What is an equation of the line that passes through the point (-4,-7) and is perpendicular to the line x+2y=4 Answer Attempt 1 out of 2 square

## Step 1<br />The given line is \(x + 2y = 4\). We can rewrite this in slope-intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. <br /><br />### \(x + 2y = 4\) can be rewritten as \(y = -\frac{1}{2}x + 2\)<br /><br />So, the slope of the given line is \(-\frac{1}{2}\).<br /><br />## Step 2<br />The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. <br /><br />### The negative reciprocal of \(-\frac{1}{2}\) is \(2\).<br /><br />So, the slope of the line perpendicular to the given line is \(2\).<br /><br />## Step 3<br />We know that the line passes through the point \((-4,-7)\). We can use the point-slope form of a line equation, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.<br /><br />### Substituting \(m = 2\) and \((x_1, y_1) = (-4,-7)\) into the equation, we get \(y - (-7) = 2(x - (-4))\).<br /><br />## Step 4<br />Simplify the equation to get the final equation of the line.<br /><br />### \(y + 7 = 2(x + 4)\) simplifies to \(y = 2x + 1\).