Question

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?

Answer 1

`23`

Explanation:

Let the equation of line be

`ax+by+c=0`

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

`a(6)+b(0)+c=0`

`6a+c=0`

`c=-6a\ .........(1)`

Now, the distance of point `(1, 3)` from the above line is given as

`\frac{|a(1)+b(3)+c|}{\sqrt{a^2+b^2}}=5`

`\frac{|a+3b-6a|}{\sqrt{a^2+b^2}}=5\ \quad (\text{from eq(1)}, c=-6a)`

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

`\frac{|3b-5a|^2}{(\sqrt{a^2+b^2})^2}=5^2`

`\frac{25a^2+9b^2-30ab}{a^2+b^2}=25`

`\frac{25\frac{a^2}{ab}+9\frac{b^2}{ab}-30\frac{ab}{ab}}{\frac{a^2}{ab}+\frac{b^2}{ab}}=25`

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

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

`16b/a=-30`

`a/b=-16/30`

`=-8/15`

For positive slope we have

`a/b=8/15`

`\therefore a+b=8+15`

`=23`

Answer 2

The eqution of the line passing through `(6,0)` and having slope `l` will be

`(y-0)=l(x-6)`

`=>-lx+y+6l=0`

The distance of this line from the point `(1,3)` is 5.

So

`(-l*1+3+6l)/sqrt(l^2+1^2)=5`

`=>(5l+3)^2=(5sqrt(l^2+1))^2`

`=>25l^2+30l+9=25(l^2+1)`

`=>30l=16`

`=l=16/30=8/15`

If the slope `l` is expressed in the form of `a/b`,where `a andb` are prime to each other then `a=8andb=15`

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