A straight line is passing through the point $A \left(1,2\right)$ with slope $\dfrac{5}{12}$. Find the points on the line which are 13 units away from $A$.
The required line passes through $A \left(1,2\right)$ with slope $\dfrac{5}{12}$
$\therefore$ $\;$ Equation of the required line is: $\hspace{1em}$ $\left(y - 2\right) = \dfrac{5}{12} \left(x - 1\right)$
i.e. $12 y - 24 = 5 x - 5$
i.e. $5x - 12 y = - 19$ $\;\;\; \cdots \; (1)$
Let $P \left(p , q\right)$ be a point on the line given by equation $(1)$.
Then we have $\hspace{1em}$ $5 p - 12 q = - 19$ $\;\;\; \cdots \; (2)$
As per question, distance $AP = 13$
Now, distance $AP = \sqrt{\left(p - 1\right)^2 + \left(q - 2\right)^2}$
$\therefore$ $\;$ We have $\hspace{1em}$ $\sqrt{\left(p - 1\right)^2 + \left(q - 2\right)^2} = 13$
i.e. $\left(p - 1\right)^2 + \left(q - 2\right)^2 = 169$ $\;\;\; \cdots \; (3)$
We have from equation $(2)$, $p = \dfrac{12 q - 19}{5}$ $\;\;\; \cdots \; (4)$
Substituting the value of $p$ from equation $(4)$ in equation $(3)$ we get,
$\left(\dfrac{12 q - 19}{5} - 1\right)^2 + \left(q - 2\right)^2 = 169$
i.e. $\left(\dfrac{12 q - 24}{5}\right)^2 + \left(q - 2\right)^2 = 169$
i.e. $\dfrac{144}{25} \left(q - 2\right)^2 + \left(q - 2\right)^2 = 169$
i.e. $\dfrac{169}{25} \left(q - 2\right)^2 = 169$
i.e. $\left(q - 2\right)^2 = 25$ $\implies$ $q - 2 = \pm 5$
i.e. $q = 7$ or $q = - 3$
Substituting the values of $q$ in equation $(4)$ we get,
when $q = 7$, $p = \dfrac{12 \times 7 - 19}{5} = 13$ and
when $q = - 3$, $p = \dfrac{12 \times \left(-3\right) - 19}{5} = -11$
$\therefore$ $\;$ The required points on the line are $\left(13, 7\right)$ and $\left(-11, -3\right)$.