ABC is an isosceles triangle in which A is -1-0- A-2-3- AB-AC and AB is along the x- axis- If BC-4-3 then the equation of the line BC is

The correct option is A x+√(3)y=3 Given,In Isosceles triangle, angles are in the ratio #160;1:1:4 then #160;respective sides are in the ratio ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Lines

ABC is an iso...

Question

$A B C$ is an isosceles triangle in which $A$ is $(- 1, 0), ∠ A = \frac{2 π}{3}, A B = A C$ and $A B$ is along the $x -$ axis. If $B C = 4 \sqrt{3}$ then the equation of the line $B C$ is

x + \sqrt{3} y = 3

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\sqrt{3} x + y = 3

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x + y = \sqrt{3}

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none of these

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Solution

The correct option is A $x + \sqrt{3} y = 3$
Given,In Isosceles triangle, angles are in the ratio $1 : 1 : 4$ then
respective sides are in the ratio $1 : 1 : \sqrt{3}$
therefore lengths of sides $A B$ and $A C$ are $4$ units
Then,coordinates of $B$ become $(3, 0)$
Now equation of line $B C$ is given by $\frac{y - 0}{x - 3} = \frac{- 1}{\sqrt{3}} ⟹ \sqrt{3} y + x = 3$

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ABC is an isosceles triangle in which A is(−1,0),∠A=2π3,AB=AC and AB is along the x− axis. If BC=4√3 then the equation of the line BC is

The correct option is A x+√3y=3Given,In Isosceles triangle, angles are in the ratio 1:1:4 then respective sides are in the ratio 1:1:√3therefore lengths of sides AB and AC are 4 unitsThen,coordinates of B become (3,0)Now equation of line BC is given by y−0x−3=−1√3⟹√3y+x=3

$A B C$ is an isosceles triangle in which $A$ is $(- 1, 0), ∠ A = \frac{2 π}{3}, A B = A C$ and $A B$ is along the $x -$ axis. If $B C = 4 \sqrt{3}$ then the equation of the line $B C$ is