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/bab where a and b are relatively prime. Then the sum of a and b is?

2 Answers

2323

Explanation:

Let the equation of line be

ax+by+c=0ax+by+c=0

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

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

6a+c=06a+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

Jul 26, 2018

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