Question

select the graph of the line that passes through ( x(1,-4) ), parallel to ( overline{yz} ) with ( y(5,2) ) and ( z(-3,-5) )

Explanation:

Step1: Calculate the slope of \(\overline{YZ}\)

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For \(Y(5,2)\) and \(Z(-3,-5)\), we have \(m_{YZ}=\frac{-5 - 2}{-3 - 5}=\frac{-7}{-8}=\frac{7}{8}\).

Step2: Use the point - slope form for the line passing through \(X(1,-4)\)

The point - slope form is \(y - y_1=m(x - x_1)\). Substituting \(m = \frac{7}{8}\), \(x_1 = 1\), and \(y_1=-4\), we get \(y+4=\frac{7}{8}(x - 1)\), which simplifies to \(y=\frac{7}{8}x-\frac{7}{8}-4=\frac{7}{8}x-\frac{7 + 32}{8}=\frac{7}{8}x-\frac{39}{8}\).

Step3: Analyze the graph

We need to check which graph has a positive slope (since \(\frac{7}{8}>0\)) and passes through the point \((1,-4)\).

Answer:

Assuming the correct graph (with positive slope and passing through \((1,-4)\) among the options, if we assume the standard coordinate - based analysis of slopes and points, we need to pick the graph with a positive slope. Without seeing the exact visual details of the options (but based on slope and point), if we assume option B (assuming in a standard multiple - choice setup where positive slope graphs are considered and the point \((1,-4)\) lies on it). But in a proper exam - like situation, one would check the rise - over - run (for the slope \(\frac{7}{8}\)) and the point \((1,-4)\) on the graph.