Question

the hypotenuse of right triangle abc is represented by segment ab, as shown on the graph. at which point could point c be located? (-3,3) (5,-3) (5,-5) (-3,2)

Explanation:

Step1: Find the coordinates of A and B

From the graph, \(A(-3,-4)\) and \(B(4,2)\)

Step2: Check the perpendicularity condition

For a right - triangle \(ABC\) with hypotenuse \(AB\), if \(C(x,y)\), then the vectors \(\overrightarrow{AC}=(x + 3,y + 4)\) and \(\overrightarrow{BC}=(x - 4,y - 2)\) should satisfy \(\overrightarrow{AC}\cdot\overrightarrow{BC}=0\) (if \(C\) is the right - angled vertex) or we can check the slope condition.
Another way is to use the property of the right - triangle in the coordinate plane. If we consider the vertical and horizontal distances.
For a right - triangle with hypotenuse \(AB\), if we fix one of the legs to be vertical and the other to be horizontal.
If we take \(x=-3\) (same \(x\) - coordinate as \(A\)), then for the right - triangle, if \(C(-3,y)\)
The length of \(AC\) is \(|y + 4|\) and if we consider the horizontal leg from \(C(-3,y)\) to \(B(4,y)\) and vertical leg from \(A(-3,-4)\) to \(C(-3,y)\)
If \(y = 2\), it's not a right - triangle. If \(y=-3\)
The vector \(\overrightarrow{AC}=(-3+3,-3 + 4)=(0,1)\) and \(\overrightarrow{BC}=(-3 - 4,-3 - 2)=(-7,-5)\) (not perpendicular)
If we take \(y = 3\), \(\overrightarrow{AC}=(-3 + 3,3 + 4)=(0,7)\) and \(\overrightarrow{BC}=(-3-4,3 - 2)=(-7,1)\) (not perpendicular)
If we take \(x = 5\) (not a good choice as it will make the distance calculation complex for a right - triangle with right - angle at \(C\))
If we consider the right - triangle with legs parallel to the axes.
The \(x\) - coordinate of \(A\) is \(-3\). If we take \(x=-3\) for point \(C\)
Let's check the distance formula. The length of \(AB=\sqrt{(4 + 3)^{2}+(2 + 4)^{2}}=\sqrt{49 + 36}=\sqrt{85}\)
If \(C(-3,3)\):
\(AC=\sqrt{(-3+3)^{2}+(3 + 4)^{2}} = 7\), \(BC=\sqrt{(-3 - 4)^{2}+(3 - 2)^{2}}=\sqrt{49+1}=\sqrt{50}\), \(AC^{2}+BC^{2}=49 + 50=99
eq85\)
If \(C(5,-3)\):
\(AC=\sqrt{(5 + 3)^{2}+(-3 + 4)^{2}}=\sqrt{64 + 1}=\sqrt{65}\), \(BC=\sqrt{(5 - 4)^{2}+(-3 - 2)^{2}}=\sqrt{1+25}=\sqrt{26}\), \(AC^{2}+BC^{2}=65 + 26=91
eq85\)
If \(C(5,-5)\):
\(AC=\sqrt{(5 + 3)^{2}+(-5 + 4)^{2}}=\sqrt{64+1}=\sqrt{65}\), \(BC=\sqrt{(5 - 4)^{2}+(-5 - 2)^{2}}=\sqrt{1 + 49}=\sqrt{50}\), \(AC^{2}+BC^{2}=65+50 = 115
eq85\)
If \(C(-3,2)\)
\(AC=\sqrt{(-3 + 3)^{2}+(2 + 4)^{2}}=6\), \(BC=\sqrt{(-3 - 4)^{2}+(2 - 2)^{2}}=7\)
By Pythagorean theorem \(AC^{2}+BC^{2}=36+49 = 85\) and \(AB^{2}=(4 + 3)^{2}+(2 + 4)^{2}=49+36=85\)

Answer:

\((-3,2)\)