Question

The slope of the base of an equilateral triangle is -1 and the vertex is the point ( 2 , -1 ). Find the equation of the remaining sides.
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Answer 1

Answer: The base of an equilateral triangle is x + y - 2 = 0 and the opposite vertex is (2, -1). Find the equation of the remaining sides.

Step-by-step explanation:

all u do is divied by 2( -1

Answer 2

Answer:

Solution Given:

let ABC be an equilateral triangle with the vertex A(2,-1) and slope =-1.

and

∡ABC=∡BAC=∡ACB=60°

slope of BC$[ m_1]=-1$

we have

$\theta$=60°

Slope of AB=$[ m_2]=a$

now

we have

angle between two lines is

$Tan\theta =±\frac{m_1-m_2}{1+m_1m_2}$

now substituting value

tan 60°= ± $\frac{-1-a}{1+-1*a}$

$\sqrt{3}=±\frac{-1-a}{1-a}$

doing criss-cross multiplication;

$(1-a)\sqrt{3}=±-(1+a)$

$\sqrt{3}-a\sqrt{3}$=±(1+a)

taking positive

$-a-a\sqrt{3}=1-\sqrt{3}$

$-a(1+\sqrt{3})=1-\sqrt{3}$

a=$\frac{1-\sqrt{3}}{-(1+\sqrt{3})}=\frac{\sqrt{3}-1}{1+\sqrt{3}}$

taking negative

$a-a\sqrt{3}=-1-\sqrt{3}$

$a(1-\sqrt{3})=-1-\sqrt{3}$

a=$\frac{-(1+\sqrt{3})}{(1-\sqrt{3})}=\frac{\sqrt{3}+1}{-1+\sqrt{3}}$

Equation of a line when a =$\frac{\sqrt{3}-1}{1+\sqrt{3}}$

and passing through (2,-1),we have

$y-y_1=m(x-x_1)$

y+1=$\frac{\sqrt{3}-1}{1+\sqrt{3}}$(x-2)

y=$\frac{\sqrt{3}-1}{1+\sqrt{3}}x-2\frac{\sqrt{3}-1}{1+\sqrt{3}}$-1

$\boxed{\bold{\green{y=\frac{\sqrt{3}-1}{1+\sqrt{3}}x-5+2\sqrt{3}}}}$ is a first side equation of line.

again

Equation of a line when a =$\frac{\sqrt{3}+1}{-1+\sqrt{3}}$

and passing through (2,-1),we have

$y-y_1=m(x-x_1)$

y+1=$\frac{\sqrt{3}+1}{-1+\sqrt{3}}$(x-2)

y+1=$\frac{\sqrt{3}+1}{-1+\sqrt{3}}x -2\frac{\sqrt{3}+1}{-1+\sqrt{3}}$

y=$\frac{\sqrt{3}+1}{-1+\sqrt{3}}x -2\frac{\sqrt{3}+1}{-1+\sqrt{3}}-1$

$\boxed{\bold{\blue{y=\frac{\sqrt{3}+1}{-1+\sqrt{3}}x -5+2\sqrt{3}}}}$ is another equation of line.