Maths M.C.Q (1 Marks)

$ \Rightarrow \,\,2x(1 + 3) + 2y(3 - 5) = (1 + 9) - (9 + 25)$

$ \Rightarrow \,\,2x - y + 6 = 0$ .....$(i)$

Perpendicular bisector of $A\,(1,\,\,3)$ and $C\,(5,\,\, - 1)$ is

$2x\,(1 - 5) + 2y(3 + 1) = (1 + 9) - (25 + 1)$

$ \Rightarrow \,\,x - y - 2 = 0$ .....$(ii)$

Point of intersection of $(i)$ and $(ii)$ is $P = ( - 8,\,\, - 10)$

Then $PA = \sqrt {{{(1 + 8)}^2} + {{(3 + 10)}^2}} = \sqrt {81 + 169} $

$ = \sqrt {250} = 5\sqrt {10} $.

$(i)$ Point $B \,(x, y)$ divides $AD$ in $1 : 2$

$\therefore \,\,\,\,x = \frac{{0 + 9}}{3} = 3$ and $y = \frac{{0 + 12}}{3} = 4$

$(ii)$ Now point $C \,(x, y)$ divides $AD$ in $2 : 1$,

Then $x = \frac{{0 + 18}}{3} = 6$ and $y = \frac{{0 + 24}}{3} = 8$.

${x_1} = 7,\,\,{x_2} = - 3,\,\,{x_3} = 1$

Similarly ${y_1},\,\,{y_2},\,\,{y_3}$ can be found.

so coordinates of $C$ will be $\left( { - \frac{1}{3},\,\frac{5}{3}} \right)$,

which is the required point.

Obviously $C$ divides the line $AB$ in $1 : 2$, therefore coordinates of $C$ are
$\left( {\frac{{1( - 3) + 2(3)}}{{1 + 2}},\,\frac{{1( - 4) + 2( - 2)}}{{1 + 2}}} \right) = \left( {1,\,\, - \frac{8}{3}} \right)$.

Similarly $D$ divides the line $AB$ in $2 : 1$, hence coordinates of $D$ are
$\left( {\frac{{2( - 3) + 1(3)}}{{2 + 1}},\,\frac{{2( - 4) + 1( - 2)}}{{2 + 1}}} \right) = \left( { - 1,\,\, - \frac{{10}}{3}} \right)$.

Now, $A{C^2} = A{B^2} + B{C^2}$

$A{C^2} = 2A{B^2} \Rightarrow 8 = A{B^2}$

$AB = BC = 2\sqrt 2 $

Now, let $B\,(x,y)$; $\therefore A{B^2} = B{C^2}$

==>${(x - 0)^2} + {(y + 1)^2} = {(x - 0)^2} + {(y - 3)^2}$

${x^2} + {y^2} + 2y + 1 = {x^2} + {y^2} - 6y + 9$

==> $y = 1;\,\,\,\therefore {x^2} + {(2)^2} = 8; \Rightarrow {x^2} = 4 \Rightarrow x = \pm \,2$

$\therefore $ other vertices are $(2, 1)\,,(-2, 1)$.

$BC = \sqrt {{{(\sqrt 3 a)}^2} + {a^2}} = 2a$

$CA = \sqrt {{{(\sqrt 3 a)}^2} + {{( - a)}^2}} = 2a$

Since $AB = BC = CA,$ hence triangle is equilateral.

Therefore, it is an acute angled triangle.

and ${x^2} + {y^2} = {(x + 2)^2} + {(y + 2)^2}\,\, \Rightarrow \,\,4x + 4y + 8 = 0$

On simplification, we get $y = - 8$ and $x = 6$.

Hence incentre is

$\left( {\frac{{11 \times 0 + 20 \times 5 + 13 \times 16}}{{11 + 20 + 13}},\,\,\frac{{11 \times 0 + 20 \times 12 + 13 \times 12}}{{11 + 20 + 13}}} \right)$i.e. $(7, 9)$.

Therefore, area $PQRS$ $ = 2.\frac{1}{2}\,\left| {\,\begin{array}{*{20}{c}}2&1&1\\4&{ - 1}&1\\3&2&1\end{array}\,} \right| = 4$.

then $ A,\, B,\, C$ are collinear if area of $\Delta ABC = 0$

$ \Rightarrow \,\,\left| {\,\begin{array}{*{20}{c}}{x + 1}&2&1\\1&{x + 2}&1\\{\frac{1}{{x + 1}}}&{\frac{2}{{x + 1}}}&1\end{array}\,} \right| = 0$ $ \Rightarrow \,\,\left| {\,\begin{array}{*{20}{c}}x&{ - x}&0\\1&{x + 2}&1\\{\frac{1}{{x + 1}}}&{\frac{2}{{x + 1}}}&1\end{array}\,} \right| = 0$

$({R_1} \to {R_1} - {R_2})$

$ \Rightarrow \,\,\left| {\,\begin{array}{*{20}{c}}x&0&0\\1&{x + 3}&1\\{\frac{1}{{x + 1}}}&{\frac{3}{{x + 1}}}&1\end{array}\,} \right| = 0$ $({C_2} \to {C_2} + {C_1})$

$ \Rightarrow \,\,x\,\left( {x + 3 - \frac{3}{{x + 1}}} \right) = 0\,\, $

$\Rightarrow \,\,x({x^2} + 3 + 4x - 3) = 0$

$ \Rightarrow \,\,{x^2}(x + 4) = 0\,\, $

$\Rightarrow \,\,x = 0,\,\, - 4$.

i.e., given triangle is equilateral.

(In centre of a triangle are same as the centriod when triangle is equilateral)

.Hence, incentre = $\left( {\frac{{7 - 1 + 3 + 2\sqrt 3 }}{3},\frac{{1 + 5 + 3 + 4\sqrt 3 }}{3}} \right)$

$= \left( {3 + \frac{2}{{\sqrt 3 }},3 + \frac{4}{{\sqrt 3 }}} \right)$.

Slope of $BC = \frac{{3 - 4}}{{1 + 2}} = \frac{{ - 1}}{3}$

Slope of $ \bot $ to $BC = 3$

Equation of altitude through $A$ is

$y + 6 = 3(x + 2)$ ==> $y + 6 = 3x + 6$ ==> $y = 3x$ .....$(i)$

Slope of $CA = \frac{{3 + 6}}{{1 + 2}} = \frac{9}{3} = 3$

Slope of $ \bot $ to $CA = \frac{{ - 1}}{3}$

Equation of altitude through $B$ is

$y - 4 = \frac{{ - 1}}{3}(x + 2)$ ==> $3y - 12 = - x - 2$

==> $x + 3y - 10 = 0$ .....$(ii)$

Solving $(i)$ and $(ii)$ $x = 1,y = 3$

Hence, orthocentre $(1, 3).$

and $x + y + 2 = 0$.....$.(ii)$

from $(i)$ and $(ii)$, $xy = 0 \Rightarrow x = y = 0$

 Vertices of triangle are $( - 2,0)\,\,(0,0)\,\,(0, - 2)$

(In a right angled triangle circumcentre is mid point of hypotenuse)

 $( - 1, - 1)$ is the circumcircle.

Co-ordinates of point of intersection of the $x = 2$ and $y + 1 = 0$ is $(2, -1)$

Co-ordinates of point of intersection of $x = 2$ and $x + 2y = 4$ is $(2, 1)$

Co-ordinates of point of intersection of $y + 1 = 0$ and $x + 2y = 4$ is $(6, -1)$

Let Co-ordinates of circumcentre is $(x, y)$

${(x - 2)^2} + {(y + 1)^2} = {(x - 2)^2} + {(y - 1)^2}$

${(y + 1)^2} = {(y - 1)^2}$; ${y^2} + 2y + 1 = {y^2} - 2y + 1$

$4y = 0$, $y = 0$ and ${(x - 2)^2} + {(y - 1)^2} = {(x - 6)^2} + {(y + 1)^2}$

In this equation put $y = 0$

${(x - 2)^2} + {(0 - 1)^2} = {(x - 6)^2} + {(0 + 1)^2}$

${(x - 2)^2} + 1 = {(x - 6)^2} + 1$; ${(x - 2)^2} - {(x - 6)^2} = 0$

$(x - 2 + x - 6)(x - 2 - x + 6) = 0$

$4(2x - 8) = 0$==> $8(x - 4) = 0$; $x - 4 = 0$ ==> $x = 4$

 Co-ordinates of circumcentre is $(4, 0)$.

Hence incentre is $\left( {\frac{{0 + 0 + 3 \times 4}}{{5 + 4 + 3}},\frac{{0 + 4 \times 3 + 0}}{{5 + 4 + 3}}} \right)$= $(1, 1)$.

Thus $c = AB = 6,\,\,a = BC = 8$ and $b = AC = 10$

Hence incentre

$ = \left( {\frac{{8 \times 0 + 10 \times 6 + 6 \times 6}}{{8 + 10 + 6}},\,\frac{{8 \times 0 + 10 \times 0 + 6 \times 8}}{{8 + 10 + 6}}} \right) = (4,\,\,2)$.

ase said to be cerlineas if Area $(\Delta A B C)=0$

$=\frac{S}{2}\left|\begin{array}{lll}2 & y & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=0$

$\therefore \quad\left|\begin{array}{lll}\lambda+1 & 1 & 1 \\ 2 \lambda+1 & 3 & 1 \\ 2 \lambda+2 & 2 \lambda & 1\end{array}\right|=0$

$\Rightarrow(\dot{x}+1)(3-2 \lambda)-1(82+1-2 \lambda-2)$

$+1\left(4 x^{2}+2 x-6 x-6\right)=0$

$\Rightarrow 3 \lambda+3-2\left(\lambda^{2}-2 \lambda+1+4 x^{2}-4 \lambda-6=0\right.$

$\Rightarrow 2 \lambda^{2}-3 \lambda-2=0$

$\Rightarrow 2 \lambda^{2}-4 \lambda+\lambda-2=0$

a) $2 x(x-2)+1(x-2)=0$

$\Rightarrow \lambda=-1 / 2$ or 2

" polsible values of

$\lambda \operatorname{are} 2$

To eliminate linear terms, we should have

$2h - 4 = 0,\,\,2k + 6 = 0\,\,\, \Rightarrow \,\,h = 2,\,\,k = - 3$

i.e., $(h,\,\,k) = (2,\,\, - 3)$.

So,${(h - a)^2} + {(k - 0)^2} = {h^2}$$ \Rightarrow \,\,\,{h^2} + {a^2} - 2ah + {k^2} = {h^2}$

Hence locus is ${y^2} - 2ax + {a^2} = 0$.

$\sqrt {{{(x - 1)}^2} + {y^2}} - \sqrt {{{(x + 1)}^2} + {y^2}} = \pm 1$

On squaring both sides, we get

$2{x^2} + 2{y^2} + 1 = 2\sqrt {{{(x - 1)}^2} + {y^2}} .\sqrt {{{(x + 1)}^2} + {y^2}} $

Again on squaring, we get $12{x^2} - 4{y^2} = 3$.

This is required locus of the point $(x, y)$.

$ \Rightarrow \,\,(1 - {k^2})\,({x^2} + {y^2}) - 2akx + 2akx + {a^2}{k^2} - {a^2} = 0$

$ \Rightarrow \,\,{x^2} + {y^2} - {a^2} = 0$.

Putting the value of t in $k = u\,\sin \alpha \,.\,t - \frac{1}{2}g{t^2},$ we get $k = h\,\tan \alpha - \frac{1}{2}g\frac{{{h^2}}}{{{u^2}{{\cos }^2}\alpha }}$

$\therefore \,\,$Locus of (h, k) is $y = x\tan \alpha - \frac{1}{2}g\frac{{{x^2}}}{{{u^2}\,{{\cos }^2}\alpha }}$, which is a parabola.

$h = \frac{{\cos \alpha + \sin \alpha + 1}}{3}$ and $k = \frac{{\sin \alpha - \cos \alpha + 2}}{3}$

$ \Rightarrow \,\,3h - 1 = \cos \alpha + \sin \alpha $ and $3k - 2 = \sin \alpha - \cos \alpha $

$ \Rightarrow \,\,{(3h - 1)^2} + {(3k - 2)^2} = 2$, (squaring and adding)

$ \Rightarrow \,9\,({h^2} + {k^2}) - 6h - 12k + 3 = 0$

$ \Rightarrow \,\,3\,({h^2} + {k^2}) - 2h - 4k + 1 = 0$

$\therefore $ Locus of $(h,\,\,k)$ is $3\,({x^2} + {y^2}) - 2x - 4y + 1 = 0$.

Given that $PA - PB = 4$

$\sqrt {{{(h - 3)}^2} + {k^2}} - \sqrt {{{(h + 3)}^2} + {k^2}} = 4$

$ \Rightarrow \,\,\sqrt {{{(h - 3)}^2} + {k^2}} = 4 + \sqrt {{{(h + 3)}^2} + {k^2}} $

Squaring both sides, we get

${(h - 3)^2} + {k^2} = 16 + {(h + 3)^2} + {k^2} + 8\sqrt {{{(h + 3)}^2} + {k^2}} $

$ \Rightarrow \,\,\,{h^2} + 9 - 6h + {k^2} = 16 + {h^2} + 9 + 6h + {k^2}$

$ + \,8\sqrt {{{(h + 3)}^2} + {k^2}} $

$ \Rightarrow \,\, - 6h = 16 + 6h + 8\sqrt {{{(h + 3)}^2} + {k^2}} $

$ \Rightarrow \,\, - 8\,\sqrt {{{(h + 3)}^2} + {k^2}} = 12h + 16$

Again, squaring both sides, we get

$64\,({h^2} + 9 + 6h + {k^2}) = 144{h^2} + 256 + 2.16.12h$

$ \Rightarrow \,\,4\,({h^2} + 9 + 6h + {k^2}) = 9{h^2} + 16 + 24h$

$ \Rightarrow \,\,4{h^2} + 36 + 24h + 4{k^2} = 9{h^2} + 16 + 24h$

$ \Rightarrow \,\,5{h^2} - 4{k^2} = 20\,\, \Rightarrow \,\,\frac{{{h^2}}}{4} - \frac{{{k^2}}}{5} = 1$

Hence, the locus of point $P$ is $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$.

In right angled triangle $CDA,\,\,\,\tan A = \frac{k}{{a - h}}$

and similarly in triangle $CDB,\,\,\tan B = \frac{k}{{a + h}}$

Also from $(i)$, $\frac{{\tan A - \tan B}}{{1 + \tan A\,.\,\tan B}} = \tan \theta $

Substituting the values of $\tan A$ and $\tan B,$ we get

${h^2} - {k^2} + 2hk\cot \theta = {a^2}$

Hence the locus is ${x^2} - {y^2} + 2xy\cot \theta = {a^2}$.

$D\,\left( {\frac{{2 \times 6 + 0}}{3},\,\frac{{ - 6 + 3}}{3}} \right)\, \equiv \,(4,\, - 1)$.

$AC$ is bisector of $\angle DAB$

$\therefore \angle DAC =\angle CAB$

and $DC \| AB$

$\therefore \angle CAB =\angle ACD \quad \ldots(2)$

$\therefore \angle DAC =\angle ACD$

Hence $AD = DC$

similarly $BC = DC$

Hence Exactly three sides are equal.

In $\triangle ABC$ and $\triangle DCB$

$AB = DC$

$BC = CB$

$AC = DB$

$\triangle ABC \cong \triangle DCB$

$\therefore \angle BAC =\angle CDB =70^{\circ}$

$\angle ABC =\angle DCB =60^{\circ}$

$\angle ACB =180^{\circ}-\angle BAC -\angle ABC$

$=180^{\circ}-70^{\circ}-60^{\circ}$

$=50^{\circ}$

$\angle DCO =60^{\circ}-\angle ACB$

$=60^{\circ}-50^{\circ}=10^{\circ}$

$\angle DOC =180^{\circ}-70^{\circ}-10^{\circ}=100^{\circ}$

Acute angle $\angle DOC =80^{\circ}$

$\frac{\operatorname{ar} \triangle ABC }{\operatorname{ar} \triangle ADE }=\frac{\frac{1}{2} \cdot AB \cdot AC \sin \theta}{\frac{1}{2} \cdot AD \cdot AE \sin \theta}$

$=\frac{ AB }{ AD } \times \frac{ AC }{ AE }$

$=\frac{5}{3} \times \frac{3}{2}=\frac{5}{2}$

correct option is $\frac{5}{2}$

Let the remaining two sides be $a$ and $b$, then

.Since, $A B C D$ is a rectangle, then

$\frac{9}{\sqrt{2}}+6+\frac{b}{\sqrt{2}}=\frac{7}{\sqrt{2}}+10+\frac{5}{\sqrt{2}}$

$\quad \frac{b}{\sqrt{2}}=4+\frac{3}{\sqrt{2}}$

$\text { Similarly, } \frac{9}{\sqrt{2}}+8+\frac{7}{\sqrt{2}}=\frac{b}{\frac{\sqrt{2}}{}}+a+\frac{5}{\sqrt{2}}$

$\quad a=4+\frac{8}{\sqrt{2}}$

$\therefore \quad a+b=7+8 \sqrt{2}=18.3$

We have, $\cos \frac{B}{2}=\frac{\frac{9}{2}+4-1}{6 \sqrt{2}}=\frac{5}{4 \sqrt{2}}$

$\therefore$ Length of angle bisector,

$B D=\frac{2 a c}{a+c} \cos \frac{B}{2}$

$\frac{3}{\sqrt{2}} =\left(\frac{4 c}{e+2}\right) \cdot \frac{5}{4 \sqrt{2}}$

$c =3$

We know that, $\frac{A B}{B C}=\frac{A D}{C D}$

$A D=\frac{3}{2}$

$\therefore$ Perimeter of $\triangle A B C=1+\frac{3}{2}+3+2=\frac{15}{2}$

If in a $\triangle A B C, A D$ is median, then by Apollonius Theorem

$A B^2+A C^2=2\left(A D^2+B D^2\right)$

Here, by using Apollonius theorem, we get

$x^2+(x-2)^2=2\left[6^2+4^2\right]$

$2 x^2-4 x+4=104$

$x^2-2 x-50=0 \Rightarrow x=1+\sqrt{51}$

The nearest integer to $b$ is $8$ .

Area of parallelogram whose diagonals are 10 and 4 and if angle between adjacent side is $\theta$ is

$A=\frac{10 \times 4}{\sin \theta}$

The area will be maximum if $\theta=\frac{\pi}{2}$, so the parallelogram must be rhombus

Perimeter of rhombus

$=4 \sqrt{2} \overline{9}=\sqrt{46} \overline{4} \text { and } \sqrt{46} \overline{4} \in(21,22]$

For an equilateral triangle $A B C$ having side length $a$. If $R$ and $r$ are radii of the circumcircle and the incircle of triangle $A B C$ respectively, then

$R=\frac{a}{2} \sec 30^{\circ}=\frac{a}{2}\left(\frac{2}{\sqrt{3}}\right)=\frac{a}{\sqrt{3}}$

and $r=\frac{a}{2} \tan 30^{\circ}=\frac{a}{2} \times \frac{1}{\sqrt{3}}=\frac{a}{2 \sqrt{3}}$

$\therefore \frac{R}{r}=\frac{\frac{a}{\sqrt{3}}}{\frac{a}{2 \sqrt{3}}}=2$, which is independent of $a$ and it is constant.

It is given that in triangle $A B C$, $\angle B A C=90^{\circ}, A D$ is the altitude from $A$ on to $B C$.

Since, $A B=15$ and $B C=25$

$\therefore \quad A C=\sqrt{B C^2-A B^2}=\sqrt{625-225}$

$=\sqrt{400}=20$

Now, since area of $\triangle A B C=\frac{1}{2}(B C)(A D)$

$=\frac{1}{2}(A B)(A C)$

$\Rightarrow \frac{1}{2}(B C)(A D)=\frac{1}{2} \times 15 \times 20$

$\Rightarrow \quad 25 \times A D=300$

$\Rightarrow \quad A D=12$

$\because A E D F$ is a rectangle, then

$E F=A D=12$

Let a convex quadrilateral $A B C D$ and $K, L, M, N$ be the mid-point of $A B$, $B C, C D, D A$ respectively.

Now, as area $\Delta A K P=$ area $\Delta B K P=x$ (let) Similarly

$\triangle B L P =\Delta C L P=y$

$\Delta C P M =\triangle D P M=z$

And  $\Delta D N P =\triangle A N P=w$

It is given that Area $(P K A N)=x+w=25$ area $(P L B K)=x+y=36$

and area $(P M D N)=z+w=41$

So $\operatorname{area}(P L C M)=y+z$

$=(x+y)+(z+w)-(x+w)$

$=\operatorname{area}(P L B K)+$ area $(P M D N)-$ area $(P K A N)$

$=36+41-25=77-25=52$

It is given that circumference of circle $C$ is $l$ and the perimeter of triangle $T$ is $l$.

Now, let the radius of circle $C$ is $r$, so

$2 \pi r=l \Rightarrow r=\frac{l}{2 \pi}$

$\therefore$ area of circle $C$ is $A_1=\pi r^2=\frac{l^2}{4 \pi}$

Now, as we know that area of triangle will be maximum for given perimeter if it is an equilateral triangle, let the length of side of equilateral triangle is ' $a$ ', then

$3 a=l \Rightarrow a=\frac{l}{3}$

and area of equilateral triangle is

$A_2=\frac{\sqrt{3}}{4} a^2$

So, $A_2=\frac{\sqrt{3}}{4}\left(\frac{l^2}{9}\right)=\frac{l^2}{12 \sqrt{3}}$

Since, as we took an equilateral triangle, which has maximum area. But we can take a triangle $T$ such that the ratio area $(C)$ is greater than any positive real area $(T)$ number $\alpha$.