Question

4. Given ABC with vertices at A(1,-4), B(-1,-2), and C(-2,-5), prove or disprove that ABC is isosceles

Answer 1

Practice Analytic Geometry

1. Find the midpoint of the line segment defined by the points: (sqrt 2, 0) and (0, -sqrt 2)

Answer is ((sqrt 2)/2, (-sqrt2)/2)

2. Find the distance of the line segment joining the two points: (5,4) and (-2,1)

Answer is sqrt 58

3. Find the distance of the line segment joining the two points: (1/2, -5/2) and (-4/3, -1/6)

Answer is (sqrt 317)/6

4. Given triangle ABC with vertices at A (1,-4), B (-1, -2), and C (-2,-5), prove or disprove that triangle ABC is an isosceles

Answer triangle ABC is isosceles since AB = sqrt 8, BC = sqrt 10, and AC = sqrt 10

Answer 2

Answer: ABC is isosceles

Explanation:

The coordinate of points given are:

A(1,-4)

B(-1, -2 )

C ( -2 , -5 )

For ABC to be isosceles , then AB must be equal to BC or AB = AC

The next thing to do is to find the value of AB , BC and AC using the formula for finding distance between two points which is given by:

D =
`\sqrt{(x_(2)-x_(1))  ^(2)+(y_(2)-y_(1))  ^(2)}`

Calculating AB first , we have

`\sqrt{(-1-1)^(2)+(-2+4)^(2)  }`

=
`√(4+4)`

=
`√(8)`

Therefore:

AB =
`√(8)`

Calculating BC , we have :

D =
`\sqrt{(x_(2)-x_(1))  ^(2)+(y_(2)-y_(1))  ^(2)}`

D =
`\sqrt{(-2+1)^(2)+(-5+2)^(2)  }`

D =
`√(10)`

Therefore:

BC =
`√(10)`

Calculating AC , we have :

D =
`\sqrt{(x_(2)-x_(1))  ^(2)+(y_(2)-y_(1))  ^(2)}`

D =
`\sqrt{(-2-1)^(2)+(-5+4)^(2)  }`

D =
`√(10)`

Therefore:

AC =
`√(10)`

Since AC = BC then triangle ABC is isosceles