\( \begin{array}{ll}\square \text { 1. Match each set of information to the correct equation: } & \\ \)\begin{tabular}{ll}\text { 1. parallel to }\( y=x-5 \text {, passes through }(2,2) \) & \text { a. }\( y=x-3 \) \\ \text { 2. parallel to }\( y=\frac{1}{3} x-4 \text {, passes through }(3,-1) \) & \text { b. }\( y=\frac{1}{3} x+3 \) \\ \text { 4. parallel to }\( y=x-5 \text {, passes through }(1,-2) \) & \text { c. }\( y=x \) \\ \hdashline & \text { d. }\( y=\frac{1}{3} x-2 \)\end{tabular}\end{array}

To find the equations, we need to recognize that parallel lines have the same slope. 1. The line parallel to \(y = x - 5\) has a slope of 1. The equation of a line in point-slope form is \(y - y_1 = m(x - x_1)\), substituting the point (2,2): \(y - 2 = 1(x - 2)\) simplifies to \(y = x - 3\) (which matches \(a\)). 2. For the line parallel to \(y = \frac{1}{3} x - 4\), the slope is \(\frac{1}{3}\). Using point (3,-1): \(y + 1 = \frac{1}{3}(x - 3)\) simplifies to \(y = \frac{1}{3}x - 2\) (which matches \(d\)). 3. The line parallel to \(y = x - 5\) from point (1,-2) also has a slope of 1: \(y + 2 = 1(x - 1)\) simplifies to \(y = x - 3\) (which we already identified as \(a\)). 4. The remaining option, for the point (1,-2), gives: \(y + 2 = 1(x - 1)\), so: \(y = x - 3\) confirming option \(a\). Thus, the matches should be: 1 → a 2 → d 4 → c