Answer :
Answer: ABC is isosceles
Step-by-step 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 = [tex]\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1}) ^{2}}[/tex]
Calculating AB first , we have
[tex]\sqrt{(-1-1)^{2}+(-2+4)^{2} }[/tex]
= [tex]\sqrt{4+4}[/tex]
=[tex]\sqrt{8}[/tex]
Therefore:
AB = [tex]\sqrt{8}[/tex]
Calculating BC , we have :
D = [tex]\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1}) ^{2}}[/tex]
D = [tex]\sqrt{(-2+1)^{2}+(-5+2)^{2} }[/tex]
D = [tex]\sqrt{10}[/tex]
Therefore:
BC = [tex]\sqrt{10}[/tex]
Calculating AC , we have :
D = [tex]\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1}) ^{2}}[/tex]
D = [tex]\sqrt{(-2-1)^{2}+(-5+4)^{2} }[/tex]
D = [tex]\sqrt{10}[/tex]
Therefore:
AC = [tex]\sqrt{10}[/tex]
Since AC = BC then triangle ABC is isosceles
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