Question

Through the point $P(4,1)$ a line is drawn to meet the line $3x-y=0$ at Q where $PQ$ = $\frac{11}{2\sqrt{2}}$. Determine the equation of the line.

Answer 1

Given $3x-y=0$ & $P(4,1)$

$\Rightarrow y=3x$

For any $Q(x,3x)$

Distance $PQ=\frac{11}{2\sqrt{2}}$

$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\frac{11}{2\sqrt{2}}$

$\Rightarrow \sqrt{{(x-4)}^2+{(3x-1)}^2}=\frac{11}{2\sqrt{2}}$

Squaring on both sides

$\Rightarrow x^2+16-8x+9x^2+1-6x=\frac{121}{8}$

$\Rightarrow 80x^2-112x+136=121$

$\Rightarrow 80x^2-112x+15=0$

$[Weknowfora{x}^2+bx+c=0rootsofequationx=\frac{-6\pm \sqrt{b^2-4ac}}{2a}]$

$x=\frac{-(-112)\pm \sqrt{{(-112)}^2-4\left(80\right)\left(15\right)}}{2\times 80}$

$=\frac{112\pm \sqrt{7788}}{160}=\frac{112\pm 88}{160}$

$=\frac{200}{160}or\frac{24}{160}$

$x=125or0.15$

$\Rightarrow y=3x=3.75ory=3x=0.45forQ(1.25,3.75)Q(4,1)$

Equation of line $(y-y_1)=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$

Equation of line $(y-1)=\left(\frac{3.75-1}{1.25-4}\right)(x-4)$

$y-1=\frac{2.75}{-2.75}(x-4)$

$\Rightarrow y-1=-x+4\Rightarrow x+y-5=0$

Equation of line when $Q(0.15,0.45)P(x-4)$

$y-1=\frac{-0.55}{-3.85}(x-4)$

$\Rightarrow 0.55x-3.85y+1.65=0$