Question 1180823
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The perpendicular from a point to a line is the shortest distance from the point to the line.<br>
The distance from the point (p,q) to the line Ax+By+C=0 is given by the formula  {{{abs((Ap+Bq+C)/(sqrt(A^2+B^2)))}}}<br>
In your problem, A=k, B=1, C=1, p=2, and q=1; and the distance is sqrt(3).<br>
{{{abs(((k)(2)+(1)(1)+1)/(sqrt(k^2+1^2)))=abs((2k+2)/sqrt(k^2+1))}}}<br>
{{{abs((2k+2)/sqrt(k^2+1))=sqrt(3)}}}<br>
Square both sides....<br>
{{{((2k+2)^2/(k^2+1))=3}}}
{{{(2k+2)^2=3(k^2+1)}}}
{{{4k^2+8k+4=3k^2+3}}}
{{{k^2+8k+1=0}}}<br>
The quadratic does not factor over the integers, so use the quadratic formula.<br>
{{{k = (-8+-sqrt(64-4))/2 = -4+-sqrt(15)}}}<br>