QUESTION IMAGE
Question
a triangle has a perimeter of 24 units. two of the vertices of the triangle are (0, 0) and (0, 6). which of these could be the third vertex of the triangle? select all that apply.
□ a) (-8, 0)
□ b) (0, 8)
□ c) (0, -8)
□ d) (8, 0)
□ e) (18, 0)
Step1: Calculate the distance between given vertices
The two given vertices are $(0,0)$ and $(0,6)$. Using the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, for $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(0,6)$, we have $d=\sqrt{(0 - 0)^2+(6 - 0)^2}=6$.
Step2: Let the third - vertex be $(x,y)$ and use the perimeter formula
The perimeter $P$ of a triangle with side - lengths $d_1$, $d_2$, and $d_3$ is $P=d_1 + d_2 + d_3$. We know $P = 24$ and $d_1 = 6$. Let the distance between $(0,0)$ and $(x,y)$ be $d_2=\sqrt{x^{2}+y^{2}}$ and the distance between $(0,6)$ and $(x,y)$ be $d_3=\sqrt{(x - 0)^2+(y - 6)^2}=\sqrt{x^{2}+(y - 6)^{2}}$. So, $6+\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(y - 6)^{2}}=24$, which simplifies to $\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(y - 6)^{2}}=18$.
Step3: Test each option
Option A: $(x=-8,y = 0)$
The distance between $(0,0)$ and $(-8,0)$ is $\sqrt{(-8 - 0)^2+(0 - 0)^2}=8$.
The distance between $(0,6)$ and $(-8,0)$ is $\sqrt{(-8 - 0)^2+(0 - 6)^2}=\sqrt{64 + 36}=\sqrt{100}=10$.
$6 + 8+10=24$.
Option B: $(x = 0,y = 8)$
The distance between $(0,0)$ and $(0,8)$ is $\sqrt{(0 - 0)^2+(8 - 0)^2}=8$.
The distance between $(0,6)$ and $(0,8)$ is $\sqrt{(0 - 0)^2+(8 - 6)^2}=2$.
$6+8 + 2=16
eq24$.
Option C: $(x = 0,y=-8)$
The distance between $(0,0)$ and $(0,-8)$ is $\sqrt{(0 - 0)^2+(-8 - 0)^2}=8$.
The distance between $(0,6)$ and $(0,-8)$ is $\sqrt{(0 - 0)^2+(-8 - 6)^2}=\sqrt{196}=14$.
$6+8 + 14=28
eq24$.
Option D: $(x = 8,y = 0)$
The distance between $(0,0)$ and $(8,0)$ is $\sqrt{(8 - 0)^2+(0 - 0)^2}=8$.
The distance between $(0,6)$ and $(8,0)$ is $\sqrt{(8 - 0)^2+(0 - 6)^2}=\sqrt{64 + 36}=\sqrt{100}=10$.
$6 + 8+10=24$.
Option E: $(x = 18,y = 0)$
The distance between $(0,0)$ and $(18,0)$ is $\sqrt{(18 - 0)^2+(0 - 0)^2}=18$.
The distance between $(0,6)$ and $(18,0)$ is $\sqrt{(18 - 0)^2+(0 - 6)^2}=\sqrt{324+36}=\sqrt{360}\approx18.97$.
$6+18+\sqrt{360}
eq24$.
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A. $(-8,0)$
D. $(8,0)$