# The slope of a line with a positive rational slope l, which passes through the point (6,0) and at a distance of 5 from (1,3). Write the slope in the form a/b where a and b are relatively prime. Then the sum of a and b is?

$23$

#### Explanation:

Let the equation of line be

$a x + b y + c = 0$

Above line passes through the point $\left(6 , 0\right)$ then the point will satisfy the above equation . Now, setting $x = 6$ & $y = 0$ in above equation as follows

$a \left(6\right) + b \left(0\right) + c = 0$

$6 a + c = 0$

$c = - 6 a \setminus \ldots \ldots \ldots \left(1\right)$

Now, the distance of point $\left(1 , 3\right)$ from the above line is given as

$\setminus \frac{| a \left(1\right) + b \left(3\right) + c |}{\setminus \sqrt{{a}^{2} + {b}^{2}}} = 5$

$\setminus \frac{| a + 3 b - 6 a |}{\setminus \sqrt{{a}^{2} + {b}^{2}}} = 5 \setminus \setminus \quad \left(\setminus \textrm{\mathfrak{o} m e q \left(1\right)} , c = - 6 a\right)$

$\setminus \frac{| 3 b - 5 a |}{\setminus \sqrt{{a}^{2} + {b}^{2}}} = 5$

$\setminus \frac{| 3 b - 5 a {|}^{2}}{{\left(\setminus \sqrt{{a}^{2} + {b}^{2}}\right)}^{2}} = {5}^{2}$

$\setminus \frac{25 {a}^{2} + 9 {b}^{2} - 30 a b}{{a}^{2} + {b}^{2}} = 25$

$\setminus \frac{25 \setminus \frac{{a}^{2}}{a b} + 9 \setminus \frac{{b}^{2}}{a b} - 30 \setminus \frac{a b}{a b}}{\setminus \frac{{a}^{2}}{a b} + \setminus \frac{{b}^{2}}{a b}} = 25$

$\setminus \frac{25 \setminus \frac{a}{b} + 9 \setminus \frac{b}{a} - 30}{\setminus \frac{a}{b} + \setminus \frac{b}{a}} = 25$

$25 \setminus \frac{a}{b} + 9 \setminus \frac{b}{a} - 30 = 25 \setminus \frac{a}{b} + 25 \setminus \frac{b}{a}$

$16 \frac{b}{a} = - 30$

$\frac{a}{b} = - \frac{16}{30}$

$= - \frac{8}{15}$

For positive slope we have

$\frac{a}{b} = \frac{8}{15}$

$\setminus \therefore a + b = 8 + 15$

$= 23$

Jul 26, 2018

The eqution of the line passing through $\left(6 , 0\right)$ and having slope $l$ will be

$\left(y - 0\right) = l \left(x - 6\right)$

$\implies - l x + y + 6 l = 0$

The distance of this line from the point $\left(1 , 3\right)$ is 5.

So

$\frac{- l \cdot 1 + 3 + 6 l}{\sqrt{{l}^{2} + {1}^{2}}} = 5$

$\implies {\left(5 l + 3\right)}^{2} = {\left(5 \sqrt{{l}^{2} + 1}\right)}^{2}$

$\implies 25 {l}^{2} + 30 l + 9 = 25 \left({l}^{2} + 1\right)$

$\implies 30 l = 16$

$= l = \frac{16}{30} = \frac{8}{15}$

If the slope $l$ is expressed in the form of $\frac{a}{b}$,where $a \mathmr{and} b$ are prime to each other then $a = 8 \mathmr{and} b = 15$

Hence the sum $a + b = 8 + 15 = 23$