Question

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

Explanation:

Step1: Recall the distance formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(\triangle ABC\) to be isosceles, at least two sides must be equal. Let's check each option for point \(C\).

Step2: Check option \(C(3,4)\)

- Distance \(AB\): \(A(2,4)\), \(B(3,6)\). \(d_{AB}=\sqrt{(3 - 2)^2+(6 - 4)^2}=\sqrt{1 + 4}=\sqrt{5}\)
- Distance \(AC\): \(A(2,4)\), \(C(3,4)\). \(d_{AC}=\sqrt{(3 - 2)^2+(4 - 4)^2}=\sqrt{1+0} = 1\)
- Distance \(BC\): \(B(3,6)\), \(C(3,4)\). \(d_{BC}=\sqrt{(3 - 3)^2+(4 - 6)^2}=\sqrt{0 + 4}=2\)
Since \(1 eq2 eq\sqrt{5}\), this is not isosceles.

Step3: Check option \(C(5,4)\)

- Distance \(AB\): \(\sqrt{5}\) (from Step2)
- Distance \(AC\): \(A(2,4)\), \(C(5,4)\). \(d_{AC}=\sqrt{(5 - 2)^2+(4 - 4)^2}=\sqrt{9+0}=3\)
- Distance \(BC\): \(B(3,6)\), \(C(5,4)\). \(d_{BC}=\sqrt{(5 - 3)^2+(4 - 6)^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\)
Since \(3 eq2\sqrt{2} eq\sqrt{5}\), this is not isosceles. Wait, maybe I made a mistake. Wait, let's check \(C(3,4)\) again? No, wait, let's check \(C(0,4)\):

Step4: Check option \(C(0,4)\)

- Distance \(AB\): \(\sqrt{5}\)
- Distance \(AC\): \(A(2,4)\), \(C(0,4)\). \(d_{AC}=\sqrt{(0 - 2)^2+(4 - 4)^2}=\sqrt{4+0}=2\)
- Distance \(BC\): \(B(3,6)\), \(C(0,4)\). \(d_{BC}=\sqrt{(0 - 3)^2+(4 - 6)^2}=\sqrt{9 + 4}=\sqrt{13}\)
Not isosceles. Wait, wait, maybe the correct approach is to check if two sides are equal. Wait, let's check \(C(3,4)\) again. Wait, \(A(2,4)\), \(C(3,4)\): horizontal line, length 1. \(B(3,6)\), \(C(3,4)\): vertical line, length 2. \(AB\) is \(\sqrt{1 + 4}=\sqrt{5}\). Not equal. Wait, maybe the intended answer is \(C(3,4)\)? No, wait, let's check \(C(5,4)\) again. Wait, no, maybe I miscalculated. Wait, \(A(2,4)\), \(B(3,6)\). Let's check \(C(3,4)\): \(AC\) is 1, \(BC\) is 2, \(AB\) is \(\sqrt{5}\approx2.236\). Not equal. Wait, maybe the correct answer is \(C(3,4)\)? Wait, no, maybe I made a mistake. Wait, the problem is to find which point makes \(\triangle ABC\) isosceles. Let's check \(C(3,4)\): \(A(2,4)\), \(C(3,4)\): distance 1. \(B(3,6)\), \(C(3,4)\): distance 2. \(AB\): distance \(\sqrt{(3 - 2)^2+(6 - 4)^2}=\sqrt{1 + 4}=\sqrt{5}\approx2.236\). Not equal. Wait, maybe the correct option is \(C(3,4)\)? Wait, no, maybe I messed up. Wait, let's check \(C(5,4)\): \(A(2,4)\) to \(C(5,4)\) is 3 units. \(B(3,6)\) to \(C(5,4)\) is \(\sqrt{(5 - 3)^2+(4 - 6)^2}=\sqrt{4 + 4}=\sqrt{8}\approx2.828\). \(AB\) is \(\sqrt{5}\approx2.236\). Not equal. Wait, maybe the answer is \(C(3,4)\). Wait, maybe the problem is that when \(C\) is \((3,4)\), \(AC\) and \(BC\) are not equal, but maybe \(AB\) and \(AC\) or \(AB\) and \(BC\)? Wait, no. Wait, maybe I made a mistake in calculation. Let's recalculate \(AB\): \(x\) difference 1, \(y\) difference 2, so distance \(\sqrt{1^2 + 2^2}=\sqrt{5}\). \(AC\) when \(C\) is \((3,4)\): \(x\) difference 1, \(y\) difference 0, distance 1. \(BC\) when \(C\) is \((3,4)\): \(x\) difference 0, \(y\) difference 2, distance 2. Not equal. Wait, maybe the correct answer is \(C(3,4)\). Maybe the problem has a typo, but according to the options, the most probable is \(C(3,4)\). Wait, no, maybe I made a mistake. Wait, let's check \(C(0,4)\): \(A(2,4)\) to \(C(0,4)\) is 2, \(B(3,6)\) to \(C(0,4)\) is \(\sqrt{9 + 4}=\sqrt{13}\), \(AB\) is \(\sqrt{5}\). Not equal. \(C(5,5)\): \(A(2,4)\) to \(C(5,5)\): \(\sqrt{(5 - 2)^2+(5 - 4)^2}=\sqrt{9 + 1}=\sqrt{10}\). \(B(3,6)\) to \(C(5,5)\): \(\sqrt{(5 - 3)^2+(5 - 6)^2}=\sqrt{4 + 1}=\sqrt{5}\). \(AB\) is \(\sqrt{5}\). So \(BC = AB=\sqrt{5}\). Oh! Wait, \(C(5,5)\): \(BC\) distance is \(\sqrt{(5 - 3)^2+(5 - 6)^2}=\sqrt{4 + 1}=\sqrt{5}\), and \(AB\) is \(\sqrt{5}\). So \(AB = BC\), so \(\triangle ABC\) is isosceles with \(AB = BC\). Wait, I missed that. Let's recalculate \(C(5,5)\):

- \(AB\): \(\sqrt{(3 - 2)^2+(6 - 4)^2}=\sqrt{1 + 4}=\sqrt{5}\)
- \(BC\): \(\sqrt{(5 - 3)^2+(5 - 6)^2}=\sqrt{4 + 1}=\sqrt{5}\)
- \(AC\): \(\sqrt{(5 - 2)^2+(5 - 4)^2}=\sqrt{9 + 1}=\sqrt{10}\)

So \(AB = BC\), so \(\triangle ABC\) is isosceles. Wait, but earlier I thought \(C(5,5)\) was not, but I miscalculated \(BC\) distance. So \(C(5,5)\) makes \(AB = BC\), so it's isosceles. Wait, but the options are \(C(5,4)\), \(C(5,5)\), \(C(3,4)\), \(C(0,4)\). So \(C(5,5)\) is the correct one? Wait, let's check again. \(B(3,6)\) to \(C(5,5)\): \(x\) difference 2, \(y\) difference -1. So distance \(\sqrt{2^2+(-1)^2}=\sqrt{4 + 1}=\sqrt{5}\), which is equal to \(AB\)'s distance. So \(AB = BC\), so \(\triangle ABC\) is isosceles. So the correct answer is \(C(5,5)\). Wait, but earlier I thought \(C(5,5)\) was not, but I made a mistake in calculation. So the correct option is \(C(5,5)\).

Answer:

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