Question

Let A = (1 , 2), B = (3 , 4) and let C = (x , y) be a point such that (x - 1)(x - 3)+(y - 2)(y - 4) = 0.If area of triangle ABC = 1, then maximum number of positions of C on the x - y plane is

Answer 1

Consider,

$\longrightarrow(x-1)(x-3)+(y-2)(y-4)=0$

$\longrightarrow x^2-4x+3+y^2-6y+8=0$

$\longrightarrow x^2-4x+4+y^2-6y+9=2$

$\longrightarrow (x-2)^2+(y-3)^2=2$

This means the point C(x, y) lies on a circle with center at (2, 3) having radius r = √2, or diameter d = 2√2.

If the area of triangle having vertices $\left(x_1,\ y_1\right),\ \left(x_2,\ y_2\right)$ and $\left(x_3,\ y_3\right)$ is A sq. units, then,

• $\dfrac{1}{2}\left|\begin{array}{cc}x-x_1&y-y_1\\x_2-x_1&y_2-y_1\end{array}\right|=\pm A$

Here the area of triangle ABC is 1 sq. units, i.e.,

$\longrightarrow\dfrac{1}{2}\left|\begin{array}{cc}x-1&y-2\\3-1&4-2\end{array}\right|=\pm 1$

$\longrightarrow\left|\begin{array}{cc}x-1&y-2\\2&2\end{array}\right|=\pm 2$

$\longrightarrow2\left|\begin{array}{cc}x-1&y-2\\1&1\end{array}\right|=\pm 2$

$\longrightarrow\left|\begin{array}{cc}x-1&y-2\\1&1\end{array}\right|=\pm1$

$\longrightarrow(x-1)-(y-2)=\pm1$

$\longrightarrow x-y+1=\pm1$

Consider,

$\longrightarrow x-y+1=1$

$\longrightarrow x-y=0$

Now we find the no. of intersection points of this lines with the circle.

The perpendicular distance of a point $\left(x_1,\ y_1\right)$ from a line $Ax+By+C=0$ is given by,

$\longrightarrow p=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

Here the perpendicular distance of the center of the circle (2, 3) from the line $x-y=0$ is,

$\longrightarrow p=\dfrac{|2-3|}{\sqrt{1^2+1^2}}$

$\longrightarrow p=\dfrac{1}{\sqrt2}<\sqrt2=r$

Note that,

• if $p=0,$ a diameter of the circle belongs to the line, and there are two points of intersection.

• if $0<p<r,$ a non - diametrical chord of the circle belongs to the line and even so the line passes through exactly two points on the circle.

• if $p=r,$ the line is a tangent to the circle and so it passes through exactly one point on the circle.

• if $p>r,$ there is no point of intersection, the line doesn't even touch the circle.

Here we see $p<r,$ which means the line passes through exactly two points on the circle.

Here we got two positions of C.

Consider,

$\longrightarrow x-y+1=-1$

$\longrightarrow x-y+2=0$

The perpendicular distance of the center of the circle (2, 3) from this line is,

$\longrightarrow p=\dfrac{|2-3+2|}{\sqrt{1^2+1^2}}$

$\longrightarrow p=\dfrac{1}{\sqrt2}<\sqrt2=r$

Here also $p<r,$ so this line also passes through exactly two points on the circle.

Here we got two more positions of C.

Hence the maximum number of positions of C on the xy plane is 4.

Answer 2

Answer:

$(x - 1)(x - 3) +(y - 2)(y - 4) = 0$

=> AC ⊥ BC => ∠ACB = 90°

∴ C is on the circle whose diameter is AB.

AB = 2√2.

As area (∆ABC) = 1,1/2, 2√2. (Altitude) = 1

∴ altitude = 1/√2 <radius.So, there are four possible position of C.

NOTE : If area (∆ABC) = 2,Altitude = radius => two position

Position of C.

Step-by-step explanation:

@GENIUS