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Byju's Answer

Standard XII

Mathematics

Pair of Lines

Two parallel ...

Question

Two parallel lines $l_{1} & l_{2}$ such that the distance between them is $\frac{6}{\sqrt{5}}$ units and origin lies in between the lines. Equation of $l_{1} : x - 2 y - 3 = 0$ . Another line $l_{3} : x + 3 y + 2 = 0$ intersect with $l_{1}$ at $A$ . Image of $l_{3}$ in $l_{1}$ is $l_{4}$ , meets with $l_{2}$ at $B$ . Image of $l_{4}$ in the line $l_{2}$ meets with the line $l_{1}$ again at $C$ and another point $D (h, k)$ is in the $x y$ plane

Equation of the circle passing through

A, B, C

(5 x - 17)^{2} + (5 y - 1)^{2} = 180

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Equation of the circle passing through

A, B, C

(x - 10)^{2} + (y - 2)^{2} = 90

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A B C D

will be a square if

(5 h, 5 k) \equiv (23, - 11)

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Equation of the line

l

for which

l_{4}

is angle bisector of

l_{2} & l

2 x + y - 7 = 0

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Solution

The correct options are
A Equation of the circle passing through $A, B, C$ is $(5 x - 17)^{2} + (5 y - 1)^{2} = 180$
C $A B C D$ will be a square if $(5 h, 5 k) \equiv (23, - 11)$
D Equation of the line $l$ for which $l_{4}$ is angle bisector of $l_{2} & l$ is $2 x + y - 7 = 0$
The coordinates of the $A (1, - 1)$

Angle between $l_{1} & l_{3}$ is
$tan θ = ∣ ∣ ∣ ∣ \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \frac{1}{3}} ∣ ∣ ∣ ∣ = 1 \Rightarrow θ = \frac{π}{4}$
$∴ l_{3} ⊥ l_{4}$ and passes through $A$
$∴$ equation of the line $l_{4} : 3 x - y - 4 = 0$
Distance between $l_{1} & l_{2}$ is $6$ units and origin lies in between them
$∴$ equation of $l_{2} : x - 2 y + 3 = 0$
Equating $l_{4} & l_{2}$
$∴ B (\frac{11}{5}, \frac{13}{5})$
$∵ l_{1} & l_{2}$ are parallel to each other so angle between $l_{4} & l_{2}$ is $\frac{π}{4}$
$∴$ image of the line $l_{4}$ in $l_{2}$ is perpendicular to $l_{4}$ and parallel to $l_{3}$

$∴ Δ A B C$ is isosceles rightangle triangle and right angle at $B$ .
Equation of the line $⊥ l_{4}$ and passing through $B$ is $x + 3 y - 10 = 0$
$∴$ coordinates of $C \equiv (\frac{29}{5}, \frac{7}{5})$
mid point of $A C$ is $(\frac{17}{5}, \frac{1}{5})$
lenght of $A C = \frac{12}{\sqrt{10}}$
Equation of the circle containing points $A, B, C$ is
$(x - \frac{17}{5})^{2} + (y - \frac{1}{5})^{2} = \frac{36}{5} \Rightarrow (5 x - 17)^{2} + (5 y - 1)^{2} = 180$
Equation of $l$
$∵$ $l$ passes through $B$ and $⊥ l_{2}$
$∴ 2 x + y - 7 = 0$
For $A B C D$ to be square the coordinate of the $D$ should be intersection of the $l & l_{2}$
$∴ D \equiv (\frac{23}{5}, - \frac{11}{5})$

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