Maths M.C.Q (1 Marks)

==> $3x - 4y + 3 = 0 \Rightarrow \frac{x}{{ - 1}} + \frac{y}{{3/4}} = 1$

Hence required length is $\sqrt {{{( - 1)}^2} + {{\left( {\frac{3}{4}} \right)}^2}} = \frac{5}{4}$.

Thus the points of intersection of the three lines on the transversal are $\left( {\frac{1}{{1 + {m_1}}},\frac{{{m_1}}}{{1 + {m_1}}}} \right),$ $\left( {\frac{1}{{1 + {m_2}}},\frac{{{m_2}}}{{1 + {m_2}}}} \right)$ and$\left( {\frac{1}{{1 + {m_3}}},\frac{{{m_3}}}{{1 + {m_3}}}} \right)$

By hypothesis, ${\left( {\frac{1}{{1 + {m_1}}} - \frac{1}{{1 + {m_2}}}} \right)^2} + {\left( {\frac{{{m_1}}}{{1 + {m_1}}} - \frac{{{m_2}}}{{1 + {m_2}}}} \right)^2}$

= ${\left( {\frac{1}{{1 + {m_2}}} - \frac{1}{{1 + {m_3}}}} \right)^2} + {\left( {\frac{{{m_2}}}{{1 + {m_2}}} - \frac{{{m_3}}}{{1 + {m_2}}}} \right)^2}$

==> $\frac{{{m_2} - {m_1}}}{{1 + {m_1}}} = \frac{{{m_3} - {m_2}}}{{1 + {m_3}}}$or $\frac{{1 + {m_2}}}{{1 + {m_1}}} - 1 = 1 - \frac{{1 + {m_2}}}{{1 + {m_3}}}$

==> $\frac{{1 + {m_2}}}{{1 + {m_1}}} + \frac{{1 + {m_2}}}{{1 + {m_3}}} = 2$==> $1 + {m_2} = \frac{{2(1 + {m_1})(1 + {m_3})}}{{(1 + {m_1}) + (1 + {m_3})}}$

==> $1 + {m_1},1 + {m_2},1 + {m_3}$ are in $H.P.$

Hence the equation of straight line is $\frac{x}{{14 - b}} + \frac{y}{b} = 1$.

Also, it passes through $(3,4)$

$\therefore $ $\frac{3}{{14 - b}} + \frac{4}{b} = 1 \Rightarrow b = 8$ or $7$

Therefore equations are $4x + 3y = 24$ and $x + y = 7$.

Trick : This question can be checked with the options as the line $4x + 3y = 24$ passes through $(3, 4)$ and also cuts the intercepts from the axes whose sum is $14$.

.${\tan ^{ - 1}}m \pm {\tan ^{ - 1}}m = {\tan ^{ - 1}}\frac{{2m}}{{1 - {m^2}}}$ or ${\tan ^{ - 1}}0$

Therefore equation of lines are $y = 0$, $y = \frac{{2mx}}{{1 - {m^2}}}$.

Therefore the equation of line $AB$ is $\frac{x}{a} + \frac{y}{b} = 1$

==> $\frac{x}{{2{x_1}}} + \frac{y}{{2{y_1}}} = 1 $

$\Rightarrow x{y_1} + y{x_1} = 2{x_1}{y_1}$.

$⇒$$\left| {\frac{{ - m}}{{\sqrt {1 + {m^2}} }}} \right| = \frac{{\sqrt 3 }}{2}$

$⇒$ $m = \pm \sqrt 3 $.

Hence the lines are $\sqrt 3 x + y - \sqrt 3 = 0$ and $\sqrt 3 x - y - \sqrt 3 = 0$.

Hence the required line is $3x - 2y = 30$.

$\therefore $ Line is $y = - \sqrt 3 x + c$

==> $\sqrt 3 x + y = c$

Now $\frac{c}{2} = |4|$

$\Rightarrow c = \pm {\rm{ }}8 $

$\Rightarrow x\sqrt 3 + y = \pm \,8$.

Thus $\tan {45^o} = \pm \frac{{m - \frac{1}{2}}}{{1 + m.\frac{1}{2}}}$ ==> $m = \pm 3$.

Hence option $(b)$ is correct.

$y - a = m{\rm{ }}(x - 0)$ or $mx - y + a = 0$ …..$(i)$

If the length of perpendicular from $(2a, 2a)$ to the line $(i)$ is ‘a’, then $a = \pm\frac{{m(2a) - 2a + a}}{{\sqrt {{m^2} + 1} }} \Rightarrow m = 0,\frac{4}{3}$.

Hence the required equations of lines are $y - a = 0$, $4x - 3y + 3a = 0$.

$\frac{x}{{ - (32/3)}} + \frac{y}{{(24/5)}} = 1$

$ \Rightarrow 9x - 20y + 96 = 0$.

Therefore, the required lines will make angles of ${90^o}$and ${210^o}$ i.e., ${30^o}$with the positive direction of $x$-axis.

Hence the lines are $x = 0$ and $y = \frac{1}{{\sqrt 3 }}x$.

So, $3a + 2b = 13$.....$(i)$

Point $Q(b,a)$is on $4x - y = 5$

So, $4b - a = 5$.....$(ii)$

By solving $(i)$ and $(ii)$, $a = 3,b = 2$

$P(a,b) \to (3,\,2)$and $Q(b,a) \to (2,\,3)$

Now, equation of $PQ$ $y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1}) $

$\Rightarrow y - 2 = \frac{{3 - 2}}{{2 - 3}}(x - 3)$

==> $y - 2 = - (x - 3)$

$\Rightarrow x + y = 5$.

$\frac{x}{{ - 3}} + \frac{y}{2} = 1$....$.(ii)$

According to the condition, $ - \frac{8}{a} = - ( - 3)$

$ \Rightarrow a = - \frac{8}{3}$ and $ - \frac{8}{b} = - (2) \Rightarrow b = 4$.

${b^3} - {c^3} = k(b - c),$ ${c^3} - {a^3} = k(c - a)$,${a^3} - {b^3} = k(a - b)$

==> $b - c = 0$or ${b^2} + {c^2} + bc = k$

==> $c - a = 0$or ${c^2} + {a^2} + ac = k$

==> $a - b = 0$or ${a^2} + {b^2} + ab = k$

==> $b = c,c = a,a = b$ or ${b^2} + {c^2} + bc = {c^2} + {a^2} + ca$

==> ${b^2} - {a^2} = c(a - b)$

$\Rightarrow b = a$ or $a + b + c = 0$.

$y = \lambda $ .....$(i)$

Solving $(i)$ with the cuve $y = \sqrt x $ .....$(ii)$

We get $P({\lambda ^2},\,\,\lambda )$ the point of intersection at $P$

$\therefore $ Slope of $(ii)$ is, m= ${\left( {\frac{{dy}}{{dx}}} \right)_{{\rm{at }}P}} = \frac{1}{{2\lambda }}$

$\therefore $ $(i)$ and $(ii)$ intersect at $P$, at $45^\circ $

$\therefore $ ${\tan ^{ - 1}}\,\left( {\frac{{m - 0}}{{1 + m.0}}} \right) = \pm 45^\circ $.

==> $m = \left( {\frac{1}{{2\lambda }}} \right) = \pm \,1$ ==> $\lambda = \pm \,\frac{1}{2}$

$\therefore $ The equation of line is $y = \frac{1}{2}$ or $y = \frac{{ - 1}}{2}$ but $y = \frac{{ - 1}}{2}$ is not given, hence the required line is $y = \frac{1}{2}$.

then intersecting point $\left( {\frac{{\lambda {x_2} + {x_1}}}{{\lambda + 1}},\,\frac{{\lambda {y_2} + {y_1}}}{{\lambda + 1}}} \right)$ lies on $ax + by + c = 0$.

Thus any point on the line joining $(3,\, - 1)$ and $(8,\,9)$ dividing it in the ratio $\lambda :1$ is $\left( {\frac{{8\lambda + 3}}{{\lambda + 1}},\,\frac{{9\lambda - 1}}{{\lambda + 1}}} \right)$ and if it lies on $y - x + 2 = 0,$ then $\frac{{9\lambda - 1}}{{\lambda + 1}} - \frac{{8\lambda + 3}}{{\lambda + 1}} + 2 = 0$

or $9\lambda - 1 - (8\lambda + 3) + 2(\lambda + 1) = 0$

or $3\lambda - 2 = 0,\,\lambda = \frac{2}{3}$ i.e. ratio is $2:3$.

and equation of $BC$ is $2x + y - 22 = 0$.....$(ii)$

Thus angle between $(i)$ and $(ii)$ is given by

${\tan ^{ - 1}}\frac{{ - \frac{1}{4} + 2}}{{1 + \left( { - \frac{1}{4}} \right){\rm{ }}( - 2)}} = {\tan ^{ - 1}}\frac{7}{6}$.

Let the angle between first and third line is ${\theta _1}$and between second and third is ${\theta _2}$, then

$\tan {\theta _1} = \frac{{3 - m}}{{1 + 3m}}$ and $\tan {\theta _2} = \frac{{m - \frac{1}{2}}}{{1 + \frac{m}{2}}}$

But ${\theta _1} = {\theta _2} \Rightarrow \frac{{3 - m}}{{1 + 3m}} = \frac{{m - \frac{1}{2}}}{{1 + \frac{m}{2}}}$

==> $7{m^2} - 2m - 7 = 0$

==> $m = \frac{{1 \pm 5\sqrt 2 }}{7}$.

 $\sin \theta + \cos \theta - 1 > $as $3 + 2 - 1 > 0$

==> $\sqrt 2 \sin \left( {\theta + \frac{\pi }{4}} \right) > 1$

==> $\sin \left( {\theta + \frac{\pi }{4}} \right) > \frac{1}{{\sqrt 2 }} $

$\Rightarrow 0 < \theta < \frac{\pi }{4}$.

${L_{(3,7)}} = 3 \times 3 - 8 \times 7 - 7 < 0$

Hence $(-1, -1)$ and $(3, 7)$ lie on the same side of line.

Distance between lines $4x - 3y = 5$ and $6y - 8x = 1$ is,

$\beta = \frac{1}{{10}} + \frac{5}{5} = \frac{{11}}{{10}}$

Therefore $\frac{\alpha }{\beta } = \frac{{2\sqrt 2 }}{{11/10}} \Rightarrow 20\sqrt 2 \beta = 11\alpha $.

or $a{\rm{ }}(x + y + 3) + b{\rm{ }}( - 2x + 3y + 4) = 0$, which represents a family of  straight lines through point of intersection of $x + y + 3 = 0$ and $ - 2x + 3y + 4 = 0 $i.e $, (-1, -2).$

Trick :Point $(-1, -2)$ satisfies the given equation of straight line.

==> $ - a + 2b - c = 0 $.

$\Rightarrow 2b = a + c$

==> $a,b,c$ are in $A. P.$

$2x + y = 18$.....$(ii)$

and $4x - 3y = 26$....$(iii)$

are equations of three lines respectively

Solving of equation $(i)$ and $(ii)$, we get $x = 8$ and $y = 2$

Put the values of $x$ and $y$ in equation $(iii)$,

L.H.S $ = 4x - 2y$ $ = 4 \times 8 - 3 \times 2$ $ = 32 - 6$= 26 = R.H.S.

$\therefore $ Point $(8,2)$ lies on line $4x - 3y = 26$, so these three lines are concurrent. Hence, these three equations have one and only one solution.

$\left( {\frac{{{1^2} \times 2 - 1 \times 1 \times 4 + 1}}{{{1^2} + {1^2}}},\frac{{{1^2} \times 4 - 1 \times 1 \times 2 + 1}}{{{1^2} + {1^2}}}} \right) = \,\left( { - \frac{1}{2},\frac{3}{2}} \right)$.

Trick : Here, in options only point $\left( { - \frac{1}{2},\frac{3}{2}} \right)$ is satisfying the given equation of line.

Equation of perpendicular is $x - y + 1 = 0$. Now the coordinates of the foot of the perpendicular are the intersection point of the lines, hence point is $(5, 6)$.

Aliter : Apply the formula given in the theory part of this book, we get required foot as
$\left( {\frac{{{1^2} \times 2 - 1 \times 1 \times 3 - 1 \times ( - 11)}}{{{1^2} + {1^2}}},\frac{{{1^2} \times 3 - 1 \times 1 \times 2 - 1( - 11)}}{{{1^2} + {1^2}}}} \right)$
$ = \left( {\frac{{2 - 3 + 11}}{2},\frac{{3 - 2 + 11}}{2}} \right) = (5,\,6)$.

Now let the image of $A(3,8)$ be $A'({x_1},{y_1}),$then point $(1, 2)$ will be the mid point of $AA'$.

==> $\frac{{{x_1} + 3}}{2} = 1$

$\Rightarrow {x_1} = - 1$ and $\frac{{{y_1} + 8}}{2} = 2$

==> ${y_1} = - 4$.

Hence the image is $(-1, -4)$.

Then the mid-point $R{\rm{ }}\left( {\frac{{a + 4}}{2},\frac{{b - 13}}{2}} \right)$ lies on $5x + y + 6 = 0$.

 $5\left( {\frac{{a + 4}}{2}} \right) + \frac{{b - 13}}{2} + 6 = 0$

$\Rightarrow 5a + b + 19 = 0$ ......$(i)$

Also $PQ$is perpendicular to $5x + y + 6 = 0$.

Therefore $\frac{{b + 13}}{{a - 4}} \times \left( { - \frac{5}{1}} \right) = - 1$

$\Rightarrow a - 5b - 69 = 0$ .....$(ii)$

Solving $(i)$ and $(ii)$, we get $a = - 1,\,\,b = - 14$.

$PQ = \frac{{2 - 6}}{{4 + 2}} = \frac{{ - 2}}{3}$

 The slope of $L = \frac{3}{2}$

Hence the equation of line which passes through point $(1,\,4)$ is $y - 4 = \frac{3}{2}(x - 1)$

==> $3x - 2y + 5 = 0$ .

${x^2} + {y^2} - hx - ky = 0$, which is the required locus of the point.

Let m be the slope of a line inclined at an angle of ${45^o}$ to $AC$, then $\tan {45^o} = \pm \frac{{m - \frac{5}{2}}}{{1 + m.\frac{5}{2}}} \Rightarrow m = - \frac{7}{3},\frac{3}{7}$.

Thus, let the slope of $AB$ or $DC$ be $3/7$ and that of $AD$ or $BC$ be $ - \frac{7}{3}$ . Then equation of $AB$ is $3x - 7y + 19 = 0$.

Also the equation of $BC$ is $7x + 3y - 4 = 0$.

On solving these equations, we get, $B\,\,\,\left( { - \frac{1}{2},\frac{5}{2}} \right)$.

Now let the coordinates of the vertex $D$ be $(h, k)$. Since the middle points of $AC$ and $BD$ are same,

therefore $\frac{1}{2}\left( {h - \frac{1}{2}} \right)\, = \frac{1}{2}(3 + 1) \Rightarrow h = \frac{9}{2}$, $\frac{1}{2}\left( {k + \frac{5}{2}} \right) = \frac{1}{2}(4 - 1)$

==> $k = \frac{1}{2}$. Hence, $D = \left( {\frac{9}{2},\,\frac{1}{2}} \right)$.

Area of $\Delta OAB = \frac{1}{2}\left( {\frac{{2p}}{{\sqrt 3 }}} \right){\rm{ }}2p = \frac{{2{p^2}}}{{\sqrt 3 }}$

By hypothesis $\frac{{2{p^2}}}{{\sqrt 3 }} = \frac{{50}}{{\sqrt 3 }} \Rightarrow p = \pm 5$.
Hence the lines are $\sqrt 3 x + y \pm 10 = 0$.

==> $\lambda = - \frac{{{p^2} + {q^2} - 1}}{{2pq - 1}}$ and the equation of median through $A$ is $(px + qy - 1) - \frac{{{p^2} + {q^2} - 1}}{{2pq - 1}}(qx + py - 1) = 0$

==> $(2pq - 1)(px + qy - 1) = ({p^2} + {q^2} - 1)(qx + py - 1)$.

Therefore the area is $\frac{1}{2}|a \times b| = 6 \Rightarrow |ab| = 12$ …..$(i)$

and hypotenuse is $5$, therefore ${a^2} + {b^2} = 25$ …..$(ii)$

On solving $(i)$ and $(ii)$, we get

$a = \pm 4$or $ \pm 3$and $b = \pm 3$or $ \pm 4$

Hence equation of line is $ \pm \frac{x}{4} \pm \frac{y}{3} = 1$or $ \pm \frac{x}{3} \pm \frac{y}{4} = 1$.

Trick: Check with options. Obviously, the line $\frac{x}{4} + \frac{y}{3} = \pm 1$ satisfies both the conditions.

(Here m is the gradient of line $AC$)

 Equation of line AC is,$y - 1 = \frac{1}{2}(x - 2)$ ==> $y = \frac{x}{2}$.

$\tan 60^\circ = \frac{{AD}}{{BD}}$

==> $\sqrt 3 = \frac{{1\,/\sqrt 5 }}{{BD}}$

==> $BD = \frac{1}{{\sqrt {15} }}$

$BC = 2BD = 2\,/\sqrt {15} $.