\displaystyle \textbf{Question 1: }\text{Find the area of a triangle whose vertices are}
\displaystyle \text{(i) }(-5,7),\,(-4,-5)\text{ and }(4,5).\hfill\text{[CBSE 2020]}
\displaystyle \text{(iv) }(1,-1),\,(-4,6)\text{ and }(-3,-5).\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|
\displaystyle =\frac{1}{2}\left|(-5)(-5-5)+(-4)(5-7)+4(7+5)\right|
\displaystyle =\frac{1}{2}\left|50+8+48\right|=53\text{ sq. units.}
\displaystyle \therefore \text{The area of the triangle is }53\text{ sq. units.}

\displaystyle \text{(iv) Area}=\frac{1}{2}\left|1(6+5)+(-4)(-5+1)+(-3)(-1-6)\right|
\displaystyle =\frac{1}{2}\left|11+16+21\right|=24\text{ sq. units.}
\displaystyle \therefore \text{The area of the triangle is }24\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the area of the quadrilaterals, the coordinates of whose vertices are}
\displaystyle \text{(i) }(-3,2),\,(5,4),\,(7,-6),\,(-5,-4).\hfill\text{[CBSE 2016]}
\displaystyle \text{(ii) }(-4,-2),\,(-3,-5),\,(3,-2),\,(2,3).\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }A(-3,2),\,B(5,4),\,C(7,-6)\text{ and }D(-5,-4)\text{ be the vertices.}
\displaystyle \text{Area of quadrilateral }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|(-3)(4+6)+5(-6-2)+7(2-4)\right|
\displaystyle =\frac{1}{2}\left|-30-40-14\right|=42\text{ sq. units.}
\displaystyle \text{Area of }\triangle ACD=\frac{1}{2}\left|(-3)(-6+4)+7(-4-2)+(-5)(2+6)\right|
\displaystyle =\frac{1}{2}\left|6-42-40\right|=38\text{ sq. units.}
\displaystyle \therefore \text{Area of quadrilateral }ABCD=42+38=80\text{ sq. units.}

\displaystyle \text{(ii) Let }A(-4,-2),\,B(-3,-5),\,C(3,-2)\text{ and }D(2,3)\text{ be the vertices.}
\displaystyle \text{Area of quadrilateral }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|(-4)(-5+2)+(-3)(-2+2)+3(-2+5)\right|
\displaystyle =\frac{1}{2}\left|12+0+9\right|=\frac{21}{2}\text{ sq. units.}
\displaystyle \text{Area of }\triangle ACD=\frac{1}{2}\left|(-4)(-2-3)+3(3+2)+2(-2+2)\right|
\displaystyle =\frac{1}{2}\left|20+15+0\right|=\frac{35}{2}\text{ sq. units.}
\displaystyle \therefore \text{Area of quadrilateral }ABCD=\frac{21}{2}+\frac{35}{2}=28\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The vertices of }\triangle ABC\text{ are }(-2,1),\,(5,4)\text{ and }(2,-3)\text{ respectively. Find the}
\displaystyle \text{area of the triangle and the length of the altitude through }A.
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|(-2)(4+3)+5(-3-1)+2(1-4)\right|
\displaystyle =\frac{1}{2}\left|-14-20-6\right|=20\text{ sq. units.}
\displaystyle BC=\sqrt{(2-5)^2+(-3-4)^2}=\sqrt{9+49}=\sqrt{58}\text{ units.}
\displaystyle \text{Let the altitude through }A\text{ be }h\text{ units.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times BC\times h
\displaystyle 20=\frac{1}{2}\times\sqrt{58}\times h
\displaystyle h=\frac{40}{\sqrt{58}}=\frac{20\sqrt{58}}{29}\text{ units.}
\displaystyle \therefore \text{The area is }20\text{ sq. units and the altitude through }A\text{ is }\frac{20\sqrt{58}}{29}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the point }P(m,3)\text{ lies on the line segment joining the points }A\left(-\frac{2}{5},6\right)
\displaystyle \text{and }B(2,8),\text{ find the value of }m.\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{The }y\text{-coordinate of every point on the line segment }AB\text{ must lie between }6\text{ and }8.
\displaystyle \text{But the }y\text{-coordinate of }P\text{ is }3.
\displaystyle \therefore P(m,3)\text{ cannot lie on the line segment }AB.
\displaystyle \text{Hence, as printed, the question has no solution.}
\displaystyle \text{If ``line segment'' is intended to mean ``line joining'', then }A,\,P\text{ and }B\text{ are collinear.}
\displaystyle \frac{3-6}{m+\frac{2}{5}}=\frac{8-6}{2+\frac{2}{5}}
\displaystyle \frac{-3}{m+\frac{2}{5}}=\frac{2}{\frac{12}{5}}=\frac{5}{6}
\displaystyle -18=5m+2
\displaystyle \therefore m=-4.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }(x,y)\text{ be on the line joining the two points }(1,-3)\text{ and }(-4,2),\text{ prove that}
\displaystyle x+y+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{Since }(x,y),\,(1,-3)\text{ and }(-4,2)\text{ are collinear, the area of the triangle formed is zero.}
\displaystyle \frac{1}{2}\left|x(-3-2)+1(2-y)+(-4)(y+3)\right|=0
\displaystyle \left|-5x+2-y-4y-12\right|=0
\displaystyle \left|-5x-5y-10\right|=0
\displaystyle -5(x+y+2)=0
\displaystyle \therefore x+y+2=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the value of }k\text{ so that the area of triangle }ABC\text{ with }
\displaystyle A(k+1,1),\,B(4,-3) \ \text{and }C(7,-k)\text{ is }6\text{ square units.}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|(k+1)(-3+k)+4(-k-1)+7(1+3)\right|
\displaystyle 6=\frac{1}{2}\left|k^2-2k-3-4k-4+28\right|
\displaystyle 12=\left|k^2-6k+21\right|
\displaystyle 12=\left|(k-3)^2+12\right|
\displaystyle \text{Since }(k-3)^2+12>0,
\displaystyle (k-3)^2+12=12
\displaystyle (k-3)^2=0
\displaystyle \therefore k=3.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }A(-3,5),\,B(-2,-7),\,C(1,-8)\text{ and }D(6,3)\text{ are the vertices of a}
\displaystyle \text{quadrilateral }ABCD,\text{ find its area.}\hfill\text{[CBSE 2014, 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Area of quadrilateral }ABCD=\text{Area of }\triangle ABC+\text{Area of }\triangle ACD.
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|(-3)(-7+8)+(-2)(-8-5)+1(5+7)\right|
\displaystyle =\frac{1}{2}\left|-3+26+12\right|=\frac{35}{2}\text{ sq. units.}
\displaystyle \text{Area of }\triangle ACD=\frac{1}{2}\left|(-3)(-8-3)+1(3-5)+6(5+8)\right|
\displaystyle =\frac{1}{2}\left|33-2+78\right|=\frac{109}{2}\text{ sq. units.}
\displaystyle \therefore \text{Area of quadrilateral }ABCD=\frac{35}{2}+\frac{109}{2}=72\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{For what value of }a\text{ are the points }(a,1),\,(1,-1)\text{ and }(11,4)\text{ collinear?}
\displaystyle \hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the three points are collinear, the area of the triangle formed by them is zero.}
\displaystyle \frac{1}{2}\left|a(-1-4)+1(4-1)+11(1+1)\right|=0
\displaystyle \left|-5a+3+22\right|=0
\displaystyle -5a+25=0
\displaystyle \therefore a=5.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the vertices of a triangle are }(1,-3),\,(4,p)\text{ and }(-9,7)\text{ and its area is}
\displaystyle 15\text{ sq. units, find the value(s) of }p.\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Area of triangle}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|
\displaystyle 15=\frac{1}{2}\left|1(p-7)+4(7+3)+(-9)(-3-p)\right|
\displaystyle 30=\left|p-7+40+27+9p\right|
\displaystyle \left|10p+60\right|=30
\displaystyle 10p+60=30\quad\text{or}\quad10p+60=-30
\displaystyle 10p=-30\quad\text{or}\quad10p=-90
\displaystyle \therefore p=-3\quad\text{or}\quad p=-9.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the area of triangle whose vertices are:}
\displaystyle \text{(i) }(at_1^2,2at_1),\,(at_2^2,2at_2)\text{ and }(at_3^2,2at_3)
\displaystyle \text{(ii) }(a,c+a),\,(a,c)\text{ and }(-a,c-a)
\displaystyle \text{Answer:}
\displaystyle \text{(i) Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|
\displaystyle =\frac{1}{2}\left|at_1^2(2at_2-2at_3)+at_2^2(2at_3-2at_1)+at_3^2(2at_1-2at_2)\right|
\displaystyle =a^2\left|t_1^2(t_2-t_3)+t_2^2(t_3-t_1)+t_3^2(t_1-t_2)\right|
\displaystyle =a^2\left|(t_1-t_2)(t_1-t_3)(t_2-t_3)\right|.
\displaystyle \therefore \text{Area}=a^2\left|(t_1-t_2)(t_1-t_3)(t_2-t_3)\right|\text{ sq. units.}

\displaystyle \text{(ii) Area}=\frac{1}{2}\left|a\{c-(c-a)\}+a\{(c-a)-(c+a)\}+(-a)\{(c+a)-c\}\right|
\displaystyle =\frac{1}{2}\left|a^2-2a^2-a^2\right|
\displaystyle =\frac{1}{2}\left|-2a^2\right|=a^2\text{ sq. units.}
\displaystyle \therefore \text{Area}=a^2\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In }\triangle ABC,\text{ the coordinates of vertex }A\text{ are }(0,-1)\text{ and }D(1,0)\text{ and }E(0,1)
\displaystyle \text{are respectively the mid-points of sides }AB\text{ and }AC.\text{ If }F\text{ is the mid-point of side }BC,
\displaystyle \text{find the area of }\triangle DEF.\hfill\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B=(x_1,y_1)\text{ and }C=(x_2,y_2).
\displaystyle \text{Since }D(1,0)\text{ is the mid-point of }AB,
\displaystyle \left(\frac{0+x_1}{2},\frac{-1+y_1}{2}\right)=(1,0)
\displaystyle \therefore x_1=2,\quad y_1=1\quad\Rightarrow\quad B=(2,1).
\displaystyle \text{Since }E(0,1)\text{ is the mid-point of }AC,
\displaystyle \left(\frac{0+x_2}{2},\frac{-1+y_2}{2}\right)=(0,1)
\displaystyle \therefore x_2=0,\quad y_2=3\quad\Rightarrow\quad C=(0,3).
\displaystyle \text{Since }F\text{ is the mid-point of }BC,
\displaystyle F=\left(\frac{2+0}{2},\frac{1+3}{2}\right)=(1,2).
\displaystyle \text{Area of }\triangle DEF=\frac{1}{2}\left|1(1-2)+0(2-0)+1(0-1)\right|
\displaystyle =\frac{1}{2}\left|-1-1\right|=1\text{ sq. unit.}
\displaystyle \therefore \text{The area of }\triangle DEF\text{ is }1\text{ sq. unit.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the area of the triangle }PQR\text{ with }Q(3,2)\text{ and the mid-points}
\displaystyle \text{of the sides through }Q\text{ being }(2,-1)\text{ and }(1,2).\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }D(2,-1)\text{ and }E(1,2)\text{ be the mid-points of }QP\text{ and }QR\text{ respectively.}
\displaystyle \text{Let }P=(x_1,y_1).\text{ Since }D\text{ is the mid-point of }QP,
\displaystyle \left(\frac{3+x_1}{2},\frac{2+y_1}{2}\right)=(2,-1)
\displaystyle \therefore x_1=1,\quad y_1=-4\quad\Rightarrow\quad P=(1,-4).
\displaystyle \text{Let }R=(x_2,y_2).\text{ Since }E\text{ is the mid-point of }QR,
\displaystyle \left(\frac{3+x_2}{2},\frac{2+y_2}{2}\right)=(1,2)
\displaystyle \therefore x_2=-1,\quad y_2=2\quad\Rightarrow\quad R=(-1,2).
\displaystyle \text{Area of }\triangle PQR=\frac{1}{2}\left|1(2-2)+3(2+4)+(-1)(-4-2)\right|
\displaystyle =\frac{1}{2}\left|0+18+6\right|=12\text{ sq. units.}
\displaystyle \therefore \text{The area of }\triangle PQR\text{ is }12\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }R(x,y)\text{ is a point on the line segment joining the points }P(a,b)\text{ and}
\displaystyle Q(b,a), \ \text{then prove that }x+y=a+b.\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }P(a,b),\,R(x,y)\text{ and }Q(b,a)\text{ are collinear, the area of the triangle formed is zero.}
\displaystyle \frac{1}{2}\left|a(y-a)+x(a-b)+b(b-y)\right|=0
\displaystyle ay-a^2+ax-bx+b^2-by=0
\displaystyle (a-b)y+(a-b)x-(a^2-b^2)=0
\displaystyle (a-b)\{x+y-(a+b)\}=0
\displaystyle \therefore x+y=a+b.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the value of }a\text{ for which the area of the triangle formed by the points}
\displaystyle A(a,2a), \ B(-2,6)\text{ and }C(3,1)\text{ is }10\text{ square units.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|a(6-1)+(-2)(1-2a)+3(2a-6)\right|
\displaystyle 10=\frac{1}{2}\left|5a-2+4a+6a-18\right|
\displaystyle \left|15a-20\right|=20
\displaystyle 15a-20=20\quad\text{or}\quad15a-20=-20
\displaystyle 15a=40\quad\text{or}\quad15a=0
\displaystyle \therefore a=\frac{8}{3}\quad\text{or}\quad a=0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }a\neq b\neq0,\text{ prove that the points }(a,a^2),\,(b,b^2),\,(0,0)\text{ are never collinear.}
\displaystyle \hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the triangle formed by the given points}
\displaystyle =\frac{1}{2}\left|a(b^2-0)+b(0-a^2)+0(a^2-b^2)\right|
\displaystyle =\frac{1}{2}\left|ab^2-a^2b\right|
\displaystyle =\frac{1}{2}\left|ab(b-a)\right|.
\displaystyle \text{Since }a\neq0,\,b\neq0\text{ and }a\neq b,\quad ab(b-a)\neq0.
\displaystyle \therefore \text{Area of the triangle}\neq0.
\displaystyle \therefore \text{The points }(a,a^2),\,(b,b^2)\text{ and }(0,0)\text{ are never collinear.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The area of a triangle is }5\text{ sq. units. Two of its vertices are at }(2,1)\text{ and}
\displaystyle (3,-2).\text{ If the third vertex is }\left(\frac{7}{2},y\right),\text{ find }y.\hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle 5=\frac{1}{2}\left|2(-2-y)+3(y-1)+\frac{7}{2}(1+2)\right|
\displaystyle 10=\left|-4-2y+3y-3+\frac{21}{2}\right|
\displaystyle 10=\left|y+\frac{7}{2}\right|
\displaystyle y+\frac{7}{2}=10\quad\text{or}\quad y+\frac{7}{2}=-10
\displaystyle y=\frac{13}{2}\quad\text{or}\quad y=-\frac{27}{2}.
\displaystyle \therefore y=\frac{13}{2}\quad\text{or}\quad y=-\frac{27}{2}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The point }A\text{ divides the join of }P(-5,1)\text{ and }Q(3,5)\text{ in the ratio }
\displaystyle k:1. \ \text{Find the two values of }k\text{ for which the area of }\triangle ABC,\text{ where }B(1,5)\text{ and }C(7,-2),
\displaystyle \text{is equal to }2\text{ units.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ divides }PQ\text{ internally in the ratio }k:1,
\displaystyle A=\left(\frac{3k-5}{k+1},\frac{5k+1}{k+1}\right).
\displaystyle \text{Area of }\triangle ABC=2
\displaystyle \frac{1}{2}\left|\frac{3k-5}{k+1}(5+2)+1\left(-2-\frac{5k+1}{k+1}\right)+7\left(\frac{5k+1}{k+1}-5\right)\right|=2
\displaystyle \left|\frac{2(7k-33)}{k+1}\right|=4
\displaystyle \left|\frac{7k-33}{k+1}\right|=2
\displaystyle \frac{7k-33}{k+1}=2\quad\text{or}\quad\frac{7k-33}{k+1}=-2
\displaystyle 7k-33=2k+2\quad\text{or}\quad7k-33=-2k-2
\displaystyle 5k=35\quad\text{or}\quad9k=31
\displaystyle \therefore k=7\quad\text{or}\quad k=\frac{31}{9}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The area of a triangle is }5.\text{ Two of its vertices are }(2,1)\text{ and }(3,-2).
\displaystyle \text{The third vertex lies on }y=x+3.\text{ Find the third vertex.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }(x,x+3).
\displaystyle 5=\frac{1}{2}\left|2(-2-x-3)+3(x+3-1)+x(1+2)\right|
\displaystyle 10=\left|-2x-10+3x+6+3x\right|
\displaystyle |4x-4|=10
\displaystyle 4x-4=10\quad\text{or}\quad4x-4=-10
\displaystyle x=\frac{7}{2}\quad\text{or}\quad x=-\frac{3}{2}.
\displaystyle \text{Since }y=x+3,\quad y=\frac{13}{2}\quad\text{or}\quad y=\frac{3}{2}.
\displaystyle \therefore \text{The third vertex is }\left(\frac{7}{2},\frac{13}{2}\right)\text{ or }\left(-\frac{3}{2},\frac{3}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find the area of a parallelogram }ABCD\text{ if three of its vertices are }A(2,4),
\displaystyle B(2+\sqrt{3},5)\text{ and }C(2,6).\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{A diagonal of a parallelogram divides it into two triangles of equal area.}
\displaystyle \therefore \text{Area of parallelogram }ABCD=2(\text{Area of }\triangle ABC).
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|2(5-6)+(2+\sqrt{3})(6-4)+2(4-5)\right|
\displaystyle =\frac{1}{2}\left|-2+4+2\sqrt{3}-2\right|=\sqrt{3}\text{ sq. units.}
\displaystyle \therefore \text{Area of parallelogram }ABCD=2\sqrt{3}\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find the value(s) of }k\text{ for which the points }(3k-1,k-2),\,
\displaystyle (k,k-7) \ \text{and }(k-1,-k-2)\text{ are collinear.}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the three points are collinear, the area of the triangle formed by them is zero.}
\displaystyle \frac{1}{2}\left|(3k-1)\{(k-7)-(-k-2)\}+k\{(-k-2)-(k-2)\}\right.
\displaystyle \left.+(k-1)\{(k-2)-(k-7)\}\right|=0
\displaystyle \left|(3k-1)(2k-5)-2k^2+5(k-1)\right|=0
\displaystyle \left|6k^2-17k+5-2k^2+5k-5\right|=0
\displaystyle \left|4k^2-12k\right|=0
\displaystyle 4k(k-3)=0
\displaystyle \therefore k=0\quad\text{or}\quad k=3.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If the points }A(-1,-4),\,B(b,c)\text{ and }C(5,-1)\text{ are collinear and }
\displaystyle 2b+c=4, \ \text{find the values of }b\text{ and }c.\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A,\,B\text{ and }C\text{ are collinear, the area of the triangle formed by them is zero.}
\displaystyle \frac{1}{2}\left|(-1)(c+1)+b(-1+4)+5(-4-c)\right|=0
\displaystyle \left|-c-1+3b-20-5c\right|=0
\displaystyle 3b-6c-21=0
\displaystyle b-2c=7 \qquad\ldots\text{(i)}
\displaystyle \text{Also, }2b+c=4 \qquad\ldots\text{(ii)}
\displaystyle \text{From (i), }b=7+2c.
\displaystyle 2(7+2c)+c=4
\displaystyle 14+5c=4\Rightarrow5c=-10\Rightarrow c=-2
\displaystyle \therefore b=7+2(-2)=3.
\displaystyle \therefore b=3,\quad c=-2.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }a\neq b\neq c,\text{ prove that the points }(a,a^2),\,(b,b^2),\,(c,c^2)\text{ can never}
\displaystyle \text{be collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the triangle formed by the given points}
\displaystyle =\frac{1}{2}\left|a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)\right|
\displaystyle =\frac{1}{2}\left|(a-b)(b-c)(c-a)\right|.
\displaystyle \text{Since }a,b,c\text{ are distinct, }(a-b)(b-c)(c-a)\neq0.
\displaystyle \therefore \text{Area of the triangle}\neq0.
\displaystyle \therefore \text{The points }(a,a^2),\,(b,b^2)\text{ and }(c,c^2)\text{ can never be collinear.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Four points }A(6,3),\,B(-3,5),\,C(4,-2)\text{ and }D(x,3x)\text{ are given such}
\displaystyle \text{that }\frac{\text{Area of }\triangle DBC}{\text{Area of }\triangle ABC}=\frac{1}{2}.\text{ Find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|6(5+2)+(-3)(-2-3)+4(3-5)\right|
\displaystyle =\frac{1}{2}\left|42+15-8\right|=\frac{49}{2}\text{ sq. units.}
\displaystyle \text{Area of }\triangle DBC=\frac{1}{2}\left|x(5+2)+(-3)(-2-3x)+4(3x-5)\right|
\displaystyle =\frac{1}{2}\left|7x+6+9x+12x-20\right|
\displaystyle =\frac{1}{2}|28x-14|=7|2x-1|.
\displaystyle \text{Given, }\frac{\text{Area of }\triangle DBC}{\text{Area of }\triangle ABC}=\frac{1}{2}
\displaystyle \frac{7|2x-1|}{49/2}=\frac{1}{2}
\displaystyle |2x-1|=\frac{7}{4}
\displaystyle 2x-1=\frac{7}{4}\quad\text{or}\quad2x-1=-\frac{7}{4}
\displaystyle \therefore x=\frac{11}{8}\quad\text{or}\quad x=-\frac{3}{8}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If three points }(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)\text{ lie on the same line, prove that}
\displaystyle \frac{y_2-y_3}{x_2x_3}+\frac{y_3-y_1}{x_3x_1}+\frac{y_1-y_2}{x_1x_2}=0.
\displaystyle \text{Answer:}
\displaystyle \text{Since the three points are collinear, the area of the triangle formed by them is zero.}
\displaystyle x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0
\displaystyle \text{Dividing throughout by }x_1x_2x_3,
\displaystyle \frac{x_1(y_2-y_3)}{x_1x_2x_3}+\frac{x_2(y_3-y_1)}{x_1x_2x_3}+\frac{x_3(y_1-y_2)}{x_1x_2x_3}=0
\displaystyle \therefore \frac{y_2-y_3}{x_2x_3}+\frac{y_3-y_1}{x_3x_1}+\frac{y_1-y_2}{x_1x_2}=0.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the points }A(1,-2),\,B(2,3),\,C(a,2)\text{ and }D(-4,-3)\text{ form a}
\displaystyle \text{parallelogram, find the value of }a\text{ and height of the parallelogram taking }AB\text{ as base.}
\displaystyle \text{Answer:}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore \text{Mid-point of }AC=\text{mid-point of }BD.
\displaystyle \left(\frac{1+a}{2},\frac{-2+2}{2}\right)=\left(\frac{2-4}{2},\frac{3-3}{2}\right)
\displaystyle \frac{1+a}{2}=-1
\displaystyle 1+a=-2
\displaystyle \therefore a=-3.
\displaystyle AB=\sqrt{(2-1)^2+(3+2)^2}=\sqrt{26}\text{ units.}
\displaystyle \text{Area of }\triangle ABD=\frac{1}{2}\left|1(3+3)+2(-3+2)+(-4)(-2-3)\right|
\displaystyle =\frac{1}{2}|6-2+20|=12\text{ sq. units.}
\displaystyle \text{Area of parallelogram }ABCD=2\times12=24\text{ sq. units.}
\displaystyle \text{Let }h\text{ be the height corresponding to base }AB.
\displaystyle 24=AB\times h=\sqrt{26}\,h
\displaystyle h=\frac{24}{\sqrt{26}}=\frac{12\sqrt{26}}{13}\text{ units.}
\displaystyle \therefore a=-3\text{ and the required height is }\frac{12\sqrt{26}}{13}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 26: }A(6,1),\,B(8,2)\text{ and }C(9,4)\text{ are three vertices of a parallelogram }ABCD.
\displaystyle \text{If }E\text{ is the mid-point of }DC,\text{ find the area of }\triangle ADE. 
\displaystyle \text{Answer:}
\displaystyle \text{Let }D=(x,y).
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore \text{Mid-point of }AC=\text{mid-point of }BD.
\displaystyle \left(\frac{6+9}{2},\frac{1+4}{2}\right)=\left(\frac{8+x}{2},\frac{2+y}{2}\right)
\displaystyle \therefore x=7,\quad y=3.
\displaystyle \therefore D=(7,3).
\displaystyle \text{Since }E\text{ is the mid-point of }DC,
\displaystyle E=\left(\frac{7+9}{2},\frac{3+4}{2}\right)=\left(8,\frac{7}{2}\right).
\displaystyle \text{Area of }\triangle ADE=\frac{1}{2}\left|6\left(3-\frac{7}{2}\right)+7\left(\frac{7}{2}-1\right)+8(1-3)\right|
\displaystyle =\frac{1}{2}\left|-3+\frac{35}{2}-16\right|=\frac{3}{4}\text{ sq. unit.}
\displaystyle \therefore \text{The area of }\triangle ADE\text{ is }\frac{3}{4}\text{ sq. unit.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }D\left(-\frac{1}{5},\frac{5}{2}\right),\,E(7,3)\text{ and }F\left(\frac{7}{2},\frac{7}{2}\right)\text{ are the mid-points}
\displaystyle \text{of sides of }\triangle ABC,\text{ find the area of }\triangle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle DEF=\frac{1}{2}\left|-\frac{1}{5}\left(3-\frac{7}{2}\right)+7\left(\frac{7}{2}-\frac{5}{2}\right)+\frac{7}{2}\left(\frac{5}{2}-3\right)\right|
\displaystyle =\frac{1}{2}\left|\frac{1}{10}+7-\frac{7}{4}\right|
\displaystyle =\frac{1}{2}\times\frac{107}{20}=\frac{107}{40}\text{ sq. units.}
\displaystyle \text{Since }D,E,F\text{ are the mid-points of the sides of }\triangle ABC,
\displaystyle \text{Area of }\triangle DEF=\frac{1}{4}\left(\text{Area of }\triangle ABC\right).
\displaystyle \therefore \text{Area of }\triangle ABC=4\times\frac{107}{40}=\frac{107}{10}\text{ sq. units.}
\displaystyle \\

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