If $a \neq 0$ and the line $2 b x + 3 c y + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y,$ then

$d^{2} + {(2 b - 3 c)}^{2} = 0$
$d^{2} + {(3 b - 2 c)}^{2} = 0$
$d^{2} + {(2 b + 3 c)}^{2} = 0$
$d^{2} + {(3 b + 2 c)}^{2} = 0$

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Show that the perpendicular bisectors of the sides of a triangle are concurrent.

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If the lines $x + q = 0, y - 2 = 0$ and $3 x + 2 y + 5 = 0$ are concurrent, then the value of q will be

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Q. The equation of an altitude of an equilateral triangle is

\sqrt{3} x + y = 2 \sqrt{3}

and one of the vertex is

(3, \sqrt{3})

. Then

The possible number of triangles is 2
Orthocentre of one of the triangle is $(1, \sqrt{3})$
Area of the triangle is $3 \sqrt{3}$ sq. units.
The possible vertices are $(0, 2 \sqrt{3}), (3, - \sqrt{3})$ and (0, 0)

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If $a + b + c = 0$ and $p \neq 0$ the lines $a x + (b + c) y = p$ , $b x + (c + a) y = p$ and $c x + (a + b) y = p$

$D o n o t i n t e r s e c t$
$I n t e r s e c t$
$A r e c o n c u r r e n t$
$N o n e o f t h e s e$

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The number of real values of $λ$ for which the lines $x - 2 y + 3 = 0, λ x + 3 y + 1 = 0$ and $4 x - λ y + 2 = 0$ are concurrent is

1
2
0
Infinite

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The point $(t^{2} + 2 t + 5, 2 t^{2} + t - 2)$ lies on the line $x + y = 2$ for

All real values of $t$
Some real values of $t$
$t = \frac{(- 3 \pm \sqrt{3})}{6}$
No real values of $t$

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Q. If the lines

2 x - p y + 1 = 0, 3 x - q y + 1 = 0

and

4 x - r y + 1 = 0

are concurrent, then

p, q, r

are in

$A . G . P .$
$G . P .$
$H . P .$
$A . P .$

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Q. The point of concurrency of the altitudes drawn from the vertices

A (a t_{1} t_{2}, a (t_{1} + t_{2})), B (a t_{2} t_{3}, a (t_{2} + t_{3}))

and

C (a t_{3} t_{1}, a (t_{3} + t_{1}))

of the triangle

A B C

(where

t_{1} \neq t_{2} \neq t_{3})

$(- a t_{1} t_{2} t_{3}, a (t_{1} + t_{2} + t_{3}))$
$(- a, a (t_{1} + t_{2} + t_{3} + t_{1} t_{2} t_{3}))$
$(- a t_{1} t_{2} t_{3}, a (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}))$
$(0, 0)$

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Q. If

x + y - 2 = 0,

2 x - y + 1 = 0

and

p x + q y - r = 0

are concurrent lines, then the slope of the member in the family of lines

2 p x + 3 q y + 4 r = 0

which is farthest from the origin is

$- 2$
$- \frac{2}{3}$
$- \frac{3}{10}$
$- \frac{1}{2}$

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Q. The point of concurrency of the altitudes drawn from the vertices

A (a t_{1} t_{2}, a (t_{1} + t_{2})), B (a t_{2} t_{3}, a (t_{2} + t_{3}))

and

C (a t_{3} t_{1}, a (t_{3} + t_{1}))

of the triangle

A B C

(where

t_{1} \neq t_{2} \neq t_{3})

$(- a t_{1} t_{2} t_{3}, a (t_{1} + t_{2} + t_{3}))$
$(- a, a (t_{1} + t_{2} + t_{3} + t_{1} t_{2} t_{3}))$
$(0, 0)$
$(- a t_{1} t_{2} t_{3}, a (t_{1} t_{2} + t_{2} t_{3} + t_{3} t_{1}))$

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If the lines $x + a y + a = 0, b x + y + b = 0$ and $c x + c y + 1 = 0$ are concurrent, then write the value of $2 a b c - a b - b c - c a .$

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If the lines $p_{1} x + q_{1} y = 1, p_{2} x + q_{2} y = 1$ and $p_{3} x + q_{3} y = 1$ be concurrent, show that the points $(p_{1}, q_{1}), (p_{2}, q_{2})$ and $p_{3}, q_{3})$ are coolinear.