\displaystyle \textbf{Question 1: }\text{If the line segment joining the points }P(x_1,y_1)\text{ and }Q(x_2,y_2)
\displaystyle \text{subtends an angle }\alpha\text{ at the origin }O,\text{ prove that}
\displaystyle OP\cdot OQ\cos\alpha=x_1x_2+y_1y_2.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle OPQ,\text{ where }\angle POQ=\alpha.
\displaystyle OP^2=x_1^2+y_1^2
\displaystyle OQ^2=x_2^2+y_2^2
\displaystyle PQ^2=(x_2-x_1)^2+(y_2-y_1)^2
\displaystyle \text{Using the cosine rule in }\triangle OPQ,
\displaystyle PQ^2=OP^2+OQ^2-2(OP)(OQ)\cos\alpha
\displaystyle \therefore (x_2-x_1)^2+(y_2-y_1)^2
\displaystyle =(x_1^2+y_1^2)+(x_2^2+y_2^2)-2(OP)(OQ)\cos\alpha
\displaystyle x_2^2+x_1^2-2x_1x_2+y_2^2+y_1^2-2y_1y_2
\displaystyle =x_1^2+y_1^2+x_2^2+y_2^2-2(OP)(OQ)\cos\alpha
\displaystyle \Rightarrow -2x_1x_2-2y_1y_2=-2(OP)(OQ)\cos\alpha
\displaystyle \Rightarrow 2(OP)(OQ)\cos\alpha=2x_1x_2+2y_1y_2
\displaystyle \Rightarrow OP\cdot OQ\cos\alpha=x_1x_2+y_1y_2
\displaystyle \therefore OP\cdot OQ\cos\alpha=x_1x_2+y_1y_2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The vertices of }\triangle ABC\text{ are }A(0,0),\ B(2,-1)\text{ and }C(9,2).
\displaystyle \text{Find }\cos B.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(0,0),\ B(2,-1)\text{ and }C(9,2).
\displaystyle \text{Using }\cos B=\frac{a^2+c^2-b^2}{2ac},\text{ where }a=BC,\ b=CA,\ c=AB.
\displaystyle a=BC=\sqrt{(2-9)^2+(-1-2)^2}=\sqrt{49+9}=\sqrt{58}
\displaystyle b=CA=\sqrt{(9-0)^2+(2-0)^2}=\sqrt{81+4}=\sqrt{85}
\displaystyle c=AB=\sqrt{(2-0)^2+(-1-0)^2}=\sqrt{4+1}=\sqrt{5}
\displaystyle \therefore \cos B=\frac{58+5-85}{2\sqrt{58}\sqrt{5}}
\displaystyle =\frac{-22}{2\sqrt{290}}=\frac{-11}{\sqrt{290}}
\displaystyle \therefore \cos B=\frac{-11}{\sqrt{290}}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Four points }A(6,3),\ B(-3,5),\ C(4,-2)\text{ and }D(x,3x)\text{ are}
\displaystyle \text{given such that }\frac{\text{Area of }\triangle DBC}{\text{Area of }\triangle ABC}=\frac{1}{2}.\text{ Find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(6,3),\ B(-3,5),\ C(4,-2)\text{ and }D(x,3x).
\displaystyle \text{The area of a triangle with vertices }(x_1,y_1),\ (x_2,y_2)\text{ and }(x_3,y_3)\text{ is}
\displaystyle \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.
\displaystyle \text{Area of }\triangle DBC
\displaystyle =\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{Area of }\triangle ABC
\displaystyle =\frac{1}{2}\left|6(5+2)-3(-2-3)+4(3-5)\right|
\displaystyle =\frac{1}{2}|42+15-8|=\frac{49}{2}
\displaystyle \text{Given, }\frac{\text{Area of }\triangle DBC}{\text{Area of }\triangle ABC}=\frac{1}{2}
\displaystyle \therefore \frac{7|2x-1|}{\frac{49}{2}}=\frac{1}{2}
\displaystyle \Rightarrow \frac{2|2x-1|}{7}=\frac{1}{2}
\displaystyle \Rightarrow |2x-1|=\frac{7}{4}
\displaystyle \therefore 2x-1=\frac{7}{4}\quad\text{or}\quad 2x-1=-\frac{7}{4}
\displaystyle \Rightarrow 2x=\frac{11}{4}\quad\text{or}\quad 2x=-\frac{3}{4}
\displaystyle \Rightarrow x=\frac{11}{8}\quad\text{or}\quad x=-\frac{3}{8}
\displaystyle \therefore x=\frac{11}{8}\text{ or }-\frac{3}{8}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The points }A(2,0),\ B(9,1),\ C(11,6)\text{ and }D(4,4)\text{ are the vertices}
\displaystyle \text{of a quadrilateral }ABCD.\text{ Determine whether }ABCD\text{ is a rhombus or not.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(2,0),\ B(9,1),\ C(11,6)\text{ and }D(4,4).
\displaystyle AB=\sqrt{(9-2)^2+(1-0)^2}=\sqrt{49+1}=\sqrt{50}
\displaystyle BC=\sqrt{(11-9)^2+(6-1)^2}=\sqrt{4+25}=\sqrt{29}
\displaystyle CD=\sqrt{(4-11)^2+(4-6)^2}=\sqrt{49+4}=\sqrt{53}
\displaystyle DA=\sqrt{(2-4)^2+(0-4)^2}=\sqrt{4+16}=\sqrt{20}
\displaystyle \text{Since }AB,\ BC,\ CD\text{ and }DA\text{ are not equal, the quadrilateral is not a rhombus.}
\displaystyle \therefore ABCD\text{ is not a rhombus.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the coordinates of the centre of the circle inscribed in a triangle}
\displaystyle \text{whose vertices are }(-36,7),\ (20,7)\text{ and }(0,-8).
\displaystyle \text{Answer:}
\displaystyle \text{Given the vertices of the triangle are }(-36,7),\ (20,7)\text{ and }(0,-8).
\displaystyle \text{The coordinates of the incentre are}
\displaystyle \left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right),
\displaystyle \text{where }a=BC,\ b=CA,\ c=AB.
\displaystyle a=BC=\sqrt{(20-0)^2+(7+8)^2}=\sqrt{400+225}=25
\displaystyle b=CA=\sqrt{(0+36)^2+(-8-7)^2}=\sqrt{1296+225}=39
\displaystyle c=AB=\sqrt{(20+36)^2+(7-7)^2}=56
\displaystyle \therefore \text{The coordinates of the incentre are}
\displaystyle \left(\frac{25(-36)+39(20)+56(0)}{25+39+56},\frac{25(7)+39(7)+56(-8)}{25+39+56}\right)
\displaystyle =\left(\frac{-900+780}{120},\frac{175+273-448}{120}\right)
\displaystyle =\left(\frac{-120}{120},\frac{0}{120}\right)=(-1,0)
\displaystyle \therefore \text{The coordinates of the incentre are }(-1,0).
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The base of an equilateral triangle with side }2a\text{ lies along the }y\text{-axis}
\displaystyle \text{such that the mid-point of the base is at the origin. Find the vertices of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertices on the }y\text{-axis be }B(0,-a)\text{ and }C(0,a).
\displaystyle \text{Let the third vertex be }A(x,0).
\displaystyle \text{Since }\triangle ABC\text{ is equilateral, }AB=BC=CA=2a.
\displaystyle AB=\sqrt{(x-0)^2+(0+a)^2}=2a
\displaystyle \Rightarrow x^2+a^2=(2a)^2
\displaystyle \Rightarrow x^2=3a^2
\displaystyle \Rightarrow x=\pm\sqrt{3}\,a
\displaystyle \therefore \text{The required vertices are }(0,a),\ (0,-a),\ (\sqrt{3}a,0)
\displaystyle \text{or }(0,a),\ (0,-a),\ (-\sqrt{3}a,0).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the distance between }P(x_1,y_1)\text{ and }Q(x_2,y_2)\text{ when}
\displaystyle \text{(i) }PQ\text{ is parallel to the }y\text{-axis}\qquad\text{(ii) }PQ\text{ is parallel to the }x\text{-axis}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(x_1,y_1)\text{ and }Q(x_2,y_2).
\displaystyle \text{We know }PQ=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.
\displaystyle \text{(i) When }PQ\text{ is parallel to the }y\text{-axis, }x_1=x_2.
\displaystyle \therefore PQ=\sqrt{(y_1-y_2)^2}=|y_1-y_2|.
\displaystyle \text{(ii) When }PQ\text{ is parallel to the }x\text{-axis, }y_1=y_2.
\displaystyle \therefore PQ=\sqrt{(x_1-x_2)^2}=|x_1-x_2|.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find a point on the }x\text{-axis which is equidistant from the points}
\displaystyle (7,6)\text{ and }(3,4).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(7,6),\ B(3,4)\text{ and let }C(x,0)\text{ be the required point.}
\displaystyle \text{Since }C\text{ is equidistant from }A\text{ and }B,
\displaystyle AC=BC\Rightarrow AC^2=BC^2.
\displaystyle \Rightarrow (7-x)^2+(6-0)^2=(3-x)^2+(4-0)^2
\displaystyle \Rightarrow 49+x^2-14x+36=9+x^2-6x+16
\displaystyle \Rightarrow 85-14x=25-6x
\displaystyle \Rightarrow 60=8x
\displaystyle \Rightarrow x=\frac{15}{2}
\displaystyle \therefore \text{The required point is }\left(\frac{15}{2},0\right).
\displaystyle \\

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