Fund the Cirucmcenter of the triangle whose sides are x+y+2=0, 5x-y-2=0, x-2y+5=0

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Fund the Cirucmcenter of the triangle whose sides are x+y+2=0, 5x-y-2=0, x-2y+5=0

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Parker 2 months 2021-10-04T22:02:38+00:00 1 Answer 0 views 0

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    2021-10-04T22:03:59+00:00

    Let us find the three vertices of the triangle.

    Find value of y in x+y+2=0 and substitute it in 5x-y-2=0 and x-2y+5=0 each.

    So,

    \longrightarrow x+y+2=0

    \longrightarrow y=-x-2

    And in,

    \longrightarrow 5x-y-2=0

    substituting value of y,

    \longrightarrow 5x-(-x-2)-2=0

    \longrightarrow 5x+x+2-2=0

    \longrightarrow6x=0

    \longrightarrow x=0

    And then,

    \longrightarrow y=-x-2

    \longrightarrow y=-0-2

    \longrightarrow y=-2

    Thus (0, -2) is a vertex.

    And in,

    \longrightarrow x-2y+5=0

    \longrightarrow x-2(-x-2)+5=0

    \longrightarrow x+2x+4+5=0

    \longrightarrow 3x+9=0

    \longrightarrow x=-3

    And then,

    \longrightarrow y=-x-2

    \longrightarrow y=-(-3)-2

    \longrightarrow y=1

    Thus (-3, 1) is a vertex.

    Consider,

    \longrightarrow 5x-y-2=0

    \longrightarrow y=5x-2

    And in,

    \longrightarrow x-2y+5=0

    substituting value of y,

    \longrightarrow x-2(5x-2)+5=0

    \longrightarrow x-10x+4+5=0

    \longrightarrow -9x+9=0

    \longrightarrow x=1

    And then,

    \longrightarrow y=5x-2

    \longrightarrow y=5(1)-2

    \longrightarrow y=3

    Thus (1, 3) is a vertex of the triangle.

    So (0, -2), (-3, 1) and (1, 3) are vertices of the triangle.

    Let (a, b) be the coordinates of the circumcenter of the triangle, which is equidistant from the points (0, -2), (-3, 1) and (1, 3) each.

    Then by distance formula,

    \longrightarrow\sqrt{(a-0)^2+(b+2)^2}=\sqrt{(a+3)^2+(b-1)^2}

    \longrightarrow a^2+b^2+4b+4=a^2+6a+9+b^2-2b+1

    \longrightarrow 4b+4=6a-2b+10

    \longrightarrow b=a+1\quad\quad\dots(1)

    And,

    \longrightarrow\sqrt{(a-0)^2+(b+2)^2}=\sqrt{(a-1)^2+(b-3)^2}

    \longrightarrow a^2+b^2+4b+4=a^2-2a+1+b^2-6b+9

    \longrightarrow 4b+4=-2a-6b+10

    \longrightarrow 10b=-2a+6

    From (1),

    \longrightarrow 10(a+1)=-2a+6

    \longrightarrow 10a+10=-2a+6

    \longrightarrow a=-\dfrac{1}{3}

    And then,

    \longrightarrow b=a+1

    \longrightarrow b=-\dfrac{1}{3}+1

    \longrightarrow b=\dfrac{2}{3}

    Hence the circumcenter of the triangle is,

    \longrightarrow\underline{\underline{(a,\ b)=\left(-\dfrac{1}{3},\ \dfrac{2}{3}\right)}}

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