14. Locus of the image of the point $(2,3)$ in the line $(x-2y+3)+\lambda (2x-3y+4)=0$ is

(a) $x^2+y^2-3x-4y-4=0$

(b) $2x^2+2y^2+2x+4y-7=0$

(c) $x^2+y^2-2x-4y+4=0$

(d) None

Show Answer

Solution: $\begin{array}{l}x-2y+3=0\\ 2x-3y+4=0\end{array}\}P(1,2)$

Let image be $\mathrm{B}(\mathrm{h},\mathrm{k})$ then $\mathrm{AP}=\mathrm{BP}$

$\Rightarrow (\mathrm{h}-1{)}^2+(\mathrm{k}-2{)}^2=1^2+1^2$

${\mathrm{h}}^2+{\mathrm{k}}^2-2\mathrm{h}-4\mathrm{k}+5=2$

${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}+3=0$

Answer : d

15. If the lines $ax+y+1=0,x+by+1=0$ and $x+y+c=0(a,b,c$ being distinct and different from 1$)$ are concurrent, then $\left({\displaystyle \frac{1}{1-a}}\right)+{\displaystyle \frac{1}{1-b}}+{\displaystyle \frac{1}{1-c}}=$

(a) 0

(b) 1

(c) ${\displaystyle \frac{1}{a+b+c}}$

(d) None

16. The set of values of ’ $b$ ’ for which the origin and the point $(1,1)$ lie on the same side of the st. line ${\mathrm{a}}^2\mathrm{x}+\mathrm{aby}+1=0\mathrm{\forall }\mathrm{a}\in \mathrm{R},\mathrm{b}>\mathrm{o}$ are

(a) $\mathrm{b}\in [2,4)$

(b) $\mathrm{b}\in (0,2)$

(c) $\mathrm{b}\in [0,2]$

(d) None of these

17. Let $P(-1,0),Q(0,0)$ and $R(3,3\sqrt{3})$ be three points. Then the equation of the bisector of the angle $\mathrm{PQR}$ is

(a) ${\displaystyle \frac{\sqrt{3}}{2}}\mathrm{x}+\mathrm{y}=0$

(b) $x+\sqrt{3}y=0$

(c) $\sqrt{3}\mathrm{x}+\mathrm{y}=0$

(d) $x+{\displaystyle \frac{\sqrt{3}}{2}}y=0$

Show Answer

Solution:

${\displaystyle \frac{\sqrt{3}-m}{1+\sqrt{3m}}}={\displaystyle \frac{m}{1+0}}$

$\begin{array}{rl}\sqrt{3}-m=m+\sqrt{3}{m}^2\Rightarrow & \sqrt{3}{m}^2+2m-\sqrt{3}=0\\ & \sqrt{3}{m}^2+3m-m-\sqrt{3}=0\\ & (\sqrt{3}m-1)(3+\sqrt{3})=0\\ & m={\displaystyle \frac{1}{\sqrt{3}}},-\sqrt{3}\end{array}$

$y=-\sqrt{3}\mathrm{x}$

$\sqrt{3}x+y=0$

18. $\mathrm{OPQR}$ is a square and $\mathrm{M},\mathrm{N}$ are the mid points of the sides $\mathrm{PQ}$ and $\mathrm{QR}$ respectively. If the ratio of the areas of the square and the triangle OMN is $\lambda :6$, then ${\displaystyle \frac{\lambda }{4}}$ is equal to

(a) 2

(b) 4

(c) 12

(d) 16

Show Answer

Solution: ar. of square $=a^2$

ar of $\mathrm{\Delta }\mathrm{OMN}={\displaystyle \frac{1}{2}}\left|\begin{array}{lll}0& 0& 1\\ \mathrm{a}& {\displaystyle \frac{\mathrm{a}}{2}}& 1\\ {\displaystyle \frac{\mathrm{a}}{2}}& \mathrm{a}& 1\end{array}\right|={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{3{\mathrm{a}}^2}{4}}={\displaystyle \frac{3{\mathrm{a}}^2}{8}}$

${\displaystyle \frac{{\mathrm{a}}^2}{{\displaystyle \frac{3{\mathrm{a}}^2}{8}}}}={\displaystyle \frac{\lambda }{6}}\Rightarrow \lambda =16$

Answer : b

19. A pair of perpendicular straight lines drawn through the origin form an isoceles triangle with line $2x+3y=6$, then area of the triangle so formed is

(a) ${\displaystyle \frac{36}{13}}$

(b) ${\displaystyle \frac{12}{17}}$

(c) ${\displaystyle \frac{13}{5}}$

(d) ${\displaystyle \frac{17}{13}}$

Show Answer

Solution: $\mathrm{OM}=\left|{\displaystyle \frac{-6}{\sqrt{4+9}}}\right|={\displaystyle \frac{6}{\sqrt{13}}},\mathrm{PQ}=2\times {\displaystyle \frac{6}{\sqrt{13}}}$, area of $\mathrm{△}\mathrm{OPQ}={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2\times 6}{\sqrt{13}}}\times {\displaystyle \frac{6}{\sqrt{13}}}={\displaystyle \frac{36}{13}}$

20. Match the column

Exercise

1. If $t_1$ and $t_2$ are roots of the equation $t^2+\lambda t+1=0$, where $\lambda$ is an arbitrary constant. Then the line joining the points $(a{t}_1{}^2,2a{t}_1)$ & $(a{t}^2,2a{t}_2)$ always passes through a fixed point

(a) $(a,0)$

(b) $(-a,0)$

(c) $(0,a)$

(d) $(0,-\mathrm{a})$

2. The equation ${\mathrm{x}}^3+{\mathrm{y}}^3=0$ represents

(a) three real straight lines

(b) three points

(c) combined equationof ast. line & a circle

(d) None of these.

3. The three lines whose combined equation is $y^3-4x^{2}y=0$ form a triangle which is

(a) isosceles

(b) equilateral

(c) right angled

(d) None of these

$\mathrm{A}(1,3)$ and $\mathrm{C}(-{\displaystyle \frac{2}{5}},-{\displaystyle \frac{2}{5}})$ are the vertices of a triangle $\mathrm{ABC}$ and the equation of the angle bisector of $\mathrm{\angle }\mathrm{ABC}$ is $\mathrm{x}+\mathrm{y}=2$

Comprehension Type

4. Equation of side $\mathrm{BC}$ is

(a) $7x+3y=4$

(b) $7x+3y+4=0$

(c) $7x-3y+4=0$

(d) $7x-3y=4$

5. Coordinates of vertex $B$ are

(a) $({\displaystyle \frac{3}{10}},{\displaystyle \frac{17}{10}})$

(b) $({\displaystyle \frac{17}{10}},{\displaystyle \frac{3}{10}})$

(c) $(-{\displaystyle \frac{5}{2}},{\displaystyle \frac{9}{2}})$

(d) $(1,1)$

6. Equation of side $\mathrm{AB}$ is

(a) $3x+7y=24$

(b) $3x+7y+24=0$

(c) $13x+7y+8=0$

(d) $13x-7y+8=0$

7. Assertion and reasoning Type

Lines ${\mathrm{L}}_1{\mathrm{L}}_2$ given by $\mathrm{y}-\mathrm{x}=0$ and $2\mathrm{x}+\mathrm{y}=0$ intersect the line ${\mathrm{L}}_3$ given by $\mathrm{y}+2=0$ at $\mathrm{P}$ and $\mathrm{Q}$, respectively. The bisector of the acute angle between ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ intersects ${\mathrm{L}}_3$ at $\mathrm{R}$.

Statement 1 : The ratio PR: RQ equals $2\sqrt{2}:\sqrt{5}$.

Statement 2 : In any triangle, bisector of an angle divides the triangle into two similar triangles.

a Statement 1 is true, Statement 2 is True; Statement 2 is a correct explanation for Statement1.

b Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1 .

c Statement 1 is True, Statement 2 is False.

d Statement 1 is False, Statement 2 is True.

8. Matrix-match

This question contains statements given in two columns which have to be matched. Statements a, b, c, d in column I have to be matched with statements p,q, r, s in column II. If the correct match is a $\to \mathrm{p},\mathrm{a}\to \mathrm{s},\mathrm{b}\to \mathrm{q}\mathrm{b}\to \mathrm{r},\mathrm{c}\to \mathrm{p},\mathrm{c}\to \mathrm{q}$ and $\mathrm{d}\to \mathrm{s}$, then the correctly dubbled $4\times 4$ matrix should be as follows :

Consider the lines given by

$\begin{array}{rl}& {\mathrm{L}}_1:\mathrm{x}+3\mathrm{y}-5=0\\ & {\mathrm{L}}_2:3\mathrm{x}-\mathrm{ky}-1=0\\ & {\mathrm{L}}_3:5\mathrm{x}+2\mathrm{y}-12=0\end{array}$