Question

the points (2,5),(4,-1),and(6,-7) are vertices of....... triangle​

Answer 1

Answer:

Points ( 2 , 5 ), ( 4 , - 1 ) and ( 6 , - 7 ) are the vertices of an isosceles triangle.

Step-by-step explanation:

Given,

Points ( 2 , 5 ), ( 4 , - 1 ) and ( 6 , - 7 ) are the vertices of a triangle.

We know that triangles are classified into three parts, while talking about the length of their sides.

Here,

We can get the length of sides, using distance formula.

Using Distance Formula : Distance between two points A and B is : $\sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}$, where $(x_1 , y_1) and (x_2 , y_2)$ are the coordinates of the points A and B respectively.

Therefore,

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = $\sqrt{(2-4)^2+\{5-(-1)\}^2}$ unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = $\sqrt{(-2)^2+(5+1)^2}$ unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = $\sqrt{(-2)^2+(6)^2}$ unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = $\sqrt{4+36}$ unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = $\sqrt{40}$ unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = $\sqrt{(4-6)^2+\{-1-(7)\}^2}$ unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = $\sqrt{(-2)^2+(-1+7)^2}$ unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = $\sqrt{(-2)^2+(6)^2}$ unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = $\sqrt{4+36}$ unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = $\sqrt{40}$ unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = $\sqrt{(6-2)^2+(-7-5)^2}$ unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = $\sqrt{(4)^2+(-12)^2}$ unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = $\sqrt{16+144}$ unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = $\sqrt{160}$ unit

Since the length of only two sides is equal, triangle is an isosceles triangle.

Answer 2

Answer:

Points ( 2 , 5 ), ( 4 , - 1 ) and ( 6 , - 7 ) are the vertices of an isosceles triangle.

Step-by-step explanation:

Given,

Points ( 2 , 5 ), ( 4 , - 1 ) and ( 6 , - 7 ) are the vertices of a triangle.

We know that triangles are classified into three parts, while talking about the length of their sides.

Here,

We can get the length of sides, using distance formula.

Using Distance Formula : Distance between two points A and B is : \sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}(x2−x1)2+(y2−y1)2 , where (x_1 , y_1) and (x_2 , y_2)(x1,y1)and(x2,y2) are the coordinates of the points A and B respectively.

Therefore,

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = \sqrt{(2-4)^2+\{5-(-1)\}^2}(2−4)2+{5−(−1)}2 unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = \sqrt{(-2)^2+(5+1)^2}(−2)2+(5+1)2 unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = \sqrt{(-2)^2+(6)^2}(−2)2+(6)2 unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = \sqrt{4+36}4+36 unit

= > Distance between ( 2 , 5 ) & ( 4 , - 1 ) = \sqrt{40}40 unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = \sqrt{(4-6)^2+\{-1-(7)\}^2}(4−6)2+{−1−(7)}2 unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = \sqrt{(-2)^2+(-1+7)^2}(−2)2+(−1+7)2 unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = \sqrt{(-2)^2+(6)^2}(−2)2+(6)2 unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = \sqrt{4+36}4+36 unit

= > Distance between ( 4 , - 1 ) & ( 6 , - 7 ) = \sqrt{40}40 unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = \sqrt{(6-2)^2+(-7-5)^2}(6−2)2+(−7−5)2 unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = \sqrt{(4)^2+(-12)^2}(4)2+(−12)2 unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = \sqrt{16+144}16+144 unit

= > Distance between ( 6 , - 7 ) & ( 2 , 5 ) = \sqrt{160}160 unit

Since the length of only two sides is equal, triangle is an isosceles triangle.