Question

The length of a line segment is 7. Its end points are (1, 3) and (k, 3). Solve for k. Is there more than one solution? Explain.

Answer 1

$Endpoints:\\\\A(1,3)\ and\ B(k,3)\\\\Formula\ for\ length\ of\ line:\\\\ |AB|=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\\\\ |AB|=7\\\\7=\sqrt{(k-1)^2+(3-3)^2}\\\\ 7=\sqrt{k^2-2k-1}\ \ |^2\\\\ 49=k^2-2k-1$$49=k^2-2k-1\\\\ -k^2+2k+1+49=0\\\\ -k^2+2k+50=0\\\\ \Delta=b^2-4ac\\\\a=-1,\ b=2,\ c=50 \\\\\Delta=(2)^2-4*(-1)*50=4-=4+200=204\\\\ \sqrt{\Delta}=2\sqrt{51}$$\\\\k_1=\frac{-b-\sqrt{\Delta}}{2a}\ \ k_1=\frac{-2-2\sqrt{51}}{2*(-1)}=1+\sqrt{51} \\\\\ or\ k_2=\frac{-b+\sqrt{\Delta}}{2a}\ \ k_1=\frac{-2+2\sqrt{51}}{2*(-1)}=1-\sqrt{51}$