Solve this question.Chapter name is Section And Mid-Point Formula.
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Let ABC be a triangle then the vertices are
A(-5,4); B(-1,-2); C(5,2)
so,
the distance between A and B is √52 and
the distance between B and C is √52 and
the distance between A and C is √104.
so, AB=BC not equal to AC
the above condition satisfied with isosceles triangle.
AC^2=AB^2+BC^2
(√104)^2=(√52)^2+(√52)^2
104=52+52
104=104
so, it is an isosceles right angle triangle.
A(-5,4); B(-1,-2); C(5,2)
so,
the distance between A and B is √52 and
the distance between B and C is √52 and
the distance between A and C is √104.
so, AB=BC not equal to AC
the above condition satisfied with isosceles triangle.
AC^2=AB^2+BC^2
(√104)^2=(√52)^2+(√52)^2
104=52+52
104=104
so, it is an isosceles right angle triangle.
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