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 equation of line BC isA- x-3 y-3-0B- -3 x-y-3-0C- x-y-3D- -3 x-y-3-0

The correct option is A x +√3y – 3 = 0 In ALB, BL=ABcosπ/6 2√(3)=AB×√(3)/2 AB=4⇒OB=2 ∴B=(3,0) Clearly∠ XBC=5 π/6∴ BC is y=tan5 π/6(x 3) ∴x+√(3)y 3=0 ...

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Standard XII

Mathematics

Equation of Line: Symmetrical Form

ABC is an iso...

Question

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

x +√3y – 3 = 0

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√3x + y – 3 = 0

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x + y =√3

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√3x – y + 3 = 0

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Solution

The correct option is A
x +√3y – 3 = 0

$I n △ A L B, B L = A B c o s \frac{π}{6} 2 \sqrt{3} = A B \times \sqrt{3} / 2 A B = 4 \Rightarrow O B = 2 ∴ B = (3, 0) Clearly ∠ X B C = \frac{5 π}{6} ∴ B C i s y = t a n \frac{5 π}{6} (x - 3) ∴ x + \sqrt{3} y - 3 = 0$

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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 equation of line BC is

The correct option is A x +√3y – 3 = 0 In△ALB,BL=ABcosπ62√3=AB×√3/2AB=4⇒OB=2∴B=(3,0)Clearly∠XBC=5π6∴ BC is y=tan5π6(x−3)∴x+√3y−3=0

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

The correct option is A
x +√3y – 3 = 0

$I n △ A L B, B L = A B c o s \frac{π}{6} 2 \sqrt{3} = A B \times \sqrt{3} / 2 A B = 4 \Rightarrow O B = 2 ∴ B = (3, 0) Clearly ∠ X B C = \frac{5 π}{6} ∴ B C i s y = t a n \frac{5 π}{6} (x - 3) ∴ x + \sqrt{3} y - 3 = 0$