Question

$\mathrm{y}=\frac{3}{4}.$The line $\mathrm{x}+\mathrm{y}=1$ meets the lines represented by the equation $y^3-x{y}^2-14x^{2}y+24x^3=0$at the points A,B,C. If O is the origin, then $O{A}^2+O{B}^2+O{C}^2$ is equal to

Answer 1

Step 1: Determine the equation of lines represented by ${\mathrm{y}}^3-{\mathrm{xy}}^2-14{\mathrm{x}}^2\mathrm{y}+24{\mathrm{x}}^3=0$

Given equation: ${\mathrm{y}}^3-{\mathrm{xy}}^2-14{\mathrm{x}}^2\mathrm{y}+24{\mathrm{x}}^3=0$

$\begin{array}{c}{\mathrm{y}}^3-{\mathrm{xy}}^2-14{\mathrm{x}}^2\mathrm{y}+24{\mathrm{x}}^3=0\\ \Rightarrow {\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^3-{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2-14\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+24=0\\ \end{array}\begin{array}{}\\ \\ \end{array}$

take $m=\frac{y}{x}$:

$\begin{array}{cc}& {\mathrm{m}}^3-{\mathrm{m}}^2-14\mathrm{m}+24=0\\ \Rightarrow & \mathrm{m}=-4,3,2.\\ \Rightarrow & \mathrm{y}=2\mathrm{x},\mathrm{y}=3\mathrm{x}\mathrm{and}\mathrm{y}=-4\mathrm{x}\end{array}$

Step 2: Compute the point name A,B,C

Hence, the three lines are:

$\mathrm{y}=2\mathrm{x}\dots \left(1\right)$

$\mathrm{y}=3\mathrm{x}\dots \left(2\right)$

$\mathrm{y}=-4\mathrm{x}\dots \left(3\right)$

Also, the intersecting these lines is $\mathrm{x}+\mathrm{y}=1\dots \left(4\right)$ at say $\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}$ respectively.

substitute$\mathrm{y}=2\mathrm{x}\mathrm{in}\left(4\right)$:

$\begin{array}{rcl}\mathrm{x}+2\mathrm{x}& =& 1\\ & \Rightarrow & 3x=1\\ & \Rightarrow & x=\frac{1}{3}\end{array}$

Hence,$\mathrm{y}=\frac{2}{3}.$

Therefore, the point where the line $\mathrm{y}=2\mathrm{x}$ intersects$\mathrm{x}+\mathrm{y}=1$ is $A\left(\frac{1}{3},\frac{2}{3}\right)$

Now substitute $\mathrm{y}=3\mathrm{x}\mathrm{in}\left(4\right)$:

$\begin{array}{rcl}\mathrm{x}+3\mathrm{x}& =& 1\\ & \Rightarrow & 4\mathrm{x}=1\\ & \Rightarrow & x=\frac{1}{4}\end{array}$

So, $\mathrm{x}=\frac{1}{4}$

Hence, the point where the line $\mathrm{y}=3\mathrm{x}$ intersects $\mathrm{x}+\mathrm{y}=1$ is $\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)$

Now, substitute $\mathrm{y}=-4\mathrm{x}\mathrm{in}\left(4\right)$

$\begin{array}{rcl}x–4x& =& 1\\ -3x& =& 1\end{array}$

$\mathrm{x}=-\frac{1}{3},\mathrm{and}\mathrm{y}=\frac{4}{3}.$

Hence, the point where the line $\mathrm{y}=-4x$ intersects $\mathrm{x}+\mathrm{y}=1$ is $\mathrm{C}\left(-\frac{1}{3},\frac{4}{3}\right)$

Step 3: Compute the value ${\mathrm{OA}}^2+{\mathrm{OB}}^2+{\mathrm{OC}}^2$

Since the Point $O$ is $(0,0)$ and the intersection are $A\left(\frac{1}{3},\frac{2}{3}\right)$,$\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)$ and$\mathrm{C}\left(-\frac{1}{3},\frac{4}{3}\right)$

Apply the distance formula to calculate the required sum ${\mathrm{OA}}^2+{\mathrm{OB}}^2+{\mathrm{OC}}^2$:

$\begin{array}{rl}& {\mathrm{OA}}^2+{\mathrm{OB}}^2+{\mathrm{OC}}^2={\left(\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2}\right)}^2+{\left(\sqrt{{\left(\frac{1}{4}\right)}^2+{\left(\frac{3}{4}\right)}^2}\right)}^2+{\left(\sqrt{{\left(\frac{-1}{3}\right)}^2+\left(\frac{4}{3}\right)}\right)}^2\\ & =\frac{1}{9}+\frac{4}{9}+\frac{1}{16}+\frac{9}{16}+\frac{1}{9}+\frac{16}{9}\\ & =\frac{22}{9}+\frac{10}{16}\\ & =\frac{442}{144}\\ & =\frac{221}{72}\\ & \end{array}$

Hence, the Value of ${\mathrm{OA}}^2+{\mathrm{OB}}^2+{\mathrm{OC}}^2$ is equal to $\frac{221}{72}$