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|=√((x_b-x_a)^2+(y_b-y_a)^2)\\\\ |AB|=7\\\\7=√((k-1)^2+(3-3)^2)\\\\ 7=√(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\\\\ √(\Delta)=2√(51)`
`\\\\k_1=(-b-√(\Delta))/(2a)\ \ k_1=(-2-2√(51))/(2*(-1))=1+√(51) \\\\\ or\ k_2=(-b+√(\Delta))/(2a)\ \ k_1=(-2+2√(51))/(2*(-1))=1-√(51)`