\displaystyle \textbf{Fill in the Blanks}

\displaystyle \textbf{1. }\text{The distance of the point }(2,3)\text{ from }x\text{-axis is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Distance from }x\text{-axis}=|y|
\displaystyle =|3|=3\text{ units.}
\displaystyle \therefore \text{The distance is }3\text{ units.}
\displaystyle \\

\displaystyle \textbf{2. }\text{The distance of the point }(-4,7)\text{ from }y\text{-axis is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Distance from }y\text{-axis}=|x|
\displaystyle =|-4|=4\text{ units.}
\displaystyle \therefore \text{The distance is }4\text{ units.}
\displaystyle \\

\displaystyle \textbf{3. }\text{If the centroid of the triangle whose vertices are }(2,4),\,(3,a),\,(4,2)\text{ is}
\displaystyle (a,3),\text{ then }a=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Using the }x\text{-coordinate of the centroid,}
\displaystyle a=\frac{2+3+4}{3}=\frac{9}{3}=3
\displaystyle \therefore a=3.
\displaystyle \\

\displaystyle \textbf{4. }\text{The distance between the points }(1,0)\text{ and }(2,\cot\theta)\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle d=\sqrt{(2-1)^2+(\cot\theta-0)^2}
\displaystyle =\sqrt{1+\cot^2\theta}=\mathrm{cosec}\theta
\displaystyle \therefore \text{The distance is }\mathrm{cosec}\theta.
\displaystyle \\

\displaystyle \textbf{5. }\text{The distance between the points }(0,5)\text{ and }(5,0)\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle d=\sqrt{(5-0)^2+(0-5)^2}
\displaystyle =\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\text{ units.}
\displaystyle \therefore \text{The distance is }5\sqrt{2}\text{ units.}
\displaystyle \\

\displaystyle \textbf{6. }\text{If the distance between the points }(2,-2)\text{ and }(-1,x)\text{ is }5, \\ \text{ then the values of }x\text{ are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{(-1-2)^2+\{x-(-2)\}^2}=5
\displaystyle 9+(x+2)^2=25\Rightarrow(x+2)^2=16
\displaystyle x+2=\pm4\Rightarrow x=2\text{ or }x=-6
\displaystyle \therefore \text{The values of }x\text{ are }2,\,-6.
\displaystyle \\

\displaystyle \textbf{7. }\text{The }x\text{-axis divides the line segment joining the points }(-4,-6) \\ \text{ and }(-1,7)\text{ in the ratio }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-axis divide the line segment in the ratio }m:n.
\displaystyle \text{Since the point of division lies on the }x\text{-axis, its }y\text{-coordinate is }0.
\displaystyle 0=\frac{7m-6n}{m+n}\Rightarrow7m=6n
\displaystyle \therefore m:n=6:7.
\displaystyle \therefore \text{The required ratio is }6:7.
\displaystyle \\

\displaystyle \textbf{8. }\text{The quadrant in which the point dividing the line segment joining the points }
\displaystyle (7,-6)\text{ and }(3,4)\text{ internally in the ratio }1:2\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle P=\left(\frac{1(3)+2(7)}{1+2},\frac{1(4)+2(-6)}{1+2}\right)
\displaystyle =\left(\frac{17}{3},-\frac{8}{3}\right)
\displaystyle \text{Since }x>0\text{ and }y<0,\text{ the point lies in the IV quadrant.}
\displaystyle \therefore \text{The required quadrant is IV quadrant.}
\displaystyle \\

\displaystyle \textbf{9. }\text{If the distance between the points }(4,p)\text{ and }(1,0)\text{ is }5,\text{ then the values of } \\ p\text{ are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{(4-1)^2+(p-0)^2}=5
\displaystyle 9+p^2=25\Rightarrow p^2=16
\displaystyle p=\pm4
\displaystyle \therefore \text{The values of }p\text{ are }4,\,-4.
\displaystyle \\

\displaystyle \textbf{10. }\text{If }P\left(\frac{a}{3},4\right)\text{ is the mid-point of the line segment joining the points }
\displaystyle Q(-6,5)\text{ and }R(-2,3),\text{ then the value of }a\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \frac{a}{3}=\frac{-6+(-2)}{2}=\frac{-8}{2}=-4
\displaystyle a=-12
\displaystyle \therefore \text{The value of }a\text{ is }-12.
\displaystyle \\

\displaystyle \textbf{11. }\text{The values of }y\text{ for which the point }(2,-4)\text{ is equidistant from the points }
\displaystyle (3,8)\text{ and }(-10,y)\text{ are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }(2,-4)\text{ is equidistant from }(3,8)\text{ and }(-10,y),
\displaystyle (3-2)^2+(8+4)^2=(-10-2)^2+(y+4)^2
\displaystyle 1+144=144+(y+4)^2
\displaystyle (y+4)^2=1
\displaystyle y+4=\pm1
\displaystyle \therefore y=-3\text{ or }y=-5.
\displaystyle \\

\displaystyle \textbf{12. }\text{The image of the point }(3,-5)\text{ in the }x\text{-axis has the coordinates }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For reflection in the }x\text{-axis, }(x,y)\rightarrow(x,-y).
\displaystyle \therefore (3,-5)\rightarrow(3,5).
\displaystyle \therefore \text{The coordinates of the image are }(3,5).
\displaystyle \\

\displaystyle \textbf{13. }\text{The coordinates of the image of the point }(-4,5)\text{ in }y\text{-axis are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For reflection in the }y\text{-axis, }(x,y)\rightarrow(-x,y).
\displaystyle \therefore (-4,5)\rightarrow(4,5).
\displaystyle \therefore \text{The coordinates of the image are }(4,5).
\displaystyle \\

\displaystyle \textbf{14. }\text{The }x\text{-coordinate of the point lying on the perpendicular bisector of the}
\displaystyle \text{line segment joining the points }A(-2,-5)\text{ and }B(2,5)\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mid-point of }AB=\left(\frac{-2+2}{2},\frac{-5+5}{2}\right)=(0,0).
\displaystyle \text{The perpendicular bisector of }AB\text{ passes through the mid-point }(0,0).
\displaystyle \therefore \text{The }x\text{-coordinate is }0.
\displaystyle \\

\displaystyle \textbf{15. }\text{If the distance between the points }A(-3,-14)\text{ and }B(a,-5)\text{ is } \\ 9\text{ units, then }a=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{\{a-(-3)\}^2+\{-5-(-14)\}^2}=9
\displaystyle (a+3)^2+81=81
\displaystyle (a+3)^2=0\Rightarrow a=-3
\displaystyle \therefore \text{The value of }a\text{ is }-3.
\displaystyle \\

\displaystyle \textbf{16. }\text{The ratio in which the point }P\left(\frac{3}{4},\frac{5}{12}\right)\text{ divides the line segment joining the points}
\displaystyle A\left(\frac{1}{2},\frac{3}{2}\right)\text{ and }B(2,-5)\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ divide }AB\text{ internally in the ratio }m:n.
\displaystyle \frac{3}{4}=\frac{2m+\frac{1}{2}n}{m+n}
\displaystyle 3(m+n)=8m+2n
\displaystyle n=5m
\displaystyle \therefore m:n=1:5.
\displaystyle \therefore \text{The required ratio is }1:5.
\displaystyle \\

\displaystyle \textbf{17. }\text{The number of points on }x\text{-axis which are at a distance of }2\sqrt{5}\text{ from the point } \\ (7,-4)\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the point on }x\text{-axis be }P(x,0).
\displaystyle \sqrt{(x-7)^2+(0+4)^2}=2\sqrt{5}
\displaystyle (x-7)^2+16=20
\displaystyle (x-7)^2=4\Rightarrow x=5\text{ or }9
\displaystyle \therefore \text{There are }2\text{ such points on the }x\text{-axis.}
\displaystyle \\

\displaystyle \textbf{18. }\text{If }(2,-2)\text{ and }(5,2)\text{ are opposite vertices of a square, then the length} \\ \text{of the side of the square is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Length of diagonal}=\sqrt{(5-2)^2+\{2-(-2)\}^2}
\displaystyle =\sqrt{9+16}=5\text{ units.}
\displaystyle \text{If the side of the square is }s,\text{ then diagonal}=s\sqrt{2}.
\displaystyle s=\frac{5}{\sqrt{2}}=\frac{5\sqrt{2}}{2}\text{ units.}
\displaystyle \therefore \text{The length of the side is }\frac{5\sqrt{2}}{2}\text{ units.}
\displaystyle \\

\displaystyle \textbf{19. }\text{The coordinates of the point which is equidistant from the vertices of the triangle}
\displaystyle \text{formed by the points }O(0,0),\,A(a,0)\text{ and }B(0,b)\text{ are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \triangle OAB\text{ is right-angled at }O.
\displaystyle \text{The point equidistant from all three vertices is the mid-point of hypotenuse }AB.
\displaystyle \text{Mid-point of }AB=\left(\frac{a+0}{2},\frac{0+b}{2}\right)
\displaystyle =\left(\frac{a}{2},\frac{b}{2}\right).
\displaystyle \therefore \text{The required point is }\left(\frac{a}{2},\frac{b}{2}\right).
\displaystyle \\

\displaystyle \textbf{20. }\text{If the distance of the point }(4,a)\text{ from }x\text{-axis is half its distance from } \\ y\text{-axis, then }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Distance from }x\text{-axis}=|a|\text{ and distance from }y\text{-axis}=|4|=4.
\displaystyle |a|=\frac{1}{2}\times4=2
\displaystyle \therefore a=\pm2.
\displaystyle \\

\displaystyle \textbf{21. }\text{The coordinates of the point equidistant from the vertices }O(0,0),\,A(6,0)\text{ and }
\displaystyle B(0,8)\text{ of }\triangle OAB\text{ are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \triangle OAB\text{ is right-angled at }O.
\displaystyle \text{The point equidistant from all three vertices is the mid-point of hypotenuse }AB.
\displaystyle \text{Mid-point of }AB=\left(\frac{6+0}{2},\frac{0+8}{2}\right)=(3,4).
\displaystyle \therefore \text{The required point is }(3,4).
\displaystyle \\

\displaystyle \textbf{22. }\text{If the centroid of the triangle formed by the points }(a,b),\,(1,a)\text{ and }(b,1)
\displaystyle \text{is at the origin, then }\frac{a^3+b^3+1}{ab}=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the centroid is at the origin,}
\displaystyle \left(\frac{a+b+1}{3},\frac{a+b+1}{3}\right)=(0,0)
\displaystyle \therefore a+b+1=0.
\displaystyle \text{Using }a^3+b^3+1-3ab=(a+b+1)(a^2+b^2+1-ab-a-b),
\displaystyle a^3+b^3+1=3ab
\displaystyle \therefore \frac{a^3+b^3+1}{ab}=3.
\displaystyle \\

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