Question 1157111
A line L1 passes through point[1,2]and has a gradient of 5. Another line L2 is perpendicular to L1 
and meets it at Point x = 4. Find the equation of L2 in the form y = mx + c
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        The solution by @mananth in his post is incorrect.

        He mistakenly assumed that line L2 passes through point (4,0),

        but the problem states something totally different, instead.


        See below my correct solution.



<pre>
Line L1: y - 2 = 5*(x-1)

         y = 5x - 5 + 2 = 5x - 3.    (1)

         Slope of L1 is 5.

         Hence, slope L2 is -1/5.

         L2 meets L1 at point x = 4,  where  y = 5*4 - 3 = 17,  according to equation (1).


 Thus, L2 has the slope -1/5 and passes through point (4,17)


Hence, an equation for L2 is

    y - 17 = {{{(-1/5)*(x - 4)}}},

or

    y - 17 = {{{(-1/5)x}}} + {{{4/5}}},

    y = {{{(-1/5)x}}} + 17{{{4/5}}}.      <<<---===  <U>ANSWER</U>
</pre>

Solved correctly.