Question

together with the pair of points (14,0) and (0,14), there are six points that could be the third vertex of an isosceles right triangle. find the coordinates of each point. the coordinates of the six points are (type ordered - pairs. use a comma to separate answers as needed.)

Explanation:

Step1: Recall distance formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let the given points be $A(14,0)$ and $B(0,14)$, and the third - point be $C(x,y)$.

Step2: Case 1: $\angle A = 90^{\circ}$

If $\angle A=90^{\circ}$, then $AB^{2}+AC^{2}=BC^{2}$. First, $AB=\sqrt{(0 - 14)^2+(14 - 0)^2}=\sqrt{196 + 196}=\sqrt{392}=14\sqrt{2}$. $AC=\sqrt{(x - 14)^2+y^{2}}$ and $BC=\sqrt{x^{2}+(y - 14)^2}$. So, $392+(x - 14)^2+y^{2}=x^{2}+(y - 14)^2$. Expanding gives $392+x^{2}-28x + 196+y^{2}=x^{2}+y^{2}-28y + 196$. Simplifying, we get $392-28x=-28y$, or $y=x - 14$. Also, since it's an isosceles right - triangle, $AC = AB$. $\sqrt{(x - 14)^2+y^{2}}=14\sqrt{2}$. Substitute $y=x - 14$ into it: $\sqrt{(x - 14)^2+(x - 14)^2}=14\sqrt{2}$. $\sqrt{2(x - 14)^2}=14\sqrt{2}$. $(x - 14)^2 = 196$. $x-14=\pm14$. If $x - 14 = 14$, then $x = 28$ and $y = 14$; if $x - 14=-14$, then $x = 0$ and $y=-14$.

Step3: Case 2: $\angle B = 90^{\circ}$

If $\angle B = 90^{\circ}$, then $AB^{2}+BC^{2}=AC^{2}$. Using the distance formula and similar steps as above, we get the equations. Since it's isosceles right - triangle, $BC = AB$. $\sqrt{x^{2}+(y - 14)^2}=14\sqrt{2}$. Also, from the right - angle condition and simplification, we find that if we substitute and solve, when $y=x + 14$. Solving $\sqrt{x^{2}+(x + 14 - 14)^2}=14\sqrt{2}$, we get $x=\pm14$. If $x = 14$, then $y = 28$; if $x=-14$, then $y = 0$.

Step4: Case 3: $\angle C = 90^{\circ}$

If $\angle C = 90^{\circ}$, then $AC^{2}+BC^{2}=AB^{2}$. And since it's isosceles, $AC = BC$. $\sqrt{(x - 14)^2+y^{2}}=\sqrt{x^{2}+(y - 14)^2}$. Squaring both sides gives $(x - 14)^2+y^{2}=x^{2}+(y - 14)^2$. Expanding: $x^{2}-28x + 196+y^{2}=x^{2}+y^{2}-28y + 196$, so $y = x$. Also, $2AC^{2}=AB^{2}$. $2[(x - 14)^2+x^{2}]=392$. Expanding: $2(x^{2}-28x + 196+x^{2})=392$. $4x^{2}-56x+392 = 392$. $4x^{2}-56x=0$. $4x(x - 14)=0$. So $x = 0,y = 0$ or $x = 14,y = 14$.

Answer:

$(0,0),(14,14),(28,14),(0, - 14),(14,28),(-14,0)$