Find the equation of a line described as follows, and express your answer in point-slope form, slope-intercept form, and standard form. Find the equation of the line perpendicular to the line $y=\frac{1}{6} x-2$ that passes through the point $(-1,2)$.

Step 1 ：First, we find the slope of the given line. The equation of the line is in the form \(y = mx + c\), where \(m\) is the slope. So, the slope of the given line is \(\frac{1}{6}\).

Step 2 ：Since the line we are looking for is perpendicular to the given line, its slope is the negative reciprocal of the slope of the given line. So, the slope of the line we are looking for is \(-6\).

Step 3 ：We know that the line we are looking for passes through the point \((-1,2)\). So, we can use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line.

Step 4 ：Substituting \(m = -6\), \(x_1 = -1\), and \(y_1 = 2\) into the point-slope form, we get \(y - 2 = -6(x + 1)\). This is the equation of the line in point-slope form.

Step 5 ：To express the equation of the line in slope-intercept form, we simplify the equation \(y - 2 = -6(x + 1)\) to get \(y = -6x - 4\).

Step 6 ：To express the equation of the line in standard form, we rearrange the equation \(y = -6x - 4\) to get \(6x + y = -4\).

Step 7 ：So, the equation of the line in point-slope form is \(y - 2 = -6(x + 1)\), in slope-intercept form is \(y = -6x - 4\), and in standard form is \(6x + y = -4\).