\displaystyle \textbf{Question 1. }L\text{ is a variable line such that the algebraic sum of the distances of the points }
\displaystyle (1,1),(2,0)\text{ and }(0,2) \text{ from the line is equal to zero. The line }L\text{ will always pass through}
\displaystyle \text{(a) }(1,1)\qquad \text{(b) }(2,1)\qquad \text{(c) }(1,2)\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line be }ax+by+c=0
\displaystyle \text{Algebraic distance of }(x_1,y_1)\text{ from the line is }\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}
\displaystyle \frac{a+b+c}{\sqrt{a^2+b^2}}+\frac{2a+c}{\sqrt{a^2+b^2}}+\frac{2b+c}{\sqrt{a^2+b^2}}=0
\displaystyle a+b+c+2a+c+2b+c=0
\displaystyle 3a+3b+3c=0
\displaystyle a+b+c=0
\displaystyle \therefore \text{The line always passes through }(1,1)
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 2. }\text{The acute angle between the medians drawn from the acute angles of} \\ \text{a right angled isosceles triangle is}
\displaystyle \text{(a) }\cos^{-1}\left(\frac{2}{3}\right)\qquad \text{(b) }\cos^{-1}\left(\frac{3}{4}\right)\qquad \text{(c) }\cos^{-1}\left(\frac{4}{5}\right)\qquad \text{(d) }\cos^{-1}\left(\frac{5}{6}\right)
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertices of the triangle be }(0,0),(a,0)\text{ and }(0,a)
\displaystyle \text{Median from }(a,0)\text{ has direction vector }\left(-a,\frac{a}{2}\right)
\displaystyle \text{Median from }(0,a)\text{ has direction vector }\left(\frac{a}{2},-a\right)
\displaystyle \cos\theta=\frac{-a\cdot\frac{a}{2}+\frac{a}{2}(-a)}{\sqrt{a^2+\frac{a^2}{4}}\sqrt{\frac{a^2}{4}+a^2}}
\displaystyle =\frac{-a^2}{\frac{5a^2}{4}}=-\frac{4}{5}
\displaystyle \text{Hence the acute angle has }\cos\theta=\frac{4}{5}
\displaystyle \therefore \theta=\cos^{-1}\left(\frac{4}{5}\right)
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 3. }\text{The distance between the orthocentre and circumcentre of the triangle} \\ \text{with vertices }(1,2),(2,1)   \text{ and }\left(\frac{3+\sqrt3}{2},\frac{3+\sqrt3}{2}\right)\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }\sqrt2\qquad \text{(c) }3+\sqrt3\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The given triangle is equilateral.}
\displaystyle \text{In an equilateral triangle, orthocentre and circumcentre coincide.}
\displaystyle \therefore \text{Distance between orthocentre and circumcentre}=0
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 4. }\text{The equation of the straight line which passes through the point }
\displaystyle (-4,3)\text{ such that the} \text{portion of the line between the axes is divided internally} \\ \text{by the point in the ratio }5:3\text{ is}
\displaystyle \text{(a) }9x-20y+96=0\qquad \text{(b) }9x+20y=24\qquad \\ \text{(c) }20x+9y+53=0\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the intercepts be }A\text{ and }B
\displaystyle \text{So the intercepts are }(A,0)\text{ and }(0,B)
\displaystyle (-4,3)\text{ divides these points internally in the ratio }5:3
\displaystyle -4=\frac{3A}{8},\quad 3=\frac{5B}{8}
\displaystyle A=-\frac{32}{3},\quad B=\frac{24}{5}
\displaystyle \text{Equation of the line is }\frac{x}{A}+\frac{y}{B}=1
\displaystyle \frac{x}{-\frac{32}{3}}+\frac{y}{\frac{24}{5}}=1
\displaystyle -\frac{3x}{32}+\frac{5y}{24}=1
\displaystyle -9x+20y=96
\displaystyle \therefore 9x-20y+96=0
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 5. }\text{The point which divides the join of }(1,2)\text{ and }(3,4)\text{ externally in the} \\ \text{ratio }1:1
\displaystyle \text{(a) lies in the III quadrant}\qquad \text{(b) lies in the II quadrant}
\displaystyle \text{(c) lies in the I quadrant}\qquad \text{(d) cannot be found}
\displaystyle \text{Answer:}
\displaystyle \text{For external division in the ratio }m:n,
\displaystyle P=\left(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n}\right)
\displaystyle \text{Here }m:n=1:1
\displaystyle \therefore m-n=1-1=0
\displaystyle \text{Hence the point cannot be found.}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 6. }\text{A line passes through the point }(2,2)\text{ and is perpendicular to} \\ \text{the line }3x+y=3.   \text{ Its }y\text{-intercept is}
\displaystyle \text{(a) }\frac{1}{3}\qquad \text{(b) }\frac{2}{3}\qquad \text{(c) }1\qquad \text{(d) }\frac{4}{3}
\displaystyle \text{Answer:}
\displaystyle 3x+y=3
\displaystyle \Rightarrow y=-3x+3
\displaystyle \text{Slope of given line}=-3
\displaystyle \text{Slope of perpendicular line}=\frac{1}{3}
\displaystyle \text{Equation of line through }(2,2)\text{ is}
\displaystyle y-2=\frac{1}{3}(x-2)
\displaystyle 3y-6=x-2
\displaystyle x-3y+4=0
\displaystyle \text{Putting }x=0,\quad -3y+4=0
\displaystyle y=\frac{4}{3}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 7. }\text{If the lines }ax+12y+1=0,\ bx+13y+1=0\text{ and } \\ cx+14y+1=0\text{ are concurrent, then }a,b,c\text{ are in}
\displaystyle \text{(a) H.P.}\qquad \text{(b) G.P.}\qquad \text{(c) A.P.}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For concurrency,}
\displaystyle \begin{vmatrix}a&12&1\\ b&13&1\\ c&14&1\end{vmatrix}=0
\displaystyle a(13-14)-12(b-c)+(14b-13c)=0
\displaystyle -a-12b+12c+14b-13c=0
\displaystyle -a+2b-c=0
\displaystyle 2b=a+c
\displaystyle \therefore a,b,c\text{ are in A.P.}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 8. }\text{The number of real values of }\lambda\text{ for which the lines } \\ x-2y+3=0,\ \lambda x+3y+1=0   \text{ and }4x-\lambda y+2=0\text{ are concurrent is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) Infinite}
\displaystyle \text{Answer:}
\displaystyle \text{For concurrency,}
\displaystyle \begin{vmatrix}1&-2&3\\ \lambda&3&1\\ 4&-\lambda&2\end{vmatrix}=0
\displaystyle 1(6+\lambda)-(-2)(2\lambda-4)+3(-\lambda^2-12)=0
\displaystyle 6+\lambda+4\lambda-8-3\lambda^2-36=0
\displaystyle -3\lambda^2+5\lambda-38=0
\displaystyle \Rightarrow 3\lambda^2-5\lambda+38=0
\displaystyle D=(-5)^2-4(3)(38)
\displaystyle =25-456=-431<0
\displaystyle \therefore \text{There is no real value of }\lambda
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 9. }\text{The equations of the sides }AB,BC\text{ and }CA\text{ of }\triangle ABC\text{ are } \\ y-x=2,\ x+2y=1   \text{ and }3x+y+5=0\text{ respectively. The equation of the} \\ \text{altitude through }B\text{ is}
\displaystyle \text{(a) }x-3y+1=0\qquad \text{(b) }x-3y+4=0\qquad \text{(c) }3x-y+2=0\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle AB:\ y-x=2\Rightarrow y=x+2
\displaystyle BC:\ x+2y=1
\displaystyle \text{Point }B\text{ is the intersection of }AB\text{ and }BC
\displaystyle x+2(x+2)=1
\displaystyle 3x+4=1
\displaystyle x=-1,\quad y=1
\displaystyle \therefore B=(-1,1)
\displaystyle CA:\ 3x+y+5=0\Rightarrow y=-3x-5
\displaystyle \text{Slope of }CA=-3
\displaystyle \text{Slope of altitude through }B=\frac{1}{3}
\displaystyle y-1=\frac{1}{3}(x+1)
\displaystyle 3y-3=x+1
\displaystyle x-3y+4=0
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 10. }\text{If }p_1\text{ and }p_2\text{ are the lengths of the perpendiculars from the origin} \\ \text{in upon the lines }x\sec\theta+y\mathrm{cosec}\theta=a   \text{ and }x\cos\theta-y\sin\theta=a\cos2\theta\text{ respectively, then}
\displaystyle \text{(a) }4p_1^2+p_2^2=a^2\qquad \text{(b) }p_1^2+4p_2^2=a^2\qquad \text{(c) }p_1^2+p_2^2=a^2\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle p_1=\frac{|a|}{\sqrt{\sec^2\theta+\mathrm{cosec}^2\theta}}
\displaystyle =\frac{|a|}{\sqrt{\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}}}
\displaystyle =|a|\sin\theta\cos\theta
\displaystyle p_1^2=a^2\sin^2\theta\cos^2\theta
\displaystyle p_2=\frac{|a\cos2\theta|}{\sqrt{\cos^2\theta+\sin^2\theta}}
\displaystyle =|a\cos2\theta|
\displaystyle p_2^2=a^2\cos^22\theta
\displaystyle 4p_1^2+p_2^2=4a^2\sin^2\theta\cos^2\theta+a^2\cos^22\theta
\displaystyle =a^2\sin^22\theta+a^2\cos^22\theta
\displaystyle =a^2
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 11. }\text{Area of the triangle formed by the points } \\ ((a+3)(a+4),a+3),   ((a+2)(a+3),a+2)\text{ and }((a+1)(a+2),a+1)\text{ is}
\displaystyle \text{(a) }25a^2\qquad \text{(b) }5a^2\qquad \text{(c) }24a^2\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=a+3,\ a+2,\ a+1
\displaystyle \text{Then the given points are of the form }(t(t+1),t)
\displaystyle \text{So the points are }((a+3)(a+4),a+3),((a+2)(a+3),a+2),((a+1)(a+2),a+1)
\displaystyle \text{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|(a+3)(a+4)(1)+(a+2)(a+3)(-2)+(a+1)(a+2)(1)\right|
\displaystyle =\frac{1}{2}\left|(a^2+7a+12)-2(a^2+5a+6)+(a^2+3a+2)\right|
\displaystyle =\frac{1}{2}|2|
\displaystyle =1
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 12. }\text{If }a+b+c=0,\text{ then the family of lines } \\ 3ax+by+2c=0\text{ pass through fixed point}
\displaystyle \text{(a) }(2,2/3)\qquad \text{(b) }(2/3,2)\qquad \text{(c) }(-2,2/3)\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle a+b+c=0
\displaystyle \Rightarrow c=-a-b
\displaystyle 3ax+by+2c=0
\displaystyle 3ax+by+2(-a-b)=0
\displaystyle a(3x-2)+b(y-2)=0
\displaystyle \text{For a fixed point, }3x-2=0\text{ and }y-2=0
\displaystyle x=\frac{2}{3},\quad y=2
\displaystyle \therefore \text{Fixed point}=\left(\frac{2}{3},2\right)
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 13. }\text{The line segment joining the points }(-3,-4)\text{ and }(1,-2) \\ \text{is divided by }y\text{-axis in the ratio}
\displaystyle \text{(a) }1:3\qquad \text{(b) }2:3\qquad \text{(c) }3:1\qquad \text{(d) }3:2
\displaystyle \text{Answer:}
\displaystyle \text{Let }y\text{-axis divide the segment in the ratio }m:n
\displaystyle \text{Since the point lies on }y\text{-axis, its }x\text{-coordinate is }0
\displaystyle 0=\frac{m(1)+n(-3)}{m+n}
\displaystyle m-3n=0
\displaystyle m:n=3:1
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 14. }\text{The area of a triangle with vertices at }(-4,-1),(1,2)\text{ and }(4,-3)\text{ is}
\displaystyle \text{(a) }17\qquad \text{(b) }16\qquad \text{(c) }15\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{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|-4(2+3)+1(-3+1)+4(-1-2)\right|
\displaystyle =\frac{1}{2}\left|-20-2-12\right|
\displaystyle =17
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 15. }\text{The line segment joining the points }(1,2)\text{ and }(-2,1) \\ \text{ is divided by the line }3x+4y=7   \text{in the ratio}
\displaystyle \text{(a) }3:4\qquad \text{(b) }4:3\qquad \text{(c) }9:4\qquad \text{(d) }4:9
\displaystyle \text{Answer:}
\displaystyle \text{Let the line divide the segment in the ratio }m:n
\displaystyle P=\left(\frac{m(-2)+n(1)}{m+n},\frac{m(1)+n(2)}{m+n}\right)
\displaystyle P=\left(\frac{-2m+n}{m+n},\frac{m+2n}{m+n}\right)
\displaystyle \text{Since }P\text{ lies on }3x+4y=7,
\displaystyle 3\left(\frac{-2m+n}{m+n}\right)+4\left(\frac{m+2n}{m+n}\right)=7
\displaystyle -6m+3n+4m+8n=7m+7n
\displaystyle -2m+11n=7m+7n
\displaystyle 4n=9m
\displaystyle m:n=4:9
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 16. }\text{If the point }(5,2)\text{ bisects the intercept of a line between the axes,} \\ \text{then its equation is}
\displaystyle \text{(a) }5x+2y=20\qquad \text{(b) }2x+5y=20\qquad \\ \text{(c) }5x-2y=20\qquad \text{(d) }2x-5y=20
\displaystyle \text{Answer:}
\displaystyle \text{Let the line cut the axes at }(a,0)\text{ and }(0,b)
\displaystyle \text{Midpoint}=\left(\frac{a}{2},\frac{b}{2}\right)=(5,2)
\displaystyle \therefore a=10,\quad b=4
\displaystyle \text{Equation of the line is }\frac{x}{10}+\frac{y}{4}=1
\displaystyle 2x+5y=20
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 17. }A(6,3),B(-3,5),C(4,-2)\text{ and }D(x,3x)\text{ are four points. If }
\displaystyle \Delta DBC:\Delta ABC=1:2, \text{ then }x\text{ is equal to}
\displaystyle \text{(a) }\frac{11}{8}\qquad \text{(b) }\frac{8}{11}\qquad \text{(c) }3\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle B(-3,5),\ C(4,-2)
\displaystyle \text{Equation of }BC\text{ is }x+y-2=0
\displaystyle \Delta DBC:\Delta ABC=d(D,BC):d(A,BC)
\displaystyle \frac{\Delta DBC}{\Delta ABC}=\frac{|x+3x-2|}{|6+3-2|}
\displaystyle \frac{1}{2}=\frac{|4x-2|}{7}
\displaystyle |4x-2|=\frac{7}{2}
\displaystyle 4x-2=\frac{7}{2}
\displaystyle 4x=\frac{11}{2}
\displaystyle x=\frac{11}{8}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 18. }\text{If }p\text{ be the length of the perpendicular from the origin on the line } \\ \frac{x}{a}+\frac{y}{b}=1,\text{ then}
\displaystyle \text{(a) }p^2=a^2+b^2\qquad \text{(b) }p^2=\frac{1}{a^2}+\frac{1}{b^2}\qquad \text{(c) }\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{x}{a}+\frac{y}{b}=1
\displaystyle \Rightarrow bx+ay-ab=0
\displaystyle p=\frac{|ab|}{\sqrt{a^2+b^2}}
\displaystyle p^2=\frac{a^2b^2}{a^2+b^2}
\displaystyle \frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}
\displaystyle \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 19. }\text{The equation of the line passing through }(1,5)\text{ and perpendicular} \\ \text{to the line }3x-5y+7=0\text{ is}
\displaystyle \text{(a) }5x+3y-20=0\qquad \text{(b) }3x-5y+7=0\qquad \\ \text{(c) }3x-5y+6=0\qquad \text{(d) }5x+3y+7=0
\displaystyle \text{Answer:}
\displaystyle 3x-5y+7=0
\displaystyle \Rightarrow y=\frac{3}{5}x+\frac{7}{5}
\displaystyle \text{Slope of given line}=\frac{3}{5}
\displaystyle \text{Slope of perpendicular line}=-\frac{5}{3}
\displaystyle y-5=-\frac{5}{3}(x-1)
\displaystyle 3y-15=-5x+5
\displaystyle 5x+3y-20=0
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 20. }\text{The figure formed by the lines }ax\pm by\pm c=0\text{ is}
\displaystyle \text{(a) a rectangle}\qquad \text{(b) a square}\qquad \text{(c) a rhombus}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle ax+by+c=0,\quad ax+by-c=0
\displaystyle \text{These two lines are parallel.}
\displaystyle ax-by+c=0,\quad ax-by-c=0
\displaystyle \text{These two lines are also parallel.}
\displaystyle \text{Distance between each pair of parallel lines is }\frac{2|c|}{\sqrt{a^2+b^2}}
\displaystyle \text{Since the distances are equal, the figure formed is a rhombus.}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 21. }\text{Two vertices of a triangle are }(-2,-1)\text{ and }(1,2)\text{ and third}
\displaystyle \text{vertex lies on the line }x+y=5. \text{ If the area of the triangle is }3\text{ square units, then the} \\ \text{third vertex is}
\displaystyle \text{(a) }(0,5)\text{ or }(4,1)\qquad \text{(b) }(5,0)\text{ or }(1,4)
\displaystyle \text{(c) }(5,0)\text{ or }(4,1)\qquad \text{(d) }(0,5)\text{ or }(1,4)
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }(x,y)
\displaystyle x+y=5
\displaystyle \text{Area}=\frac{1}{2}\left|-2(2-y)+1(y+1)+x(-1-2)\right|
\displaystyle 3=\frac{1}{2}\left|-4+2y+y+1-3x\right|
\displaystyle 6=\left|3y-3x-3\right|
\displaystyle 2=\left|y-x-1\right|
\displaystyle y=5-x
\displaystyle 2=\left|5-x-x-1\right|
\displaystyle 2=\left|4-2x\right|
\displaystyle 4-2x=\pm2
\displaystyle x=1\text{ or }3
\displaystyle y=4\text{ or }2
\displaystyle \text{This gives }(1,4)\text{ or }(3,2).
\displaystyle \text{Hence, to get option (b), change the third vertex line to }x+y=5\text{ and first two vertices to } \\ (-2,-1),(3,2)\text{ is not sufficient.}
\\

\displaystyle \textbf{Question 22. }\text{The inclination of the straight line passing through the point }(-3,6)
\displaystyle \text{ and the mid-point of the line joining }\text{point }(4,-5)\text{ and }(-2,9)\text{ is}
\displaystyle \text{(a) }\frac{\pi}{4}\qquad \text{(b) }\frac{\pi}{6}\qquad \text{(c) }\frac{\pi}{3}\qquad \text{(d) }\frac{3\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Mid-point of }(4,-5)\text{ and }(-2,9)\text{ is}
\displaystyle \left(\frac{4-2}{2},\frac{-5+9}{2}\right)=(1,2)
\displaystyle \text{Slope of line through }(-3,6)\text{ and }(1,2)\text{ is}
\displaystyle m=\frac{2-6}{1+3}=-1
\displaystyle \tan\theta=-1
\displaystyle \text{Since }0<\theta<\pi,\quad \theta=\frac{3\pi}{4}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 23. }\text{Distance between the lines }5x+3y-7=0\text{ and } \\ 15x+9y+14=0\text{ is}
\displaystyle \text{(a) }\frac{35}{\sqrt{34}}\qquad \text{(b) }\frac{1}{3\sqrt{34}}\qquad \text{(c) }\frac{35}{3\sqrt{34}}\qquad \text{(d) }\frac{35}{2\sqrt{34}}
\displaystyle \text{Answer:}
\displaystyle 15x+9y+14=0
\displaystyle \Rightarrow 5x+3y+\frac{14}{3}=0
\displaystyle \text{Distance}=\frac{\left|-7-\frac{14}{3}\right|}{\sqrt{5^2+3^2}}
\displaystyle =\frac{\left|-\frac{35}{3}\right|}{\sqrt{34}}
\displaystyle =\frac{35}{3\sqrt{34}}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 24. }\text{The angle between the lines }2x-y+3=0\text{ and } \\ x+2y+3=0\text{ is}
\displaystyle \text{(a) }90^\circ\qquad \text{(b) }60^\circ\qquad \text{(c) }45^\circ\qquad \text{(d) }30^\circ
\displaystyle \text{Answer:}
\displaystyle 2x-y+3=0\Rightarrow y=2x+3
\displaystyle \therefore m_1=2
\displaystyle x+2y+3=0\Rightarrow y=-\frac{1}{2}x-\frac{3}{2}
\displaystyle \therefore m_2=-\frac{1}{2}
\displaystyle m_1m_2=2\left(-\frac{1}{2}\right)=-1
\displaystyle \therefore \text{The lines are perpendicular.}
\displaystyle \therefore \text{Angle}=90^\circ
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 25. }\text{The value of }\lambda\text{ for which the lines }3x+4y=5,\ 5x+4y=4
\displaystyle \text{ and }\lambda x+4y=6 \text{ meet at a point is}
\displaystyle \text{(a) }2\qquad \text{(b) }1\qquad \text{(c) }4\qquad \text{(d) }3
\displaystyle \text{Answer:}
\displaystyle 3x+4y=5\qquad \text{and}\qquad 5x+4y=4
\displaystyle \text{Subtracting, }-2x=1
\displaystyle x=-\frac{1}{2}
\displaystyle 3\left(-\frac{1}{2}\right)+4y=5
\displaystyle 4y=\frac{13}{2}
\displaystyle y=\frac{13}{8}
\displaystyle \text{Since }\lambda x+4y=6\text{ also passes through this point,}
\displaystyle \lambda\left(-\frac{1}{2}\right)+4\left(\frac{13}{8}\right)=6
\displaystyle -\frac{\lambda}{2}+\frac{13}{2}=6
\displaystyle -\lambda+13=12
\displaystyle \lambda=1
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 26. }\text{Three vertices of a parallelogram taken in order are }
\displaystyle (-1,-6),(2,-5)\text{ and }(7,2). \text{ The fourth vertex is}
\displaystyle \text{(a) }(1,4)\qquad \text{(b) }(4,1)\qquad \text{(c) }(1,1)\qquad \text{(d) }(4,4)
\displaystyle \text{Answer:}
\displaystyle A=(-1,-6),\quad B=(2,-5),\quad C=(7,2)
\displaystyle \text{Fourth vertex }D=A+C-B
\displaystyle D=(-1,-6)+(7,2)-(2,-5)
\displaystyle D=(4,1)
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 27. }\text{The centroid of a triangle is }(2,7)\text{ and two of its vertices are }
\displaystyle (4,8)\text{ and }(-2,6). \text{ The third vertex is}
\displaystyle \text{(a) }(0,0)\qquad \text{(b) }(4,7)\qquad \text{(c) }(7,4)\qquad \text{(d) }(7,7)
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }(x,y)
\displaystyle \left(\frac{4-2+x}{3},\frac{8+6+y}{3}\right)=(2,7)
\displaystyle \frac{x+2}{3}=2,\quad \frac{y+14}{3}=7
\displaystyle x=4,\quad y=7
\displaystyle \therefore \text{Third vertex}=(4,7)
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 28. }\text{If the lines }x+q=0,\ y-2=0\text{ and }3x+2y+5=0\text{ are concurrent,} \\ \text{then the value of }q\text{ will be}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }5
\displaystyle \text{Answer:}
\displaystyle x+q=0\Rightarrow x=-q
\displaystyle y-2=0\Rightarrow y=2
\displaystyle \text{Since }3x+2y+5=0\text{ also passes through }(-q,2),
\displaystyle 3(-q)+2(2)+5=0
\displaystyle -3q+9=0
\displaystyle q=3
\displaystyle \therefore \text{Correct option is (c)}
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\displaystyle \textbf{Question 29. }\text{The medians }AD\text{ and }BE\text{ of a triangle with vertices }A(0,b),B(0,0)
\displaystyle \text{ and }C(a,0)\text{ are perpendicular to each other, if}
\displaystyle \text{(a) }a=\frac{b}{2}\qquad \text{(b) }b=\frac{a}{2}\qquad \text{(c) }ab=1\qquad \text{(d) }a=\pm\sqrt2 b
\displaystyle \text{Answer:}
\displaystyle D\text{ is the midpoint of }BC=\left(\frac{a}{2},0\right)
\displaystyle E\text{ is the midpoint of }AC=\left(\frac{a}{2},\frac{b}{2}\right)
\displaystyle \text{Slope of }AD=\frac{0-b}{\frac{a}{2}-0}=-\frac{2b}{a}
\displaystyle \text{Slope of }BE=\frac{\frac{b}{2}-0}{\frac{a}{2}-0}=\frac{b}{a}
\displaystyle \text{Since }AD\perp BE,
\displaystyle \left(-\frac{2b}{a}\right)\left(\frac{b}{a}\right)=-1
\displaystyle \frac{2b^2}{a^2}=1
\displaystyle a^2=2b^2
\displaystyle a=\pm\sqrt2 b
\displaystyle \therefore \text{Correct option is (d)}
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\displaystyle \textbf{Question 30. }\text{The equation of the line with slope }-\frac{3}{2}\text{ and which is concurrent with the lines }
\displaystyle 4x+3y-7=0 \text{ and }8x+5y-1=0\text{ is}
\displaystyle \text{(a) }3x+2y-63=0\qquad \text{(b) }3x+2y-2=0\qquad \text{(c) }2y-3x-2=0\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 4x+3y-7=0\qquad \text{and}\qquad 8x+5y-1=0
\displaystyle \text{Multiplying the first equation by }2,
\displaystyle 8x+6y-14=0
\displaystyle \text{Subtracting }8x+5y-1=0\text{ from it,}
\displaystyle y-13=0
\displaystyle y=13
\displaystyle 4x+3(13)-7=0
\displaystyle 4x+32=0
\displaystyle x=-8
\displaystyle \text{So, the point of concurrence is }(-8,13)
\displaystyle \text{Required line has slope }-\frac{3}{2}
\displaystyle y-13=-\frac{3}{2}(x+8)
\displaystyle 2y-26=-3x-24
\displaystyle 3x+2y-2=0
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 31. }\text{The vertices of a triangle are }(6,0),(0,6)\text{ and }(6,6).\text{ The distance }
\displaystyle \text{between its circumcentre and centroid is}
\displaystyle \text{(a) }2\sqrt2\qquad \text{(b) }2\qquad \text{(c) }\sqrt2\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{The triangle is right-angled at }(6,6)
\displaystyle \therefore \text{Circumcentre is the midpoint of the hypotenuse joining }(6,0)\text{ and }(0,6)
\displaystyle O=\left(\frac{6+0}{2},\frac{0+6}{2}\right)=(3,3)
\displaystyle G=\left(\frac{6+0+6}{3},\frac{0+6+6}{3}\right)=(4,4)
\displaystyle OG=\sqrt{(4-3)^2+(4-3)^2}
\displaystyle =\sqrt2
\displaystyle \therefore \text{Correct option is (c)}
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\displaystyle \textbf{Question 32. }\text{A point equidistant from the lines }4x+3y+10=0, \\ 5x-12y+26=0\text{ and }7x+24y-50=0\text{ is}
\displaystyle \text{(a) }(1,-1)\qquad \text{(b) }(1,1)\qquad \text{(c) }(0,0)\qquad \text{(d) }(0,1)
\displaystyle \text{Answer:}
\displaystyle \text{Checking }(1,-1),
\displaystyle d_1=\frac{|4(1)+3(-1)+10|}{\sqrt{4^2+3^2}}=\frac{11}{5}
\displaystyle d_2=\frac{|5(1)-12(-1)+26|}{\sqrt{5^2+12^2}}=\frac{43}{13}
\displaystyle \text{Hence }(1,-1)\text{ is not equidistant.}
\displaystyle \text{Checking }(1,1),
\displaystyle d_1=\frac{|4(1)+3(1)+10|}{5}=\frac{17}{5}
\displaystyle d_2=\frac{|5(1)-12(1)+26|}{13}=\frac{19}{13}
\displaystyle \text{Hence }(1,1)\text{ is not equidistant.}
\displaystyle \text{Checking }(0,0),
\displaystyle d_1=\frac{|10|}{5}=2
\displaystyle d_2=\frac{|26|}{13}=2
\displaystyle d_3=\frac{|-50|}{\sqrt{7^2+24^2}}=\frac{50}{25}=2
\displaystyle \therefore (0,0)\text{ is equidistant from the three lines.}
\displaystyle \therefore \text{Correct option is (c)}
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\displaystyle \textbf{Question 33. }\text{The ratio in which the line }3x+4y+2=0\text{ divides the distance} \\ \text{between the lines }3x+4y+5=0  \text{ and }3x+4y-5=0\text{ is}
\displaystyle \text{(a) }1:2\qquad \text{(b) }3:7\qquad \text{(c) }2:3\qquad \text{(d) }2:5
\displaystyle \text{Answer:}
\displaystyle \text{All three lines are parallel.}
\displaystyle \text{Required ratio}=\left|2-5\right|:\left|2+5\right|
\displaystyle =3:7
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 34. }\text{The coordinates of the foot of the perpendicular from the point } \\ (2,3)\text{ on the line }x+y-11=0\text{ are}
\displaystyle \text{(a) }(-6,5)\qquad \text{(b) }(5,6)\qquad \text{(c) }(-5,6)\qquad \text{(d) }(6,5)
\displaystyle \text{Answer:}
\displaystyle \text{Foot of perpendicular from }(x_1,y_1)\text{ to }ax+by+c=0\text{ is}
\displaystyle \left(x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2},y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2}\right)
\displaystyle a=1,\ b=1,\ c=-11,\quad (x_1,y_1)=(2,3)
\displaystyle ax_1+by_1+c=2+3-11=-6
\displaystyle \text{Foot}=\left(2-\frac{1(-6)}{2},3-\frac{1(-6)}{2}\right)
\displaystyle =(5,6)
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 35. }\text{The reflection of the point }(4,-13)\text{ about the line } \\ 5x+y+6=0\text{ is}
\displaystyle \text{(a) }(-1,-14)\qquad \text{(b) }(3,4)\qquad \text{(c) }(0,0)\qquad \text{(d) }(1,2)
\displaystyle \text{Answer:}
\displaystyle \text{Image of }(x_1,y_1)\text{ in }ax+by+c=0\text{ is}
\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=5,\ b=1,\ c=6,\quad (x_1,y_1)=(4,-13)
\displaystyle ax_1+by_1+c=20-13+6=13
\displaystyle a^2+b^2=25+1=26
\displaystyle x'=4-\frac{2(5)(13)}{26}=-1
\displaystyle y'=-13-\frac{2(1)(13)}{26}=-14
\displaystyle \therefore \text{Reflection}=(-1,-14)
\displaystyle \therefore \text{Correct option is (a)}
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