Question

point a lies at (2,4) and point b lies at (3,6). which point in needed so that \\( \triangle abc \\) is isosceles? (1 point)
\\( \bigcirc \\) c (3,4)
\\( \bigcirc \\) c (5.4)
\\( \bigcirc \\) c (0,4)
\\( \bigcirc \\) c (5,5)

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 \(C(3,4)\)

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

Step3: Calculate distance for \(C(5,4)\)

For \(A(2,4)\) and \(C(5,4)\):
\(d_{AC}=\sqrt{(5 - 2)^2+(4 - 4)^2}=\sqrt{9+0}=3\)
For \(B(3,6)\) and \(C(5,4)\):
\(d_{BC}=\sqrt{(5 - 3)^2+(4 - 6)^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\)

Step4: Calculate distance for \(C(0,4)\)

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

Step5: Calculate distance for \(C(5,5)\)

For \(A(2,4)\) and \(C(5,5)\):
\(d_{AC}=\sqrt{(5 - 2)^2+(5 - 4)^2}=\sqrt{9+1}=\sqrt{10}\)
For \(B(3,6)\) and \(C(5,5)\):
\(d_{BC}=\sqrt{(5 - 3)^2+(5 - 6)^2}=\sqrt{4 + 1}=\sqrt{5}\)
Since \(d_{AB}=\sqrt{5}\) and \(d_{BC}=\sqrt{5}\), \(\triangle ABC\) is isosceles when \(C(5,5)\)

Answer:

C (5,5)