Question

pre-algebra ic sem 2 fall 2025
square roots and right triangles
the hypotenuse of a right triangle has endpoints a(4, 1) and b(-1, -2).
at which points could the vertex of the right angle in the triangle be located? choose two correct answers.
(2, -2) (-1, 4)
(1, 1) (-1, 1)
(4, -2) (4, -1)

Explanation:

Step1: Recall the property of a right - angled triangle

In a right - angled triangle, if the endpoints of the hypotenuse are \(A(x_1,y_1)\) and \(B(x_2,y_2)\), then the vertex of the right - angle \(C(x,y)\) satisfies the following: If \(A=(4,1)\) and \(B = (-1,-2)\), for a point \(C(x,y)\) to be the right - angle vertex, the slopes of \(AC\) and \(BC\) should be perpendicular (product of slopes \(m_{AC}\times m_{BC}=- 1\)) or we can use the distance formula \(AB^{2}=AC^{2}+BC^{2}\) (by Pythagoras theorem \(c^{2}=a^{2}+b^{2}\), where \(c = AB\), \(a = AC\), \(b = BC\)).

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\), so \(AB^{2}=(4 + 1)^{2}+(1+2)^{2}=25 + 9=34\)

Step2: Calculate for each point

• For point \((2,-2)\):

\(AC^{2}=(4 - 2)^{2}+(1 + 2)^{2}=4+9 = 13\), \(BC^{2}=(-1 - 2)^{2}+(-2 + 2)^{2}=9+0 = 9\), \(AC^{2}+BC^{2}=13 + 9=22
eq34\)

• For point \((-1,4)\):

\(AC^{2}=(4 + 1)^{2}+(1 - 4)^{2}=25 + 9=34\), \(BC^{2}=(-1+1)^{2}+(-2 - 4)^{2}=0 + 36=36\), \(AC^{2}+BC^{2}=34+36 = 70
eq34\)

• For point \((1,1)\):

\(AC^{2}=(4 - 1)^{2}+(1 - 1)^{2}=9+0 = 9\), \(BC^{2}=(-1 - 1)^{2}+(-2 - 1)^{2}=4 + 9=13\), \(AC^{2}+BC^{2}=9+13 = 22
eq34\)

• For point \((-1,1)\):

\(AC^{2}=(4 + 1)^{2}+(1 - 1)^{2}=25+0 = 25\), \(BC^{2}=(-1+1)^{2}+(-2 - 1)^{2}=0 + 9=9\), \(AC^{2}+BC^{2}=25 + 9=34\)

• For point \((4,-2)\):

\(AC^{2}=(4 - 4)^{2}+(1 + 2)^{2}=0+9 = 9\), \(BC^{2}=(-1 - 4)^{2}+(-2 + 2)^{2}=25+0 = 25\), \(AC^{2}+BC^{2}=9+25 = 34\)

• For point \((4,-1)\):

\(AC^{2}=(4 - 4)^{2}+(1 + 1)^{2}=0+4 = 4\), \(BC^{2}=(-1 - 4)^{2}+(-2 + 1)^{2}=25+1 = 26\), \(AC^{2}+BC^{2}=4+26 = 30
eq34\)

Answer:

\((-1,1)\), \((4,-2)\)