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?

2 Answers
Harish Chandra Rajpoot
Jul 26, 2018

#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#

P dilip_k
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#

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