Question

point a lies at (2,4) and point b lies at (3,6). which point in needed so that △abc is isosceles? (1 point)
c (5,4)
c (0,4)
c (5,5)
c (3,4)

Explanation:

Step1: Calculate distance between two points

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(A(2,4)\) and \(B(3,6)\):
\(d_{AB}=\sqrt{(3 - 2)^2+(6 - 4)^2}=\sqrt{1 + 4}=\sqrt{5}\)

Step2: Calculate distance for each option

• For \(C(5,4)\) and \(A(2,4)\):

\(d_{AC}=\sqrt{(5 - 2)^2+(4 - 4)^2}=\sqrt{9+0}=3\)
For \(C(5,4)\) and \(B(3,6)\):
\(d_{BC}=\sqrt{(5 - 3)^2+(4 - 6)^2}=\sqrt{4 + 4}=2\sqrt{2}\)

• For \(C(0,4)\) and \(A(2,4)\):

\(d_{AC}=\sqrt{(0 - 2)^2+(4 - 4)^2}=\sqrt{4+0}=2\)
For \(C(0,4)\) and \(B(3,6)\):
\(d_{BC}=\sqrt{(0 - 3)^2+(4 - 6)^2}=\sqrt{9 + 4}=\sqrt{13}\)

• For \(C(5,5)\) and \(A(2,4)\):

\(d_{AC}=\sqrt{(5 - 2)^2+(5 - 4)^2}=\sqrt{9+1}=\sqrt{10}\)
For \(C(5,5)\) and \(B(3,6)\):
\(d_{BC}=\sqrt{(5 - 3)^2+(5 - 6)^2}=\sqrt{4+1}=\sqrt{5}\)

• For \(C(3,4)\) and \(A(2,4)\):

\(d_{AC}=\sqrt{(3 - 2)^2+(4 - 4)^2}=\sqrt{1+0}=1\)
For \(C(3,4)\) and \(B(3,6)\):
\(d_{BC}=\sqrt{(3 - 3)^2+(4 - 6)^2}=\sqrt{0 + 4}=2\)

Answer:

C. \(C(5,5)\)