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The equations of two equal sides AB and AC of an isosceles $△ ABC$ are $x + y = 5$ and $7 x - y = 3$ respectively. Then the equations of the side BC, if area of $△ ABC = 5$ square units, can be

(The question has multiple correct options)

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x - 3 y + 1 = 0

x - 3 y - 21 = 0

3 x + y + 2 = 0

3 x + y - 12 = 0

answer is A.

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Detailed Solution

Given isosceles

△ ABC

in which,
AB=AC
Here, side BC is perpendicular to bisectors of

∠ BAC

.
Here, the equations of the bisectors of lines AB and AC is given as,

\frac{(x + y - 5)}{\sqrt{2}} = \pm \frac{(7 x - y - 3)}{5 \sqrt{2}}

\Rightarrow x - 3 y + 11 = 0

3 x + y - 7 = 0

Suppose that, AD= altitude from vertex A.
When equation of AD is

3 x + y - 7 = 0

, then equation of the side BC can be given as,

x - 3 y + λ = 0

Suppose, θ is the angle between AD and AC.
So,

\tan θ = |\frac{- 3 + 1}{1 + 3}| = \frac{1}{2}

Now,

AD = λ

, Here, the area of

ΔABC

= \frac{1}{2} \times [BC]

= \frac{1}{2} \times λ

= 2 λ |\tan θ|

= λ^{2} | \tan θ |

.
So,

λ^{2} \times \frac{1}{2} = 5

\Rightarrow λ^{2} = 10

\Rightarrow (11 - λ)^{2} = 100

\Rightarrow 11 - λ = \pm

\Rightarrow λ = 1, 21

So, the equation of BC is

x - 3 y + 1 = 0

x - 3 y + 21 = 0

.
Now, when the equation of BC =

3 x + y + k = 0

Then the equation of AD =

x - 3 y + 11 = 0

,
So,,

|\tan θ| = |\frac{7 - \frac{1}{2}}{1 + \frac{7}{3}}| = 2

\Rightarrow λ^{2} |\tan θ| = 2 λ^{2} = 5

\Rightarrow λ^{2} = \frac{5}{2}

Also,

\frac{5}{2} = \frac{(3 + 4 + λ)^{2}}{10}

\Rightarrow 7 + λ = \pm 5

\Rightarrow λ = 2, - 12

So, we get the equation of BC is

3 x + y + 2 = 0

3 x + y - 12 = 0

.
Here total 4 equations of side BC we get as,

x - 3 y + 1 = 0, x - 3 y - 21 = 0, 3 x + y + 2 = 0

3 x + y - 12 = 0

.
So, option 1, 2, 3, 4 are correct.

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