Question

The orthocentre of a triangle formed by the line $x+y=3$ and pairs of lines $6x^2-5xy-6y^2+x+5y-1=0$ is

• A. $(\frac{13}{169},\frac{65}{169})$
• B. $(\frac{13}{169},\frac{-65}{169})$
• C. $(\frac{-13}{169},\frac{-65}{169})$
• D. $(\frac{-13}{169},\frac{65}{169})$

Answer 1

The correct option is A $(\frac{13}{169},\frac{65}{169})$

Given:$x+y-3=0$ is a line and pair of lines $6x^2-5xy-6y^2+x+5y-1=0$

$6x^2-5xy-6y^2+x+5y-1=0$

$6x^2-x(5y-1)+(-6y^2+5y-1)=0$ is quadratic in $x$

$\Rightarrow x=\frac{(5y-1)\pm \sqrt{{(5y-1)}^2-4\times 6\times (-6y^2+5y-1)}}{12}$

$\Rightarrow x=\frac{(5y-1)\pm \sqrt{25y^2+1-10y+144y^2-120y+24}}{12}$

$\Rightarrow x=\frac{(5y-1)\pm \sqrt{169y^2-130y+25}}{12}$

$\Rightarrow x=\frac{(5y-1)\pm \sqrt{{(13y-5)}^2}}{12}$

$\Rightarrow x=\frac{(5y-1)\pm (13y-5)}{12}$

$\Rightarrow 12x=(5y-1)\pm (13y-5)$

$\Rightarrow 12x=5y-1+13y-5,5y-1-13y+5$

$\Rightarrow 12x=18y-6,-8y+4$

Thus,$2x-3y=-1$ and $3x+2y=1$

Let $H(h,k)$ be the orthocentre.

Slope of $2x-3y=-1$ is $-\frac{2}{-3}=\frac{2}{3}$

Slope of $3x+2y=1$ is $-\frac{3}{2}=\frac{-3}{2}$

Slope of $x+y-3=0$ is $-1$

From $△ABC,$

Solving the equations $3x+2y=1$ ........$\left(1\right)$

$x+y=3$ ........$\left(2\right)$

$2x-3y=-1$ .......$\left(3\right)$

$3\times \left(2\right)-1\times \left(1\right)\Rightarrow 3x+3y-3x-2y=9-1$

$\Rightarrow y=8$

Put $y=8$ in $\left(2\right)$

$\Rightarrow x=3-y=3-8=-5$

$\therefore B$ is $(-5,8)$

$2\times \left(2\right)-1\times \left(3\right)\Rightarrow 2x+2y-2x+3y=6+1$

$\Rightarrow 5y=7$

Put $y=\frac{7}{5}$ in $\left(2\right)$

$\Rightarrow x=3-y=3-\frac{7}{5}=\frac{15-17}{5}=\frac{8}{5}$

$\therefore C$ is $(\frac{8}{5},\frac{7}{5})$

$3\times \left(2\right)-2\times \left(1\right)\Rightarrow 6x-9y-6x-4y=-3-2$

$\Rightarrow -13y=-5$

Put $y=\frac{5}{13}$ in $\left(1\right)$

$\Rightarrow 3x=1-2y=1-2\times \frac{5}{13}=\frac{13-10}{13}=\frac{3}{13}$

$\Rightarrow x=\frac{1}{13}$

$\therefore A$ is $(\frac{1}{13},\frac{5}{13})$

Let $H(h,k)$ be the orthocentre.

$\Rightarrow \left(SlopeofAH\right)\left(SlopeofBC\right)=-1$

$\Rightarrow \frac{k-\frac{5}{13}}{h-\frac{1}{13}}\times -1=-1$

$\Rightarrow k-\frac{5}{13}=h-\frac{1}{13}$

$\Rightarrow k-h=\frac{-1+5}{13}=\frac{4}{13}$ ........$\left(1\right)$

$\Rightarrow \left(SlopeofBH\right)\left(SlopeofAC\right)=-1$

$\Rightarrow \frac{k-8}{h+5}\times \frac{2}{3}=-1$

$\Rightarrow 2k-16=-3h-15$

$\Rightarrow 2k+3h=1$ ........$\left(2\right)$

Put $k=\frac{4}{13}+h$ from $\left(1\right)$ in $\left(2\right)$ we get

$\Rightarrow 2(\frac{4}{13}+h)+3h=1$

$\Rightarrow \frac{8}{13}+2h+3h=1$

$\Rightarrow 2h+3h=1-\frac{8}{13}$

$\Rightarrow 5h=\frac{13-8}{13}$

$\Rightarrow 5h=\frac{5}{13}$

$\therefore h=\frac{1}{13}$

Put $h=\frac{1}{13}$ in $k=\frac{4}{13}+h$ we get

$k=\frac{4}{13}+\frac{1}{13}=\frac{5}{13}$

$\therefore$ the orthocentre $H$ is $(\frac{1}{13},\frac{5}{13})=(\frac{1\times 13}{13\times 13},\frac{5\times 13}{13\times 13})=(\frac{13}{169},\frac{65}{169})$