Question

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​

Answer 1

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,

$AB=\sqrt{(5-1)^2+(5-2)^2}\;units$

$AB=\sqrt{4^2+3^2}\;units$

$AB=\sqrt{16+9}\;units$

$AB=\sqrt{25}\;units$

$AB=5\;units$

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,

$Area=\dfrac{1}{2} \times (Product\:of\:length\:of\:arms)$

$6.5=\dfrac{1}{2} \times x \times y$

$13=xy$

$x=\dfrac{13}{y}$

Substituting value of 'x' in equation (1),

$x^2+y^2=5^2$

$\left( \dfrac{13}{y} \right)^2+y^2=5^2$

$\dfrac{169}{y^2} +y^2=25$

$\dfrac{169+y^4}{y^2}=25$

Cross Multiply,

$169+y^4=25y^2$

Rearranging it,

$y^4-25y^2+169=0$

Substitite y² = t,

$(y^2)^2-25y^2+169=0$

$t^2-25t+169=0$

Calculating value of Discriminant,

$D=b^2-4ac$

Here,

a = 1

b = -25

c = 169

$D=(-25)^2-4(1)(169)$

$D=625-676$

$D=-51$

$D<0$

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.

Answer 2

So, there are Zero (0) possible points for point C.

Hence, option C is correct.

0 is your answer.

Hope it will help you..