QUESTION IMAGE
Question
in the xy - plane, two vertices of isosceles triangle xyz are x(1,2) and y(7,2), and xz = yz. which ordered pair could represent the location of point z? a (3,5) b (3,4) c (4,2) d (4,5)
Step1: Recall distance formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let $Z=(x,y)$, $X=(1,2)$ and $Y=(7,2)$. Then $XZ=\sqrt{(x - 1)^2+(y - 2)^2}$ and $YZ=\sqrt{(x - 7)^2+(y - 2)^2}$.
Step2: Set up the equation
Since $XZ = YZ$, we have $\sqrt{(x - 1)^2+(y - 2)^2}=\sqrt{(x - 7)^2+(y - 2)^2}$. Square both sides to get $(x - 1)^2+(y - 2)^2=(x - 7)^2+(y - 2)^2$.
Step3: Expand the equation
Expand the squares: $x^{2}-2x + 1+(y - 2)^2=x^{2}-14x + 49+(y - 2)^2$. Subtract $x^{2}$ and $(y - 2)^2$ from both sides. We obtain $-2x + 1=-14x+49$.
Step4: Solve for x
Add $14x$ to both sides: $-2x+14x + 1=-14x+14x + 49$, which simplifies to $12x+1 = 49$. Subtract 1 from both sides: $12x=48$, so $x = 4$. Since the $y$-coordinate of $X$ and $Y$ is 2 and the mid - point of the base of an isosceles triangle (where the equal sides are $XZ$ and $YZ$) has the same $y$-coordinate as the endpoints of the base for a horizontal base, $y$ can be any value. Among the options, when $x = 4$, a possible point is $(4,5)$.
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D. $(4,5)$