Question 3 of 10 Identify an equation in point-slope form for the line perpendicular to y=-(1)/(3)x-6 that passes through (-1,5) A. y-5=3(x+1) B. y+5=(1)/(3)(x-1) C y-5=-(1)/(3)(x+1) D y+1=3(x-5)

## Step 1<br />The given equation is \(y=-\frac{1}{3}x-6\). The slope of this line is \(-\frac{1}{3}\).<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. Therefore, the slope of the line perpendicular to the given line is \(-\frac{1}{-\frac{1}{3}} = 3\).<br /><br />## Step 3<br />The point-slope form of a line is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.<br /><br />## Step 4<br />Substituting the slope \(m = 3\) and the point \((-1, 5)\) into the point-slope form, we get \(y - 5 = 3(x - (-1))\), which simplifies to \(y - 5 = 3(x + 1)\).