1.

Show that A(6,4),B(5,-2) and C(7,-2)are the vertices of an isosceles triangle​

Answer»

POINTS given:- A(6,4), B(5,-2) and C(7,-2)

\longrightarrow An isosceles TRIANGLE is the ONE in which two sides are equal.

Here, we will find the DISTANCE between two points one by one. If the two distance are equal, then it is an isosceles triangle​. Else, not.

Distance formula - \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

1. Distance between AB

Here,

x_1 = 6, \:\:\:\:\:\:y_1=4,\\x_2 = 5, \:\:\:\:\:\:y_2=-2

Substituting the values in the formula.

=> \sqrt{(5-6)^2+(-2-4)^2} UNITS

=> \sqrt{(-1)^2+(-6)^2} units

=> \sqrt{1+36} units

=> \sqrt{37} units

2. Distance between BC

Here,

x_1 = 5, \:\:\:\:\:\:y_1=-2,\\x_2 = 7, \:\:\:\:\:\:y_2=-2

Substituting the values in the formula.

=> \sqrt{(7-2)^2+(-2-(-2))^2} units

=> \sqrt{(2)^2+(-2+2)^2} units

=> \sqrt{(2)^2+(0)^2} units

=> \sqrt{4+0} units

=> \sqrt{4} units

=> 2 units

3. Distance between AC

Here,

x_1 = 6, \:\:\:\:\:\:y_1=4,\\x_2 = 7, \:\:\:\:\:\:y_2=-2

Substituting the values in the formula.

=> \sqrt{(7-6)^2+(-2-4)^2} units

=> \sqrt{(1)^2+(-6)^2} units

=> \sqrt{1+36} units

=> \sqrt{37} units

Here, distance between AB and AC are equal, but distance between BC is odd. Therefore, A(6,4), B(5,-2) and C(7,-2) are the vertices of an isosceles triangle​.


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