\frac{9}{2}\right).
\displaystyle \text{Let }H\text{ be the mid-point of }DA.
\displaystyle H=\left(\frac{-1+4}{2},\frac{-9+1}{2}\right)=\left(\frac{3}{2},-4\right).
\displaystyle \text{Slope of }EF\ (m_1)=\frac{\frac{7}{2}-4}{-\frac{5}{2}-\frac{5}{2}}=\frac{-\frac{1}{2}}{-5}=\frac{1}{10}.
\displaystyle \text{Slope of }FG\ (m_2)=\frac{-\frac{9}{2}-\frac{7}{2}}{-\frac{7}{2}-\left(-\frac{5}{2}\right)}=\frac{-8}{-1}=8.
\displaystyle \text{Slope of }GH\ (m_3)=\frac{-4-\left(-\frac{9}{2}\right)}{\frac{3}{2}-\left(-\frac{7}{2}\right)}=\frac{\frac{1}{2}}{5}=\frac{1}{10}.
\displaystyle \text{Slope of }HE\ (m_4)=\frac{4-(-4)}{\frac{5}{2}-\frac{3}{2}}=\frac{8}{1}=8.
\displaystyle \text{Since }m_1=m_3,\ EF\parallel GH.
\displaystyle \text{Also, }m_2=m_4,\ FG\parallel HE.
\displaystyle \therefore EFGH\text{ is a parallelogram.}
\displaystyle \\

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