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 line.

Answer 1

Any point on the $3x-y=0$ is of the form $(a,3a)$.

$\therefore$ Distance between $(a,3a)$ and $(4,1)$ is $\frac{11}{2\sqrt{2}}$.

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

$(a^2-8a+16+9a^2-6a+1)=\frac{121}{8}$

$10a^2-14a+17=\frac{121}{8}$

$10a^2-14a+\frac{15}{8}=0$

$80a^2-112a+15=0$

$(4a-5)(20a-3)=0$

$a=\frac{5}{4},\frac{3}{20}$

$\therefore$ Equation of line passing through $(4,1)$ and $(\frac{5}{4},\frac{15}{4})$ is

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

$(y-1)=\left(\frac{\frac{15}{4}-1}{5-4}\right)(x-4)$

$(y-1)=\frac{11}{4}(x-4)$

$4y-4=11x-44$

$11x-4y=40$

$\therefore$ Equation of line passing through $(4,1)$ and $(\frac{3}{20},\frac{9}{20})$ is

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

$(y-1)=\left(\frac{\frac{9}{20}-1}{\frac{3}{20}-4}\right)(x-4)$

$(y-1)=\frac{11}{77}(x-4)$

$77y-77=11x-44$

$11x-77y+33=0$

$x-7y+3=0$