Straight Lines and Pair of Straight Lines
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 :
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 $
$\Rightarrow c \le {{{{225}^2}} \over {56}} \approx 904$