Straight Lines and Pair of Straight Lines
From the point $(-1,-1)$, two rays are sent making angles of $45^{\circ}$ with the line $x+y=0$. These rays get reflected from the mirror $x+2 y=1$. If the equations of the reflected rays are $\mathrm{a} x+\mathrm{b} y=9$ and $c x+d y=7, a, b, c, d \in \mathbf{Z}$, then the value of $a d+b c$ is $\_\_\_\_$ .
Explanation:
Line $x+y=0$ has slope $-1$. If a ray has slope $m$ and makes an angle $45^{\circ}$ with this line, then by the angle formula between two lines we use:
$\tan 45^{\circ} = \left|\frac{m+1}{1-m}\right|$
Now $\tan 45^{\circ}=1$, so we get:
$\rightarrow \pm 1 = \frac{m+1}{1-m}$
Solving this gives two possible directions. One solution gives a horizontal line ($m=0$), and the other corresponds to a vertical line (perpendicular to the $x$-axis).
$\rightarrow m = 0$ & one line is perpendicular to x axis.
Both rays start from $(-1,-1)$, so the two incident rays are:
$L_1$: $x=-1$, $L_2$: $y=-1$
Next, each ray reflects from the mirror line $x+2 y=1$. To get the reflected ray, we take the mirror image of the incident line in the mirror.
First, take the mirror image of $x+1=0$ in $x+2y=1$.
Now taking mirror image of $x+1=0$ in $x+2y=1$
$A': \frac{x+1}{1} = \frac{y+1}{2} = \frac{-2(-1-2-1)}{1^2+2^2}$
From this, we get the image point:
$x=\frac{3}{5}, y=\frac{11}{5} \qquad A'\left(\frac{3}{5}, \frac{11}{5}\right)$
So the corresponding reflected line is written as:
line $= 3x-4y+7=0$
Similarly, take the mirror image of $y+1=0$ in $x+2y=1$.
Now taking mirror image of $y+1=0$ in $x+2y=1$
$A'': \frac{x+1}{1} = \frac{y+1}{2} = \frac{-2(-1-2-1)}{1^2+2^2}$
This again gives:
$x=\frac{3}{5}, y=\frac{11}{5}$
$A''\left(\frac{3}{5}, \frac{11}{5}\right)$
Now write the equation of the reflected line using point-slope form:
line :
$ (y+1)=\frac{\left(\frac{11}{5}+1\right)}{\left(\frac{3}{5}-3\right)}(x-3) $
Simplifying gives one reflected ray as:
$4x+3y=9$
Compare $4x+3y=9$ with $\mathrm{a}x+\mathrm{b}y=9$. Then:
comparing $4x+3y=9$ with $ax+by=9$ &
$a=4$, $b=3$
The other reflected ray is written as:
by line
$-3x+4y=7$
Compare $-3x+4y=7$ with $cx+dy=7$. Then:
comparing with
$cx+dy=7$
$\therefore c=-3$, $d=4$
Now calculate $ad+bc$:
So, $ad+bc$
$= 4 \times 4 + 3 \times (-3)$
$= 16-9=7$
Let $\mathrm{A}, \mathrm{B}$ be points on the two half-lines $x-\sqrt{3}|y|=\alpha, \alpha>0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $A B$ meets the angle bisector of the given half-lines at the point $Q$. If $P Q=\frac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle \mathrm{PAB}$, then $\frac{\alpha^2}{R}$ is equal to $\_\_\_\_$
Explanation:
The two half-lines are given by
$ x-\sqrt{3}|y|=\alpha,\qquad \alpha>0 $
We will first understand their geometry.
1. Find the two half-lines and their point of intersection
Since $|y|$ is present, we split into two cases:
Case 1: $y \ge 0$
$ x-\sqrt{3}y=\alpha $
Case 2: $y \le 0$
$ x+\sqrt{3}y=\alpha $
These are two lines meeting at the point $P$.
To find their intersection, put $y=0$ in either equation:
$ x=\alpha $
So,
$ P=(\alpha,0) $
Thus the two half-lines start from $P$ and go outward along these two lines.
2. Find points $A$ and $B$
Points $A$ and $B$ lie on these two half-lines and are each at distance $\alpha$ from $P$.
Let $A$ be on the upper line $x-\sqrt{3}y=\alpha$ and $B$ on the lower line $x+\sqrt{3}y=\alpha$.
The slope of the upper line is
$ y=\frac{x-\alpha}{\sqrt{3}} \Rightarrow m=\frac{1}{\sqrt{3}} $
So this line makes angle $30^\circ$ with the positive $x$-axis.
Similarly, the lower line makes angle $-30^\circ$ with the positive $x$-axis.
Hence the angle between the two half-lines is
$ 60^\circ $
Since $PA=\alpha$ and $PB=\alpha$, the points are obtained by moving distance $\alpha$ from $P=(\alpha,0)$ along directions $\pm 30^\circ$.
Therefore,
$ A=\left(\alpha+\alpha\cos30^\circ,\ \alpha\sin30^\circ\right) $
$ B=\left(\alpha+\alpha\cos30^\circ,\ -\alpha\sin30^\circ\right) $
Using $\cos30^\circ=\frac{\sqrt{3}}{2}$ and $\sin30^\circ=\frac12$,
$ A=\left(\alpha+\frac{\sqrt{3}\alpha}{2},\ \frac{\alpha}{2}\right) $
$ B=\left(\alpha+\frac{\sqrt{3}\alpha}{2},\ -\frac{\alpha}{2}\right) $
3. Find the angle bisector and point $Q$
The figure is symmetric about the $x$-axis, so the angle bisector of the two half-lines is the positive $x$-axis.
Also, $A$ and $B$ have the same $x$-coordinate, so $AB$ is a vertical segment.
Hence $AB$ meets the $x$-axis at
$ Q=\left(\alpha+\frac{\sqrt{3}\alpha}{2},\ 0\right) $
So,
$ PQ = \left(\alpha+\frac{\sqrt{3}\alpha}{2}\right)-\alpha = \frac{\sqrt{3}\alpha}{2} $
Given that
$ PQ=\frac92 $
therefore,
$ \frac{\sqrt{3}\alpha}{2}=\frac92 $
So,
$ \sqrt{3}\alpha=9 $
$ \alpha=\frac{9}{\sqrt{3}}=3\sqrt{3} $
Hence,
$ \alpha^2 = (3\sqrt{3})^2=27 $
4. Find circumradius $R$ of $\triangle PAB$
In $\triangle PAB$,
$ PA=PB=\alpha $
and the included angle
$ \angle APB=60^\circ $
So the triangle is actually equilateral, because
$ AB^2=\alpha^2+\alpha^2-2\alpha^2\cos60^\circ =2\alpha^2-\alpha^2 =\alpha^2 $
Thus,
$ AB=\alpha $
Hence $\triangle PAB$ is equilateral with side $\alpha$.
For an equilateral triangle of side $a$, circumradius is
$ R=\frac{a}{\sqrt{3}} $
So here,
$ R=\frac{\alpha}{\sqrt{3}} $
Using $\alpha=3\sqrt{3}$,
$ R=\frac{3\sqrt{3}}{\sqrt{3}}=3 $
5. Compute $\dfrac{\alpha^2}{R}$
$ \frac{\alpha^2}{R}=\frac{27}{3}=9 $
Therefore, the required value is
$ \boxed{9} $
Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $P Q R$ is formed such that $Q$ lies on one of the parallel lines, while R lies on the other. Then $(Q R)^2$ is equal to _________.
Explanation:
We set up a coordinate system so that the two parallel lines are given by
$ y = 0 \quad \text{and} \quad y = 5, $
since their distance is 5 units. Choose point
$ P = (0,1) $
so that the distance from $P$ to the line $y=0$ is 1 unit (and its distance to the line $y=5$ is 4 units).
Let point
$ Q = (a,0) $
be on the line $y = 0$, and let point
$ R = (b,5) $
be on the line $y = 5$. Since triangle $PQR$ is equilateral with side length $s$, we require:
$ PQ = PR = QR = s. $
A convenient method is to “rotate” $Q$ about $P$ by an angle of $60^\circ$ to obtain $R$. In complex-number (or vector) terms, if we translate so that $P$ is at the origin, then the rotation is given by
$ e^{i60^\circ} = \cos 60^\circ + i \sin 60^\circ = \frac{1}{2} + i \frac{\sqrt{3}}{2}. $
Thus, writing $Q$ in vector form relative to $P$, we have
$ Q - P = (a, -1). $
Rotating this by $60^\circ$ gives
$ R - P = \left(a\cos60^\circ - (-1)\sin60^\circ,\; a\sin60^\circ + (-1)\cos60^\circ\right). $
Substituting the values $\cos60^\circ = \frac{1}{2}$ and $\sin60^\circ = \frac{\sqrt{3}}{2}$, we obtain
$ \begin{aligned} R - P &= \left(\frac{a}{2} + \frac{\sqrt{3}}{2},\; \frac{a\sqrt{3}}{2} - \frac{1}{2}\right), \\ \text{so} \quad R &= \left( \frac{a+\sqrt{3}}{2},\; 1 + \frac{a\sqrt{3}}{2} - \frac{1}{2} \right) = \left( \frac{a+\sqrt{3}}{2},\; \frac{a\sqrt{3}+1}{2} \right). \end{aligned} $
Since $R$ lies on $y = 5$, its $y$-coordinate must equal 5:
$ \frac{a\sqrt{3}+1}{2} = 5. $
Solve for $a$:
$ \begin{aligned} a\sqrt{3} + 1 &= 10, \\ a\sqrt{3} &= 9, \\ a &= \frac{9}{\sqrt{3}} = 3\sqrt{3}. \end{aligned} $
Now, the side length $s$ (which is the distance $PQ$) is given by
$ \begin{aligned} s^2 &= PQ^2 = \left(3\sqrt{3} - 0\right)^2 + \left(0 - 1\right)^2 \\ &= (3\sqrt{3})^2 + 1^2 \\ &= 27 + 1 \\ &= 28. \end{aligned} $
Thus, the square of side $QR$ is
$ (QR)^2 = s^2 = 28. $
Let a ray of light passing through the point $(3,10)$ reflects on the line $2 x+y=6$ and the reflected ray passes through the point $(7,2)$. If the equation of the incident ray is $a x+b y+1=0$, then $a^2+b^2+3 a b$ is equal to _________.
Explanation:
Equation of incident ray : $a x+b y+1=0$
Using mirror image,
$\frac{m-7}{2}=\frac{n-2}{1}=\frac{-2(14+2-6)}{5}$
$\begin{array}{l|l} \frac{m-7}{2}=-4 & n-2=-4 \\ m=-8+7 & n=-2 \\ m=-1 & \end{array}$
${ }^*$ Note: It can be observed from diagram $A, P, B$' are collinear.
Equation of Incident Ray,
Using two-point form,
$\begin{aligned} & (y-10)=\frac{10+2}{3+1}(x-3) \\ & (y-10)=\frac{12}{4}(x-3) \\ & y-10=+3(x-3) \\ & y-10=+3 x-9 \\ & 3 x-y+1=0 \end{aligned}$
On comparing,
$\begin{aligned} & a=3 \\ & b=-1 \end{aligned}$
If the orthocentre of the triangle formed by the lines $2 x+3 y-1=0, x+2 y-1=0$ and $a x+b y-1=0$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $(3,4)$ and $(-6,-8)$, then the value of $|a-b|$ is _________.
Explanation:
Let circumcentre, orthocentre and centroid of a triangle $P Q R$ are $C_1, H_1$ and $G_1$ respectively
$\Rightarrow G_1 \equiv(0,0)$ orthocentre of $\triangle A B C$ is $(0,0)$
$\begin{aligned} & m_{A H_2}=+\frac{b}{a} \Rightarrow a+b=0 \\ & \text { eq }{ }^{\text {n }} \text { of lines } H_2 C \text { is } y=\frac{3}{2} x \\ & \Rightarrow \text { point } C \equiv\left(\frac{1}{4}, \frac{3}{8}\right) \text { lies on } a x+b y-1=0 \\ & \Rightarrow \frac{a}{4}+\frac{3}{8} b-1=0 \Rightarrow \frac{a}{4}-\frac{3}{8} a-1=0 \\ & \Rightarrow a=-8, b=8 \\ & |a-b|=16 \end{aligned}$
Explanation:
$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}[$ By sine rule $]$
$ 2 c=8 \Rightarrow c=4 $
$\begin{gathered}\mathrm{AB}=|(\mathrm{b}+1)|=4 \\\\ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 \\\\ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} \\\\ \mathrm{BC}:-\mathrm{y}=\frac{-1}{\sqrt{3}}(\mathrm{x}-3) \\\\ \sqrt{3} \mathrm{y}+\mathrm{x}=3\end{gathered}$
Point of intersection : $y=x+3, \sqrt{3} y+x=3$
$\begin{aligned} & ({\sqrt{3}+1}) y=6 \\\\ & y=\frac{6}{\sqrt{3}+1} \\\\ & x=\frac{6}{\sqrt{3}+1}-3 \\\\ & =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} \\\\ & =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2}\end{aligned}$
$\frac{\beta^4}{\alpha^2}=36$
Explanation:
To find the maximum number of points of intersection of pairs of lines from the given set, we need to consider how the lines are arranged based on the given conditions.
Firstly, there are 10 lines (${L}_1, {L}_3, ..., {L}_{19}$) that are parallel to each other. Since parallel lines do not intersect with each other, these 10 lines will not contribute to the number of intersection points among themselves.
Secondly, there are 10 lines (${L}_2, {L}_4, ..., {L}_{20}$) that all pass through a given point $P$. Although these lines intersect at $P$, they only contribute one unique point of intersection to the total count.
To calculate the maximum number of intersection points, we need to consider the total number of ways to pick pairs of lines from the 20 lines available without restrictions, then subtract the combinations that do not result in intersections, which includes the combinations of parallel lines among themselves and the concurrent lines through point $P$.
This calculation is represented as:
$Total = ^{20}C_2 - ^{10}C_2 - ^{10}C_2 + 1$
Here, $^{20}C_2$ calculates the total number of ways to pick any two lines out of 20, which includes intersecting and non-intersecting lines. $^{10}C_2$ is subtracted twice: once for the set of parallel lines (${L}_1, {L}_3, ..., {L}_{19}$) that don't intersect among themselves and once more for the set of concurrent lines (${L}_2, {L}_4, ..., {L}_{20}$) intersecting only at point $P$. Since all the concurrent lines intersect at the same point, we add 1 back to include this intersection point.
Carrying out this calculation gives us the total number of distinct intersection points as $101$.
Let $A(-2,-1), B(1,0), C(\alpha, \beta)$ and $D(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to ___________.
Explanation:
$\begin{aligned} & \mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right) \\ & \frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} \\ & \Rightarrow \alpha-\gamma=3 \ldots .(1), \beta-\delta=1 \ldots \ldots (2) \end{aligned}$
Also, $(\gamma, \delta)$ lies on $3 x-2 y=6$
$3 \gamma-2 \delta=6$ ..... (3)
and $(\alpha, \beta)$ lies on $2 x-y=5$
$\Rightarrow 2 \alpha-\beta=5 \text {. }$
Solving (1), (2), (3), (4)
$\begin{aligned} & \alpha=-3, \beta=-11, \gamma=-6, \delta=-12 \\ & |\alpha+\beta+\gamma+\delta|=32 \end{aligned}$
If the sum of squares of all real values of $\alpha$, for which the lines $2 x-y+3=0,6 x+3 y+1=0$ and $\alpha x+2 y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $p$ is _________.
Explanation:
$\begin{aligned} & 2 x-y+3=0 \\ & 6 x+3 y+1=0 \\ & \alpha x+2 y-2=0 \end{aligned}$
Will not form a $\Delta$ if $\alpha x+2 y-2=0$ is concurrent with $2 x-y+3=0$ and $6 x+3 y+1=0$ or parallel to either of them so
Case-1: Concurrent lines
$\left|\begin{array}{ccc} 2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2 \end{array}\right|=0 \Rightarrow \alpha=\frac{4}{5}$
Case-2 : Parallel lines
$\begin{aligned} & -\frac{\alpha}{2}=\frac{-6}{3} \text { or }-\frac{\alpha}{2}=2 \\ & \Rightarrow \alpha=4 \text { or } \alpha=-4 \\ & P=16+16+\frac{16}{25} \\ & {[P]=\left[32+\frac{16}{25}\right]=32} \end{aligned}$
If the line $l_{1}: 3 y-2 x=3$ is the angular bisector of the lines $l_{2}: x-y+1=0$ and $l_{3}: \alpha x+\beta y+17=0$, then $\alpha^{2}+\beta^{2}-\alpha-\beta$ is equal to _________.
Explanation:
Point of intersection of $L_1 $ and $ L_2$ is $(0,1)$, should lie on $L_3 \Rightarrow \beta=-17$
Any point, say $\left(\frac{-3}{2}, 0\right)$ on $L_1$ should be equidistant from the lines $L_2 $ and $ L_3$
$ \begin{aligned} & \Rightarrow\left|\frac{\frac{-3}{2}-0+1}{\sqrt{1^2+1^2}}\right|=\left|\frac{\frac{-3 \alpha}{2}+0+17}{\sqrt{\alpha^2+(-17)^2}}\right| \\\\ & \Rightarrow (\alpha-7)(\alpha-17)=0 \end{aligned} $
For $\alpha=17, L_2 $ and $ L_3$ coincides $\Rightarrow \alpha=7$
$ \begin{aligned} \alpha^2+\beta^2-\alpha-\beta & =(7)^2+(-17)^2-7+17 \\\\ & =348 \end{aligned} $
Let the equations of two adjacent sides of a parallelogram $\mathrm{ABCD}$ be $2 x-3 y=-23$ and $5 x+4 y=23$. If the equation of its one diagonal $\mathrm{AC}$ is $3 x+7 y=23$ and the distance of A from the other diagonal is $\mathrm{d}$, then $50 \mathrm{~d}^{2}$ is equal to ____________.
Explanation:
We have, $A B C D$ is a parallelogram
Let equation of $A B$ be $2 x-3 y=-23 \ldots$ (i)
and equation of $B C$ be $5 x+4 y=23\ldots$ (ii)
Equation of $A C$ is $3 x+7 y=23 \ldots$ (iii)
Solving Eqs. (i) and (ii), we get
$x=-1$, and $y=7$
$\therefore$ Co-ordinate of $B$ is $(-1,7)$
On solving Eqs. (ii) and (iii), we get
$ x=3, y=2 $
$\therefore$ Co-ordinate of $C$ is $(3,2)$
On solving Eqs. (i) and (iii), we get $x=-4$ and $y=5$
$\therefore$ Co-ordinate of $A$ is $(-4,5)$.
Let $E$ be the intersection point of diagonal co-ordinate of
$E$ is $\left(\frac{-4+3}{2}, \frac{5+2}{2}\right)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$
$\because E$ is mid-point of $A C$
$ \begin{aligned} & \text { Equation of } B D \text { is } y-7=\left(\frac{7-\frac{7}{2}}{-1+\frac{1}{2}}\right)(x+1) \\\\ & \Rightarrow 7 x+y=0 \end{aligned} $
Distance of $A$ from diagonal $B D=\frac{|7 \times(-4)+5|}{\sqrt{7^2+1^2}}$
$ \therefore d=\frac{23}{\sqrt{50}} $
Hence, $50 d^2=(23)^2=529$
The equations of the sides AB, BC and CA of a triangle ABC are : $2x+y=0,x+py=21a,(a\pm0)$ and $x-y=3$ respectively. Let P(2, a) be the centroid of $\Delta$ABC. Then (BC)$^2$ is equal to ___________.
Explanation:
$ \begin{aligned} & \therefore 4 p^{2}-21 a p+8 p+42 a-5=0\quad...(1) \end{aligned} $
And $\frac{-42 a}{1-2 p}-2+\frac{21 a-3}{p+1}=3 a$
$ \therefore 4 p^{2}-81 a p+6 a p^{2}-24 a+8 p-5=0 \quad...(2) $
From equation (1) - equation (2) we get;
$ 60 a p+66 a-6 a p^{2}=0 $
$ \begin{aligned} \because a \neq 0 \Rightarrow p^{2}-10 p-11=0 \\\\ p=-1 \text { or } 11 \Rightarrow p=11 . \end{aligned} $
When $p=11$ then $a=3$
Coordinate of $B=(-3,6)$
And coordinate of $C=(8,5)$
$\therefore B C^{2}=122$
The equations of the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ of a triangle $\mathrm{ABC}$ are $2 x+y=0, x+\mathrm{p} y=15 \mathrm{a}$ and $x-y=3$ respectively. If its orthocentre is $(2, a),-\frac{1}{2}<\mathrm{a}<2$, then $\mathrm{p}$ is equal to ______________.
Explanation:
Slope of $AH = {{a + 2} \over 1}$
Slope of $BC = - {1 \over p}$
$\therefore$ $p = a + 2$ ...... (i)
Coordinate of $C = \left( {{{18p - 30} \over {p + 1}},\,{{15p - 33} \over {p + 1}}} \right)$
Slope of $HC = {{{{15P - 33} \over {p + 1}} - a} \over {{{18p - 30} \over {p + 1}} - 2}} = {{15p - 33 - (p - 2)(p + 1)} \over {18p - 30 - 2p - 2}}$
$ = {{16p - {p^2} - 31} \over {16p - 32}}$
$\because$ ${{16p - {p^2} - 31} \over {16p - 32}} \times - 2 = - 1$
$\therefore$ ${p^2} - 8p + 15 = 0$
$\therefore$ $p = 3$ or $5$
But if $p = 5$ then $a = 3$ not acceptable
$\therefore$ $p = 3$
A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be ($\alpha$, $\beta$). Then, the value of 7$\alpha$ + 3$\beta$ is equal to ____________.
Explanation:
${4 \over {5 - \alpha }} = {3 \over {\alpha - 2}} \Rightarrow 4\alpha - 8 = 15 - 3\alpha $
$\alpha = {{23} \over 7}$
$A = \left( {{{23} \over 7},0} \right)\,Q = (5,4)$
$R = \left( {{{10 + {{23} \over 7}} \over 3},{8 \over 3}} \right)$
$ = \left( {{{31} \over 7},{8 \over 3}} \right)$
Bisector of angle PAQ is $X = {{23} \over 7}$
$ \Rightarrow M = \left( {{{23} \over 7},{8 \over 3}} \right)$
So, $7\alpha + 3\beta = 31$
Let $A\left( {{3 \over {\sqrt a }},\sqrt a } \right),\,a > 0$, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If $D(3\cos \theta ,a\sin \theta )$ is a point in the fourth quadrant such that the maximum area of $\Delta$ACD is 12 square units, then a is equal to ____________.
Explanation:
$ \begin{aligned} &\text { Area of } \triangle A C D=\frac{1}{2}\left|\begin{array}{ccc} \frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=|3 \sqrt{a} \sin \theta+3 \sqrt{a} \cos \theta|=3 \sqrt{a}|\sin \theta+\cos \theta| \\\\ &\Rightarrow \quad \Delta_{\max }=3 \sqrt{a} \cdot \sqrt{2}=12 \Rightarrow a=(2 \sqrt{2})^{2}=8 \end{aligned} $
Explanation:
$\Delta = {1 \over 2}\left| {\matrix{ 1 & 2 & 1 \cr 7 & 5 & 1 \cr 2 & 3 & 1 \cr } } \right|$
$ = {1 \over 2}[1(5 - 3) - 2(7 - 2) + 1(21 - 10)]$
$ = {1 \over 2}[2 - 10 + 11]$
$\Delta$DEF $ = {1 \over 2}(3) = {3 \over 2}$
$\Delta$ABC = 4$\Delta$DEF $ = 4\left( {{3 \over 2}} \right) = 6$
Explanation:
For minimum $(P R+R Q)$
$R$ lies on $P Q^{\prime}$ (where $Q^{\prime}$ is image of $Q$ in $X$-axis)
$\Rightarrow$ Equation on $P Q^{\prime}$ is
$ 2 x+y+2=0 \Rightarrow R(-1,0) $
$ \therefore $ 50(PR2 + RQ2)
= 50(20 + 5)
= 50(25)
= 1250
Explanation:
${\left( {\sqrt {50} } \right)^2} = {\left( {\sqrt {45} } \right)^2} + {\left( {\sqrt 5 } \right)^2}$
$\angle B = 90^\circ $
Circum-center $ = \left( {{1 \over 2},{{11} \over 2}} \right)$
Mid point of BC $ = \left( {2,{{17} \over 2}} \right)$
Line : $\left( {y - {{11} \over 2}} \right) = 2\left( {x - {1 \over 2}} \right) \Rightarrow y = 2x + {9 \over 2}$
Passing through $\left( {0,{\alpha \over 2}} \right)$
${\alpha \over 2} = {9 \over 2} \Rightarrow \alpha = 9$
Explanation:
$ \Rightarrow $ y2 = ${1 \over {{x^2}}}$
$ \Rightarrow $ y = $ \pm {1 \over x}$
Graph of this equation,
$OA \bot OB$
$ \Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$
$ \Rightarrow {p^2}{q^2} = 1$
$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$ midpoint of AB lies
On ${x^2}{y^2} = 1$
$ \Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$
$ \Rightarrow {(p + q)^2}{(p - q)^2} = 16$
$ \Rightarrow {({p^2} - {q^2})^2} = 16$
$ \Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$
$ \Rightarrow {p^4} \pm 4{p^2} - 1 = 0$
$ \Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5 $
$ \Rightarrow {p^2} = 2 + \sqrt 5 $ or $ - 2 + \sqrt 5 $
$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$ or $ - 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5 $
Area $ = 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5 $
Explanation:
$ \Rightarrow $ Centroid also lies on y-axis.
$ \Rightarrow $ $\sum {\cos \alpha = 0} $
cos$\alpha$ + cos$\beta$ + cos$\gamma$ = 0
$ \Rightarrow $ cos3 $\alpha$ + cos3 $\beta$ + cos3 $\gamma$ = 3cos$\alpha$cos$\beta$cos$\gamma$
$ \therefore $ ${{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}$
$ = {{4({{\cos }^3}\alpha + {{\cos }^3}\beta + {{\cos }^3}\gamma ) - 3(\cos \alpha + \cos \beta + \cos \gamma )} \over {\cos \alpha \cos \beta \cos \gamma }} = 12$
then, ${\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2} = 144$
Explanation:
3x + 4y $ \le $ 100
4x + 3y $ \le $ 75
x $ \ge $ 0, y $ \ge $ 0
Feasible region is shown in the graph
Let maximum value of 6xy + y2 = c
For a solution with feasible region,
6xy + y2 = c and 4x + 3y = 75 must have at least one positive solution.
${y^2} + 6y\left( {{{75 - 3y} \over 4}} \right) - c = 0 $
$\Rightarrow {7 \over 2}{y^2} - {{225} \over 2}y + c = 0$
$ \Rightarrow {\left( {{{225} \over 2}} \right)^2} \ge 4.{7 \over 2}.c $
$\Rightarrow c \le {{{{225}^2}} \over {56}} \approx 904$
${L_1}:x\sqrt 2 + y - 1 = 0$ and ${L_2}:x\sqrt 2 - y + 1 = 0$
For a fixed constant $\lambda$, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is $\lambda$2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is $\sqrt {270} $. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.
The value of $\lambda$2 is __________.
Explanation:
$C:\left| {{{x\sqrt 2 + y - 1} \over {\sqrt 3 }}} \right|\left| {{{x\sqrt 2 - y + 1} \over {\sqrt 3 }}} \right| = {\lambda ^2}$
$ \Rightarrow C:{{\left| {{{(x\sqrt 2 )}^2} - {{(y - 1)}^2}} \right|} \over {\sqrt 3 \times \sqrt 3 }} = {\lambda ^2}$
$ \Rightarrow C:\left| {2{x^2} - {{(y - 1)}^2}} \right| = 3{\lambda ^2}$
Let R $\equiv$ (x1, y1) and S(x2, y2)
$\because$ C cuts y $-$ 1 = 2x at R and S.
So, $\left| {2{x^2} - 4{x^2}} \right| = 3{\lambda ^2}$
$ \Rightarrow x = \pm \sqrt {{3 \over 2}} \left| \lambda \right|$
$\therefore$ $\left| {{x_1} - {x_2}} \right| = \sqrt 6 \left| \lambda \right|$
and $\left| {{y_1} - {y_2}} \right| = 2\left| {{x_1} - {x_2}} \right| = 2\sqrt 6 \left| \lambda \right|$
$\because$ RS2 = 270 (given)
$ \Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270$
$ \Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270$
$ \Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9$
${L_1}:x\sqrt 2 + y - 1 = 0$ and ${L_2}:x\sqrt 2 - y + 1 = 0$
For a fixed constant $\lambda$, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is $\lambda$2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is $\sqrt {270} $. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.
The value of D is __________.
Explanation:
$C:\left| {{{x\sqrt 2 + y - 1} \over {\sqrt 3 }}} \right|\left| {{{x\sqrt 2 - y + 1} \over {\sqrt 3 }}} \right| = {\lambda ^2}$
$ \Rightarrow C:{{\left| {{{(x\sqrt 2 )}^2} - {{(y - 1)}^2}} \right|} \over {\sqrt 3 \times \sqrt 3 }} = {\lambda ^2}$
$ \Rightarrow C:\left| {2{x^2} - {{(y - 1)}^2}} \right| = 3{\lambda ^2}$
Let R $\equiv$ (x1, y1) and S(x2, y2)
$\because$ C cuts y $-$ 1 = 2x at R and S.
So, $\left| {2{x^2} - 4{x^2}} \right| = 3{\lambda ^2}$
$ \Rightarrow x = \pm \sqrt {{3 \over 2}} \left| \lambda \right|$
$\therefore$ $\left| {{x_1} - {x_2}} \right| = \sqrt 6 \left| \lambda \right|$
and $\left| {{y_1} - {y_2}} \right| = 2\left| {{x_1} - {x_2}} \right| = 2\sqrt 6 \left| \lambda \right|$
$\because$ RS2 = 270 (given)
$ \Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270$
$ \Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270$
$ \Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9$
Now, mid-point of RS is $\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right) \equiv (0,1)$ and slope of RS = 2 and slope of $R'S' = {{ - 1} \over 2}$
$\therefore$ Equation of $R'S':y - 1 = - {1 \over 2}x$ i.e. 2y $-$ 2 = $-$x
$ \Rightarrow x + 2y - 2 = 0$
On solving x + 2y $-$ 2 = 0 with C, we get
${x^2} = {{12} \over 7}{\lambda ^2} \Rightarrow \left| {{x_1} - {x_2}} \right| = 2\sqrt {{{12} \over 7}} \left| \lambda \right|$
and $\left| {{y_1} - {y_2}} \right| = {1 \over 2}\left| {{x_1} - {x_2}} \right| = \sqrt {{{12} \over 7}} \left| \lambda \right|$
Hence, $D \equiv {(R'S')^2} = {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2}$
$ = {{12} \over 7} \times 9 \times 5 = {{12 \times 45} \over 7} \approx 77.14$
${1 \over {\sqrt 5 }}$ and ${2 \over {\sqrt 5 }}$ from the lines 4x - 2y + $\alpha $ = 0
and 6x - 3y + $\beta $ = 0, respectively, then the sum of all possible values of $\alpha $ and $\beta $ is :
Explanation:
$4x - 2y + \alpha = 0$
$4x - 2y + 6 = 0$
$\left| {{{\alpha - 6} \over {25}}} \right| = {1 \over {55}}$
$|\alpha - 6|\, = 2 \Rightarrow \alpha = 8,4$
sum = 12
Again
$6x - 3y + \beta = 0$
$6x - 3y + 9 = 0$
$\left| {{{\beta - 9} \over {3\sqrt 5 }}} \right| = {2 \over {\sqrt 5 }}$
$|\beta - 9|\, = 6 \Rightarrow \beta = 15,3$
sum = 18
$ \therefore $ Sum of all values of $\alpha$ and $\beta$ is = 30
Explanation:
P is centroid of the triangle ABC.
P = $\left( {{{1 + 6 + {3 \over 2}} \over 3},{{0 + 2 + 6} \over 3}} \right)$
= $\left( {{{17} \over 6},{8 \over 3}} \right)$
Given Q $\left( { - {7 \over 6}, - {1 \over 3}} \right)$.
$ \therefore $ PQ = $\sqrt {{{\left( {{{24} \over 6}} \right)}^2} + {{\left( {{9 \over 3}} \right)}^2}} $ = 5
Explanation:
Let $P(X, Y)$ is the point in first quadrant.
Now, $2 \leq\left|\frac{x-y}{\sqrt{2}}\right|+\left|\frac{x+y}{\sqrt{2}}\right| \leq 4$
$2 \sqrt{2} \leq|x-y|+|x+y| \leq 4 \sqrt{2}$
Case I : $x \geq y$
$\begin{array}{rlrl}2 \sqrt{2} \leq(x-y)+(x+y) \leq 4 \sqrt{2} \\\\ \Rightarrow x \in[\sqrt{2}, 2 \sqrt{2}]\end{array}$
Case II : $x < y$
$\begin{aligned} 2 \sqrt{2} & \leq y-x+(x+y) \leq 4 \sqrt{2} \\\\ y & \in[\sqrt{2}, 2 \sqrt{2}]\end{aligned}$
$\Rightarrow A=(2 \sqrt{2})^2-(\sqrt{2})^2=6$ sq units
the equation $\left| {\matrix{ {ax - by - c} & {bx + ay} & {cx + a} \cr {bx + ay} & { - ax + by - c} & {cy + b} \cr {cx + a} & {cy + b} & { - ax - by + c} \cr } } \right| = 0$
represents a straight line.
[area $\left( {\Delta {P_1},{P_2},{P_3}} \right)$]/[area $\left( {{P_2},{P_3},{P_4}} \right)$]
