Determine the tangent and normal lines to the function below at a specified point.
${(x+y)}^2=4x+6$, $(1,3)$

Tangent line: $y=-\frac{1}{2}x+\frac{3}{2}$

Normal line: $y=-2x+1$

Tangent line: $y=-2x+\frac{1}{2}$

Normal line: $y=2x-1$

Tangent line: $y=-\(\frac\)12x+\(\frac\)72$

Normal line: $y=2x+1$

Tangent line: $y=\frac{1}{2}x+\frac{7}{2}$

Normal line: $y=-2x+2$