Question

together with the pair of points (13,0) and (0,13), 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 properties of isosceles right - triangle

In an isosceles right - triangle, if two points are \(A(13,0)\) and \(B(0,13)\), and let the third point be \(C(x,y)\). The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Also, for an isosceles right - triangle, two sides are equal in length and the Pythagorean theorem \(a^{2}+a^{2}=c^{2}\) (where \(a\) are the legs and \(c\) is the hypotenuse) holds.

Step2: Case 1: \(AC = BC\) and \(\angle C=90^{\circ}\)

The mid - point of \(AB\) is \((\frac{13 + 0}{2},\frac{0+13}{2})=(6.5,6.5)\). The slope of \(AB\) is \(m_{AB}=\frac{13 - 0}{0 - 13}=- 1\). The slope of the line perpendicular to \(AB\) is \(m = 1\). The equation of the line passing through the mid - point \((6.5,6.5)\) with slope \(m = 1\) is \(y-6.5=x - 6.5\) or \(y=x\). Let \(C(x,x)\), then \(AC^{2}=(x - 13)^{2}+x^{2}\) and \(BC^{2}=x^{2}+(x - 13)^{2}\). Using the Pythagorean theorem \(2[(x - 13)^{2}+x^{2}]=(13 - 0)^{2}+(0 - 13)^{2}=2\times13^{2}\). Expanding \((x - 13)^{2}+x^{2}=13^{2}\), \(x^{2}-26x + 169+x^{2}=169\), \(2x^{2}-26x=0\), \(2x(x - 13)=0\), \(x = 0\) or \(x = 13\). So the points are \((0,0)\) and \((13,13)\).

Step3: Case 2: \(AC=AB\) and \(\angle A = 90^{\circ}\)

\(AB=\sqrt{(13 - 0)^{2}+(0 - 13)^{2}}=\sqrt{13^{2}+(-13)^{2}}=\sqrt{2\times13^{2}} = 13\sqrt{2}\). Let \(C(x,y)\), \(AC^{2}=(x - 13)^{2}+y^{2}=2\times13^{2}\). Since \(\angle A=90^{\circ}\), the slope of \(AC\) times the slope of \(AB=-1\). The slope of \(AB=-1\), so the slope of \(AC = 1\). The equation of the line passing through \(A(13,0)\) with slope \(1\) is \(y-0=x - 13\) or \(y=x - 13\). Substitute \(y=x - 13\) into \((x - 13)^{2}+y^{2}=2\times13^{2}\), \((x - 13)^{2}+(x - 13)^{2}=2\times13^{2}\), \(2(x - 13)^{2}=2\times13^{2}\), \((x - 13)^{2}=13^{2}\), \(x-13=\pm13\). When \(x-13 = 13\), \(x = 26\) and \(y = 13\); when \(x-13=-13\), \(x = 0\) and \(y=-13\).

Step4: Case 3: \(BC=AB\) and \(\angle B=90^{\circ}\)

The slope of \(BC\) is \(1\) (since the slope of \(AB=-1\)). The equation of the line passing through \(B(0,13)\) with slope \(1\) is \(y-13=x-0\) or \(y=x + 13\). Let \(C(x,y)\), \(BC^{2}=x^{2}+(y - 13)^{2}=2\times13^{2}\). Substitute \(y=x + 13\) into \(x^{2}+(y - 13)^{2}=2\times13^{2}\), \(x^{2}+x^{2}=2\times13^{2}\), \(2x^{2}=2\times13^{2}\), \(x=\pm13\). When \(x = 13\), \(y = 26\); when \(x=-13\), \(y = 0\).

Answer:

\((0,0),(13,13),(26,13),(0,-13),(13,26),(-13,0)\)