Question 1200868
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For angle ABC to be 90 degrees, the segments AB and BC must be perpendicular, so the product of the slopes of the two segments must be negative reciprocals.<br>
slope of AB: (2k-3)/(k-11)
slope of BC: (2k+11)/(k+1)<br>
{{{((2k-3)/(k-11))((2k+11)/(k+1))=-1}}}<br>
{{{4k^2+16k-33=-1(k^2-10k-11)}}}<br>
{{{5k^2+6k-44=0}}}<br>
The solutions are irrational; you can use the quadratic formula to find them.<br>
The idea of the problem is good; however, it would have been far more educational if the solutions had been rational so a realistic diagram could be drawn.<br>