Question

b. passing through (-2, -11) and perpendicular to x + 4y = 8

Explanation:

Step1: Find slope of given line

Rewrite \(x + 4y = 8\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Subtract \(x\) from both sides: \(4y=-x + 8\).
Divide by 4: \(y=-\frac{1}{4}x + 2\). So the slope of the given line \(m_1=-\frac{1}{4}\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes \(m_1\times m_2=- 1\). Let the slope of the perpendicular line be \(m_2\).
We have \(-\frac{1}{4}\times m_2=-1\).
Solve for \(m_2\): \(m_2 = 4\) (multiply both sides by - 4).

Step3: Use point - slope form to find equation

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(-2,-11)\) and \(m = 4\).
Substitute the values: \(y-(-11)=4(x - (-2))\).
Simplify: \(y + 11=4(x + 2)\).
Expand the right - hand side: \(y+11 = 4x+8\).
Subtract 11 from both sides: \(y=4x+8 - 11\).
So \(y = 4x-3\).

Answer:

The equation of the line is \(y = 4x-3\)