Sum
A(−3, 2), B(3, 2) and C(−3, −2) are the vertices of the right triangle, right angled at A. Show that the mid-point of the hypotenuse is equidistant from the vertices
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Solution
Mid−point of a line = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
Mid−point of BC (D) = `((3 - 3)/2, (2 - 2)/2)`
= `(0/2, 0/2)`
= (0, 0)
Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
AD = `sqrt((0 + 3)^2 + (0 - 2)^2`
= `sqrt(3^2 + (-2)^2`
= `sqrt(9 + 4)`
= `sqrt(13)`
CD = `sqrt((0 + 3)^2 + (0 + 2)^2`
= `sqrt(3^2 + 2^2)`
= `sqrt(9 + 4)`
= `sqrt(13)`
BD = `sqrt((0 - 3)^2 + (0 - 2)^2`
= `sqrt((-3)^2 + (-2)^2`
= `sqrt(9 + 4)`
= `sqrt(13)`
AD = CD = BD = `sqrt(13)`
Mid-point of BC is equidistance from the centre.
Concept: The Mid-point of a Line Segment (Mid-point Formula)
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