\displaystyle \textbf{Question 1. }\text{Write an equation representing a pair of lines through the point }(a,b) \\ \text{and parallel to the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Lines through }(a,b)\text{ parallel to coordinate axes are }x=a\text{ and }y=b
\displaystyle \therefore (x-a)(y-b)=0
\\

\displaystyle \textbf{Question 2. }\text{Write the coordinates of the orthocentre of the triangle formed by the} \\ \text{lines }x^2-y^2=0\text{ and }x+6y=18.
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=0
\displaystyle (x-y)(x+y)=0
\displaystyle \therefore y=x\quad \text{and}\quad y=-x
\displaystyle \text{These two lines are perpendicular and meet at }(0,0)
\displaystyle \therefore \text{The triangle is right-angled at }(0,0)
\displaystyle \therefore \text{Orthocentre}=(0,0)
\\

\displaystyle \textbf{Question 3. }\text{If the centroid of a triangle formed by the points }(0,0),(\cos\theta,\sin\theta) \\ \text{and }(\sin\theta,-\cos\theta)\text{ lies on the line }y=2x,\text{ then write the value of }\tan\theta.
\displaystyle \text{Answer:}
\displaystyle G=\left(\frac{0+\cos\theta+\sin\theta}{3},\frac{0+\sin\theta-\cos\theta}{3}\right)
\displaystyle G=\left(\frac{\cos\theta+\sin\theta}{3},\frac{\sin\theta-\cos\theta}{3}\right)
\displaystyle \text{Since }G\text{ lies on }y=2x,
\displaystyle \frac{\sin\theta-\cos\theta}{3}=2\left(\frac{\cos\theta+\sin\theta}{3}\right)
\displaystyle \sin\theta-\cos\theta=2\cos\theta+2\sin\theta
\displaystyle -\sin\theta-3\cos\theta=0
\displaystyle \sin\theta=-3\cos\theta
\displaystyle \therefore \tan\theta=-3
\\

\displaystyle \textbf{Question 4. }\text{Write the value of }\theta\in\left(0,\frac{\pi}{2}\right)\text{ for which area of the triangle} \\ \text{formed by points }O(0,0),A(a\cos\theta,b\sin\theta)\text{ and }B(a\cos\theta,-b\sin\theta)\text{ is maximum.}
\displaystyle \text{Answer:}
\displaystyle \text{Area}=\frac{1}{2}\left|a\cos\theta(-b\sin\theta)-b\sin\theta(a\cos\theta)\right|
\displaystyle =\frac{1}{2}\left|-2ab\sin\theta\cos\theta\right|
\displaystyle =ab\sin\theta\cos\theta
\displaystyle =\frac{ab}{2}\sin2\theta
\displaystyle \text{Area is maximum when }\sin2\theta=1
\displaystyle 2\theta=\frac{\pi}{2}
\displaystyle \therefore \theta=\frac{\pi}{4}
\\

\displaystyle \textbf{Question 5. }\text{Write the distance between the lines }4x+3y-11=0\text{ and } \\ 8x+6y-15=0.
\displaystyle \text{Answer:}
\displaystyle 8x+6y-15=0
\displaystyle \Rightarrow 4x+3y-\frac{15}{2}=0
\displaystyle \text{Distance}=\frac{\left|-11+\frac{15}{2}\right|}{\sqrt{4^2+3^2}}
\displaystyle =\frac{\left|-\frac{7}{2}\right|}{5}
\displaystyle =\frac{7}{10}
\\

\displaystyle \textbf{Question 6. }\text{Write the coordinates of the orthocentre of the triangle formed} \\ \text{by the lines }xy=0\text{ and }x+y=1.
\displaystyle \text{Answer:}
\displaystyle xy=0
\displaystyle \Rightarrow x=0\quad \text{or}\quad y=0
\displaystyle \text{The triangle is formed by }x=0,\ y=0\text{ and }x+y=1
\displaystyle \text{Since }x=0\text{ and }y=0\text{ are perpendicular, the triangle is right-angled at }(0,0)
\displaystyle \therefore \text{Orthocentre}=(0,0)
\\

\displaystyle \textbf{Question 7. }\text{If the lines }x+ay+a=0,\ bx+y+b=0\text{ and }cx+cy+1=0\text{ are concurrent, then write the value of }2abc-ab-bc-ca.
\displaystyle \text{Answer:}
\displaystyle \text{For concurrency,}
\displaystyle \begin{vmatrix}1&a&a\\ b&1&b\\ c&c&1\end{vmatrix}=0
\displaystyle 1-ab-bc-ca+2abc=0
\displaystyle \therefore 2abc-ab-bc-ca=-1
\\

\displaystyle \textbf{Question 8. }\text{Write the area of the triangle formed by the coordinate axes and} \\ \text{the line }(\sec\theta-\tan\theta)x+(\sec\theta+\tan\theta)y=2.
\displaystyle \text{Answer:}
\displaystyle (\sec\theta-\tan\theta)x+(\sec\theta+\tan\theta)y=2
\displaystyle x\text{-intercept}=\frac{2}{\sec\theta-\tan\theta}
\displaystyle y\text{-intercept}=\frac{2}{\sec\theta+\tan\theta}
\displaystyle \text{Area}=\frac{1}{2}\cdot \frac{2}{\sec\theta-\tan\theta}\cdot \frac{2}{\sec\theta+\tan\theta}
\displaystyle =\frac{2}{\sec^2\theta-\tan^2\theta}
\displaystyle =\frac{2}{1}
\displaystyle =2
\\

\displaystyle \textbf{Question 9. }\text{If the diagonals of the quadrilateral formed by the lines }
\displaystyle l_1x+m_1y+n_1=0,\ l_2x+m_2y+n_2=0,l_1x+m_1y+n_1'=0\text{ and }l_2x+m_2y+n_2'=0 \text{ are perpendicular, then write the value of }l_1^2-l_2^2+m_1^2-m_2^2.
\displaystyle \text{Answer:}
\displaystyle \text{For perpendicular diagonals,}
\displaystyle l_1^2+m_1^2=l_2^2+m_2^2
\displaystyle \therefore l_1^2-l_2^2+m_1^2-m_2^2=0
\\

\displaystyle \textbf{Question 10. }\text{Write the coordinates of the image of the point }(3,8)\text{ in the line } \\ x+3y-7=0.
\displaystyle \text{Answer:}
\displaystyle \text{Image of }(x_1,y_1)\text{ in }ax+by+c=0\text{ is given by}
\displaystyle x'=x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2},\quad y'=y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2}
\displaystyle a=1,\ b=3,\ c=-7,\ x_1=3,\ y_1=8
\displaystyle ax_1+by_1+c=3+24-7=20
\displaystyle a^2+b^2=1^2+3^2=10
\displaystyle x'=3-\frac{2(1)(20)}{10}=-1
\displaystyle y'=8-\frac{2(3)(20)}{10}=-4
\displaystyle \therefore \text{Image}=(-1,-4)
\\

\displaystyle \textbf{Question 11. }\text{Write the integral values of }m\text{ for which the }x\text{-coordinate} \\ \text{of the point of intersection of the lines }y=mx+1 \text{and }3x+4y=9\text{ is an integer.}
\displaystyle \text{Answer:}
\displaystyle y=mx+1
\displaystyle 3x+4y=9
\displaystyle 3x+4(mx+1)=9
\displaystyle (4m+3)x=5
\displaystyle x=\frac{5}{4m+3}
\displaystyle \text{For }x\text{ to be an integer, }4m+3\text{ must divide }5
\displaystyle 4m+3=\pm1,\ \pm5
\displaystyle \text{Since }m\text{ is an integer, }4m+3=-1\text{ or }-5
\displaystyle \therefore m=-1,\ -2
\\

\displaystyle \textbf{Question 12. }\text{If }a\neq b\neq c,\text{ write the condition for which the equations }
\displaystyle (b-c)x+(c-a)y+(a-b)=0 \text{and }(b^3-c^3)x+(c^3-a^3)y+(a^3-b^3)=0 \text{ represent the same line.}
\displaystyle \text{Answer:}
\displaystyle \text{For the two equations to represent the same line,}
\displaystyle \frac{b^3-c^3}{b-c}=\frac{c^3-a^3}{c-a}=\frac{a^3-b^3}{a-b}
\displaystyle b^2+bc+c^2=c^2+ca+a^2=a^2+ab+b^2
\displaystyle b^2+bc+c^2=c^2+ca+a^2
\displaystyle \Rightarrow b^2+bc=ca+a^2
\displaystyle \Rightarrow (b-a)(a+b+c)=0
\displaystyle \text{Since }a\neq b,\quad a+b+c=0
\\

\displaystyle \textbf{Question 13. }\text{If }a,b,c\text{ are in G.P., write the area of the triangle formed by the line }
\displaystyle ax+by+c=0 \text{with the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle ax+by+c=0
\displaystyle x\text{-intercept}=-\frac{c}{a},\quad y\text{-intercept}=-\frac{c}{b}
\displaystyle \text{Area}=\frac{1}{2}\left|\frac{c^2}{ab}\right|
\displaystyle \text{Since }a,b,c\text{ are in G.P., }b^2=ac
\displaystyle \therefore c=\frac{b^2}{a}
\displaystyle \text{Area}=\frac{1}{2}\left|\frac{c^2}{ab}\right|
\displaystyle =\frac{1}{2}\left|\frac{c}{a}\cdot\frac{c}{b}\right|
\\

\displaystyle \textbf{Question 14. }\text{Write the area of the figure formed by the lines } \\ a|x|+b|y|+c=0.
\displaystyle \text{Answer:}
\displaystyle a|x|+b|y|+c=0
\displaystyle \Rightarrow a|x|+b|y|=-c
\displaystyle x\text{-intercepts are }\pm\frac{-c}{a}\text{ and }y\text{-intercepts are }\pm\frac{-c}{b}
\displaystyle \text{The figure formed is a rhombus.}
\displaystyle \text{Length of one diagonal}=\frac{2|c|}{|a|}
\displaystyle \text{Length of other diagonal}=\frac{2|c|}{|b|}
\displaystyle \text{Area}=\frac{1}{2}\cdot\frac{2|c|}{|a|}\cdot\frac{2|c|}{|b|}
\displaystyle =\frac{2c^2}{|ab|}
\\

\displaystyle \textbf{Question 15. }\text{Write the locus of a point the sum of whose distances from the} \\ \text{coordinate axes is unity.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the point be }(x,y)
\displaystyle \text{Distance from }x\text{-axis}=|y|
\displaystyle \text{Distance from }y\text{-axis}=|x|
\displaystyle \text{Given }|x|+|y|=1
\displaystyle \therefore \text{Required locus is }|x|+|y|=1
\\

\displaystyle \textbf{Question 16. }\text{If }a,b,c\text{ are in A.P., then the line }ax+by+c=0\text{ passes through}
\displaystyle \text{a fixed point. Write the coordinates of that point.}
\displaystyle \text{Answer:}
\displaystyle a,b,c\text{ are in A.P.}
\displaystyle \therefore 2b=a+c
\displaystyle \Rightarrow a-2b+c=0
\displaystyle \text{Comparing with }ax+by+c=0,
\displaystyle x=1,\quad y=-2
\displaystyle \therefore \text{Fixed point}=(1,-2)
\\

\displaystyle \textbf{Question 17. }\text{Write the equation of the line passing through the point }(1,-2)\text{ and} \\ \text{cutting off equal intercepts from the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Equation of line cutting equal intercepts from the axes is}
\displaystyle \frac{x}{a}+\frac{y}{a}=1
\displaystyle \Rightarrow x+y=a
\displaystyle \text{Since it passes through }(1,-2),
\displaystyle 1-2=a
\displaystyle \Rightarrow a=-1
\displaystyle \therefore x+y=-1
\displaystyle \Rightarrow x+y+1=0
\\

\displaystyle \textbf{Question 18. }\text{Find the locus of the mid-points of the portion of the line } \\ x\sin\theta+y\cos\theta=p\text{ intercepted between the axes.}
\displaystyle \text{Answer:}
\displaystyle x\sin\theta+y\cos\theta=p
\displaystyle x\text{-intercept}=\frac{p}{\sin\theta},\quad y\text{-intercept}=\frac{p}{\cos\theta}
\displaystyle \text{Let the midpoint be }(h,k)
\displaystyle h=\frac{p}{2\sin\theta},\quad k=\frac{p}{2\cos\theta}
\displaystyle \sin\theta=\frac{p}{2h},\quad \cos\theta=\frac{p}{2k}
\displaystyle \sin^2\theta+\cos^2\theta=1
\displaystyle \frac{p^2}{4h^2}+\frac{p^2}{4k^2}=1
\displaystyle \therefore \frac{p^2}{x^2}+\frac{p^2}{y^2}=4
\displaystyle \text{Required locus is }\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}
\\

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