Answer: The correct answer is (5, 3)
Step-by-step explanation: I got it right on the exam
(5,3) will make a rhombus.
A Quadrilateral with all sides equal but with nonequal diagonals is know as Rhombus.
Let D have coordinate (x,y)
The length of all the sides of the rhombus is equal.
length of AB = [tex]\sqrt{({2-3})^2 + ({2-4})^2} = \sqrt{1+4} = \sqrt{5}[/tex]
Similarly BD = [tex]\sqrt{({3-x})^2 + ({4-y})^2} = \sqrt{25+{x}^2+{y}^2-6x-8y}[/tex] -------------(1)
CD = [tex]\sqrt{({4-x})^2 + ({1-y})^2} = \sqrt{17+{x}^2+{y}^2-8x-2y}[/tex] -------------(2)
We know BD = CD gives,
[tex]\sqrt{17+{x}^2+{y}^2-8x-2y} = \sqrt{25+{x}^2+{y}^2-6x-8y}[/tex]
By squaring both sides and solving we get,
3y - x = 4 ----------------------(3)
Since BCD is a right-angled triangle,
[tex]{BC}^2 = {BD}^2 + {CD}^2[/tex]
BC = [tex]\sqrt{10}[/tex]
10 = 5 + [tex]\sqrt{17+{x}^2+{y}^2-8x-2y}[/tex]
8 = [tex]{x}^2+{y}^2-8x-2y[/tex] -------------------(4)
solving equation (3) and (4) we get x = 5 and y = 3
TO Learn more about Rhombus visit:
https://brainly.com/question/2456096
#SPJ2
