Straight Lines and Pair of Straight Lines
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
$\begin{aligned} & -\frac{\alpha}{2}=\frac{-6}{3} \text { or }-\frac{\alpha}{2}=2 \\ & \Rightarrow \alpha=4 \text { or } \alpha=-4 \\ & P=16+16+\frac{16}{25} \\ & {[P]=\left[32+\frac{16}{25}\right]=32} \end{aligned}$
Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15 x-y=82,6 x-5 y=-4$ and $9 x+4 y=17$. Then $\alpha+2 \beta$ and $2 \alpha-\beta$ are the roots of the equation :
If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$ between the co-ordinates axes, then $\alpha$ is equal to :
Let $C(\alpha, \beta)$ be the circumcenter of the triangle formed by the lines
$4 x+3 y=69$
$4 y-3 x=17$, and
$x+7 y=61$.
Then $(\alpha-\beta)^{2}+\alpha+\beta$ is equal to :
The straight lines $\mathrm{l_{1}}$ and $\mathrm{l_{2}}$ pass through the origin and trisect the line segment of the line L : $9 x+5 y=45$ between the axes. If $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ are the slopes of the lines $\mathrm{l_{1}}$ and $\mathrm{l_{2}}$, then the point of intersection of the line $\mathrm{y=\left(m_{1}+m_{2}\right)}x$ with L lies on :
The combined equation of the two lines $ax+by+c=0$ and $a'x+b'y+c'=0$ can be written as
$(ax+by+c)(a'x+b'y+c')=0$.
The equation of the angle bisectors of the lines represented by the equation $2x^2+xy-3y^2=0$ is :
If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $(\alpha,\beta)$, then the quadratic equation whose roots are $\alpha+4\beta$ and $4\alpha+\beta$, is :
Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y-2 x=2$ such that $\triangle A B C$ is an equilateral triangle. Then, the area of the $\triangle A B C$ is :
A light ray emits from the origin making an angle 30$^\circ$ with the positive $x$-axis. After getting reflected by the line $x+y=1$, if this ray intersects $x$-axis at Q, then the abscissa of Q is :
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 $C=(8,5)$
$\therefore B C^{2}=122$
Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side a such that $a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220$. If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$, where $\alpha \in\left(0, \frac{\pi}{2}\right)$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10$, then $72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$ is equal to :
Let $\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6)$ and $\mathrm{C}\left(\frac{\alpha}{4},-2\right)$ be vertices of a $\triangle \mathrm{ABC}$. If $\left(5, \frac{\alpha}{4}\right)$ is the circumcentre of $\triangle \mathrm{ABC}$, then which of the following is NOT correct about $\triangle \mathrm{ABC}$?
Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q$\left(k_{1}, k_{2}\right)$, then $k_{1}+k_{2}$ is equal to :
The equations of the sides $\mathrm{AB}, \mathrm{BC}$ and CA of a triangle ABC are $2 x+y=0, x+\mathrm{p} y=39$ and $x-y=3$ respectively and $\mathrm{P}(2,3)$ is its circumcentre. Then which of the following is NOT true?
Let $A(1,1), B(-4,3), C(-2,-5)$ be vertices of a triangle $A B C, P$ be a point on side $B C$, and $\Delta_{1}$ and $\Delta_{2}$ be the areas of triangles $A P B$ and $A B C$, respectively. If $\Delta_{1}: \Delta_{2}=4: 7$, then the area enclosed by the lines $A P, A C$ and the $x$-axis is :
A point $P$ moves so that the sum of squares of its distances from the points $(1,2)$ and $(-2,1)$ is 14. Let $f(x, y)=0$ be the locus of $\mathrm{P}$, which intersects the $x$-axis at the points $\mathrm{A}$, $\mathrm{B}$ and the $y$-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to :
Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L_{1}: 3 x-4 y+12=0$, and $L_{2}: 8 x+6 y+11=0$. If $P$ lies below $L_{1}$ and above ${ }{L_{2}}$, then $100(\alpha+\beta)$ is equal to :
A line, with the slope greater than one, passes through the point $A(4,3)$ and intersects the line $x-y-2=0$ at the point B. If the length of the line segment $A B$ is $\frac{\sqrt{29}}{3}$, then $B$ also lies on the line :
Let $\alpha$1, $\alpha$2 ($\alpha$1 < $\alpha$2) be the values of $\alpha$ fo the points ($\alpha$, $-$3), (2, 0) and (1, $\alpha$) to be collinear. Then the equation of the line, passing through ($\alpha$1, $\alpha$2) and making an angle of ${\pi \over 3}$ with the positive direction of the x-axis, is :
The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2y + 1 = 0$ and $2x - y - 1 = 0$ and whose orthocenter is $\left( {{7 \over 3},{7 \over 3}} \right)$ is :
The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle ${\pi \over 4}$ at the origin, is equal to :
Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $-$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :
In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($\alpha$, $\beta$) is the centroid of $\Delta$ABC, then 15($\alpha$ + $\beta$) is equal to :
Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $\Delta$PQR is :
Let the area of the triangle with vertices A(1, $\alpha$), B($\alpha$, 0) and C(0, $\alpha$) be 4 sq. units. If the points ($\alpha$, $-$$\alpha$), ($-$$\alpha$, $\alpha$) and ($\alpha$2, $\beta$) are collinear, then $\beta$ is equal to :
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} $
x cosec $\alpha$ $-$ y sec $\alpha$ = k cot 2$\alpha$ and
x sin$\alpha$ + y cos$\alpha$ = k sin2$\alpha$
respectively, then k2 is equal to :
(a) reflection about the line y = x.
(b) translation through 2 units along the positive direction of x-axis.
(c) rotation through angle ${\pi \over 4}$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point P are $\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right)$, then the value of 2a + b is equal to :
A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of $\Delta$ABC and $\Delta$PQC respectively, such that A1 = 3A2, then the value of m is equal to :
