In a ΔABC , A = (1,2) ; B =(5,5) In angleACB = 90° If area of ΔABC is to be 6.5 squnits the possible number of points for C are Options A ) 1 B) 2 C) 0 D) 4 About the author Eloise
Topic :- Coordinate Geometry Given :- In a ΔABC, A ≡ (1, 2); B ≡ (5, 5) and ∠ACB = 90°. Area of ΔABC is to be 6.50 sq. units. To Find :- Possible number of points for C. Solution :- ΔABC is a right angle triangle as ∠ACB = 90°. Side AB opposite to ∠ACB will act as hypotenuse for the triangle. Calculating length AB from Distance Formula, [tex]AB=\sqrt{(5-1)^2+(5-2)^2}\;units[/tex] [tex]AB=\sqrt{4^2+3^2}\;units[/tex] [tex]AB=\sqrt{16+9}\;units[/tex] [tex]AB=\sqrt{25}\;units[/tex] [tex]AB=5\;units[/tex] Assuming length of arms of triangle, Let length of arms of given triangle be x and y. Applying Pythagoras Theorem, x² + y² = 5² . . . . equation (1) Area of Right Angle Triangle, [tex]Area=\dfrac{1}{2} \times (Product\:of\:length\:of\:arms)[/tex] [tex]6.5=\dfrac{1}{2} \times x \times y[/tex] [tex]13=xy[/tex] [tex]x=\dfrac{13}{y}[/tex] Substituting value of ‘x’ in equation (1), [tex]x^2+y^2=5^2[/tex] [tex]\left( \dfrac{13}{y} \right)^2+y^2=5^2[/tex] [tex]\dfrac{169}{y^2} +y^2=25[/tex] [tex]\dfrac{169+y^4}{y^2}=25[/tex] Cross Multiply, [tex]169+y^4=25y^2[/tex] Rearranging it, [tex]y^4-25y^2+169=0[/tex] Substitite y² = t, [tex](y^2)^2-25y^2+169=0[/tex] [tex]t^2-25t+169=0[/tex] Calculating value of Discriminant, [tex]D=b^2-4ac[/tex] Here, a = 1 b = -25 c = 169 [tex]D=(-25)^2-4(1)(169)[/tex] [tex]D=625-676[/tex] [tex]D=-51[/tex] [tex]D<0[/tex] which means Real ‘t’ doesn’t exist which means Real ‘y’ doesn’t exist. Thus, there are no possible point for point C as y doesn’t exist. Answer :- So, there are Zero (0) possible points for point C. Hence, option C is correct. Reply
So, there are Zero (0) possible points for point C. Hence, option C is correct. 0 is your answer. Hope it will help you.. Reply
Topic :-
Coordinate Geometry
Given :-
In a ΔABC, A ≡ (1, 2); B ≡ (5, 5) and ∠ACB = 90°.
Area of ΔABC is to be 6.50 sq. units.
To Find :-
Possible number of points for C.
Solution :-
ΔABC is a right angle triangle as ∠ACB = 90°.
Side AB opposite to ∠ACB will act as hypotenuse for the triangle.
Calculating length AB from Distance Formula,
[tex]AB=\sqrt{(5-1)^2+(5-2)^2}\;units[/tex]
[tex]AB=\sqrt{4^2+3^2}\;units[/tex]
[tex]AB=\sqrt{16+9}\;units[/tex]
[tex]AB=\sqrt{25}\;units[/tex]
[tex]AB=5\;units[/tex]
Assuming length of arms of triangle,
Let length of arms of given triangle be x and y.
Applying Pythagoras Theorem,
x² + y² = 5² . . . . equation (1)
Area of Right Angle Triangle,
[tex]Area=\dfrac{1}{2} \times (Product\:of\:length\:of\:arms)[/tex]
[tex]6.5=\dfrac{1}{2} \times x \times y[/tex]
[tex]13=xy[/tex]
[tex]x=\dfrac{13}{y}[/tex]
Substituting value of ‘x’ in equation (1),
[tex]x^2+y^2=5^2[/tex]
[tex]\left( \dfrac{13}{y} \right)^2+y^2=5^2[/tex]
[tex]\dfrac{169}{y^2} +y^2=25[/tex]
[tex]\dfrac{169+y^4}{y^2}=25[/tex]
Cross Multiply,
[tex]169+y^4=25y^2[/tex]
Rearranging it,
[tex]y^4-25y^2+169=0[/tex]
Substitite y² = t,
[tex](y^2)^2-25y^2+169=0[/tex]
[tex]t^2-25t+169=0[/tex]
Calculating value of Discriminant,
[tex]D=b^2-4ac[/tex]
Here,
a = 1
b = -25
c = 169
[tex]D=(-25)^2-4(1)(169)[/tex]
[tex]D=625-676[/tex]
[tex]D=-51[/tex]
[tex]D<0[/tex]
which means
Real ‘t’ doesn’t exist which means Real ‘y’ doesn’t exist.
Thus, there are no possible point for point C as y doesn’t exist.
Answer :-
So, there are Zero (0) possible points for point C.
Hence, option C is correct.
So, there are Zero (0) possible points for point C.
Hence, option C is correct.
0 is your answer.
Hope it will help you..