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

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 =±(m_1-m_2)/(1+m_1m_2)`

now substituting value

tan 60°= ±
`(-1-a)/(1+-1*a)`

`√(3)=±(-1-a)/(1-a)`

doing criss-cross multiplication;

`(1-a)√(3)=±-(1+a)`

`√(3)-a√(3)`=±(1+a)

taking positive

`-a-a√(3)=1-√(3)`

`-a(1+√(3))=1-√(3)`

a=
`(1-√(3))/(-(1+√(3)))=(√(3)-1)/(1+√(3))`

taking negative

`a-a√(3)=-1-√(3)`

`a(1-√(3))=-1-√(3)`

a=
`(-(1+√(3)))/((1-√(3)))=(√(3)+1)/(-1+√(3))`

Equation of a line when a =
`(√(3)-1)/(1+√(3))`

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

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

y+1=
`(√(3)-1)/(1+√(3))`(x-2)

y=
`(√(3)-1)/(1+√(3))x-2(√(3)-1)/(1+√(3))`-1

`\boxed{\bold{\green{y=(√(3)-1)/(1+√(3))x-5+2√(3)}}}` is a first side equation of line.

again

Equation of a line when a =
`(√(3)+1)/(-1+√(3))`

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

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

y+1=
`(√(3)+1)/(-1+√(3))`(x-2)

y+1=
`(√(3)+1)/(-1+√(3))x -2(√(3)+1)/(-1+√(3))`

y=
`(√(3)+1)/(-1+√(3))x -2(√(3)+1)/(-1+√(3))-1`

`\boxed{\bold{\blue{y=(√(3)+1)/(-1+√(3))x -5+2√(3)}}}` is another equation of line.