Question

An equilateral triangle has two vertices at the points (3, 4) and (-2, 3), find the coordinates of the third vertex.

Answer 1

Answer:

Let 3rd vertex of an equilatral triangle is (x, y)

Let 3rd vertex of an equilatral triangle is (x, y)we know,

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+9

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=12

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)again ,

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)again ,(x+2)^2+(y-3)^2=25+1

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)again ,(x+2)^2+(y-3)^2=25+1x^2+y^2+4x-6y=13--------------(2)

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)again ,(x+2)^2+(y-3)^2=25+1x^2+y^2+4x-6y=13--------------(2)solve equation (1) and (2)

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)again ,(x+2)^2+(y-3)^2=25+1x^2+y^2+4x-6y=13--------------(2)solve equation (1) and (2)x= {1+root3}/2 and {1-root3 }/2

Let 3rd vertex of an equilatral triangle is (x, y)we know, all side of equilatral triangle are same so,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2=(5)^2+(1)^2now,(x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2-6x-8y+9+16=4x-6y+4+910x+2y=125x+y=6 -----------------(1)again ,(x+2)^2+(y-3)^2=25+1x^2+y^2+4x-6y=13--------------(2)solve equation (1) and (2)x= {1+root3}/2 and {1-root3 }/2and y=6-5x

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