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The resulting rho and theta are indeed one step away from the line(s) you are looking for in the image. They represent a line passing through the origin that is perpendicular to the line you want to find. This page has a great introduction to Hough transform concepts, and explains this point.

You might want to use cvHoughLines2 instead, which will allow finding segments instead of lines and a couple of other improvements.

In this example you find some code to draw the lines from rho and theta; the idea is simple: calculate a point of that line, namely x0 = rho cos(theta), y0 = rho sin(theta), and notice that the slope of the line is -theta, cos(-theta)=cos(theta), sin(-theta)=-sin(theta), and the very large numbers are there to use integer arithmetic.

for( size_t i = 0; i < lines.size(); i++ )
{
    float rho = lines[i][0];
    float theta = lines[i][1];
    double a = cos(theta), b = sin(theta);
    double x0 = a*rho, y0 = b*rho;
    Point pt1(cvRound(x0 + 1000*(-b)),
              cvRound(y0 + 1000*(a)));
    Point pt2(cvRound(x0 - 1000*(-b)),
              cvRound(y0 - 1000*(a)));
    line( color_dst, pt1, pt2, Scalar(0,0,255), 3, 8 );
}

You might want to use cvHoughLines2 instead, which will allow finding segments instead of lines and a couple of other improvements.

In this example you find some code to draw the lines from rho and theta; the idea is simple: calculate a point of that line, namely x0 = rho cos(theta), y0 = rho sin(theta), and notice that the slope of the line is ~~-theta,~~ (-theta), cos(-theta)=cos(theta), sin(-theta)=-sin(theta), and the very large numbers are there to use integer arithmetic.

for( size_t i = 0; i < lines.size(); i++ )
{
    float rho = lines[i][0];
    float theta = lines[i][1];
    double a = cos(theta), b = sin(theta);
    double x0 = a*rho, y0 = b*rho;
    Point pt1(cvRound(x0 + 1000*(-b)),
              cvRound(y0 + 1000*(a)));
    Point pt2(cvRound(x0 - 1000*(-b)),
              cvRound(y0 - 1000*(a)));
    line( color_dst, pt1, pt2, Scalar(0,0,255), 3, 8 );
}