Math

QuestionFind the coordinates of point CC in right isosceles triangle ABC\triangle ABC where A(2,3)A(2,3) and B(7,3)B(7,3).

Studdy Solution

STEP 1

Assumptions1. AB\overline{\mathrm{AB}} is the hypotenuse of the right isosceles ABC\triangle \mathrm{ABC}. . The coordinates of point A are (,3).
3. The coordinates of point B are (7,3).
4. In a right isosceles triangle, the two legs are of equal length and the angles are45°,45°, and90°.

STEP 2

First, we need to find the length of the hypotenuse AB\overline{\mathrm{AB}}. We can do this using the distance formula.
AB=(x2x1)2+(y2y1)2AB = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}

STEP 3

Now, plug in the given values for the coordinates of A and B to calculate the length of AB.
AB=(72)2+(33)2AB = \sqrt{(7 -2)^2 + (3 -3)^2}

STEP 4

Calculate the length of AB.
AB=()2+(0)2=25=AB = \sqrt{()^2 + (0)^2} = \sqrt{25} =

STEP 5

In a right isosceles triangle, the length of the hypotenuse is 2\sqrt{2} times the length of each leg. Therefore, the length of AC and BC (the legs of the triangle) is the length of the hypotenuse divided by 2\sqrt{2}.
AC=BC=AB2AC = BC = \frac{AB}{\sqrt{2}}

STEP 6

Plug in the value for the length of AB to calculate the length of AC and BC.
AC=BC=52AC = BC = \frac{5}{\sqrt{2}}

STEP 7

implify the length of AC and BC.
AC=BC=522AC = BC = \frac{5\sqrt{2}}{2}

STEP 8

Since ABC\triangle ABC is a right isosceles triangle and AB is the hypotenuse, the coordinates of C will be either above or below the line AB. We can find these coordinates by adding or subtracting the length of AC (or BC) from the y-coordinate of A (or B).
C1=(x1,y1±AC)C1 = (x1, y1 \pm AC)C2=(x2,y2±BC)C2 = (x2, y2 \pm BC)

STEP 9

Plug in the values for the coordinates of A and B and the length of AC and BC to calculate the coordinates of C.
C=(2,3±522)C = (2,3 \pm \frac{5\sqrt{2}}{2})C2=(7,3±522)C2 = (7,3 \pm \frac{5\sqrt{2}}{2})

STEP 10

Calculate the coordinates of C.
C=(2,3+522),(2,3522)C = (2,3 + \frac{5\sqrt{2}}{2}), (2,3 - \frac{5\sqrt{2}}{2})C2=(7,3+522),(7,3522)C2 = (7,3 + \frac{5\sqrt{2}}{2}), (7,3 - \frac{5\sqrt{2}}{2})However, since AB\overline{\mathrm{AB}} is the hypotenuse, we discard the points (2,3+522)(2,3 + \frac{5\sqrt{2}}{2}) and (7,3+522)(7,3 + \frac{5\sqrt{2}}{2}).
The possible coordinates of C are (2,3522)(2,3 - \frac{5\sqrt{2}}{2}) and (7,3522)(7,3 - \frac{5\sqrt{2}}{2}).

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