Respuesta :
Answer:
[tex](-1,1),(4,-2)[/tex]
Step-by-step explanation:
Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).
To find: coordinates of vertex of the right angle
Solution:
Let C be point [tex](x,y)[/tex]
Distance between points [tex](x_1,y_1),(x_2,y_2)[/tex] is given by [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}[/tex]
ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)
[tex]34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ][/tex]
Put [tex](x,y)=(-1,1)[/tex]
[tex]34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34[/tex]
which is true. So, [tex](-1,1)[/tex] can be a vertex
Put [tex](x,y)=(4,-2)[/tex]
[tex]34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34[/tex]
which is true. So, [tex](4,-2)[/tex] can be a vertex
Put [tex](x,y)=(1,1)[/tex]
[tex]34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22[/tex]
which is not true. So, [tex](1,1)[/tex] cannot be a vertex
Put [tex](x,y)=(2,-2)[/tex]
[tex]34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22[/tex]
which is not true. So, [tex](2,-2)[/tex] cannot be a vertex
Put [tex](x,y)=(4,-1)[/tex]
[tex]34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30[/tex]
which is not true. So, [tex](4,-1)[/tex] cannot be a vertex
Put [tex](x,y)=(-1,4)[/tex]
[tex]34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70[/tex]
which is not true. So, [tex](-1,4)[/tex] cannot be a vertex
So, possible points for the vertex are [tex](-1,1),(4,-2)[/tex]
