Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

2 thoughts on “Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.”

1. Answer: yes it is an isosceles triangle

   Step-by-step explanation:

   Let the vertices be A(5,-2); B(6,4) and C(7,-2)

   For the triangle to be an isosceles triangle, 2 of the sides should be equal to each other in magnitude

   This implies that either AB = BC (or) BC = AC (or) AB = AC

   We first need to find the length of the sides

   Formula for finding the length of the side is

   [tex]\sqrt{ (x2 – x1)^{2} + (y2 – y1)^{2} }[/tex]

   Where (x1, y1) and (x2, y2) are the points between which the distance is to be found

   Substituting values of the vertices and solving, we get

   AB = [tex]\sqrt{ (6-5)^{2} + (-4-2)^{2} }[/tex]

   => AB = [tex]\sqrt{ 1^{2} +6^{2} } = \sqrt{1+36} = \sqrt{37}[/tex]

   => AB = [tex]\sqrt{37}[/tex]

   Similarly while solving by substituting, we get

   BC = [tex]\sqrt{37}[/tex]

   And

   AC = [tex]\sqrt{2^{2}} = 2[/tex]

   Therefore, AB = BC

   Thus ABC is an isosceles triangle with AB = BC

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2. Answer:

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   Step-by-step explanation:

   Let ABC be the triangle where A(5,−2),B(6,4),C(7,−2)

   [tex]AB = \sqrt{(6 – 5) {}^{2} + (4 + 2) {}^{2} } = \sqrt{1 + 36} = \sqrt{37} \\ \\ BC = \sqrt{(7 – 6) {}^{2} + ( – 2 – 4) {}^{2} } = \sqrt{1 + 36} = \sqrt{37} \\ \\ AC = \sqrt{(7 – 5) {}^{2} + ( – 2 + 2) {}^{2} } = \sqrt{4 + 0} = 2[/tex]

   Here AB = BC

   Hence (5,−2),(6,4) & (7,−2) are vertices of isosceles triangle

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