Question

write an equation of the line passing through the given point and satisfying the given condition. give the equation (a) in slope - intercept form and (b) in standard form. (12, 3), perpendicular to 4x - y = 6 (a) write the equation of the line in slope - intercept form. (simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find slope of given line

Rewrite \(4x - y = 6\) in slope - intercept form \(y=mx + b\) (where \(m\) is slope).
\(y = 4x-6\), so slope of given line \(m_1 = 4\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let slope of required line be \(m_2\).
\(m_1\times m_2=-1\), \(4\times m_2=-1\), so \(m_2=-\frac{1}{4}\).

Step3: Use point - slope form to find equation

Point - slope form is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(12,3)\) and \(m =-\frac{1}{4}\).
\(y - 3=-\frac{1}{4}(x - 12)\)

Step4: Simplify to slope - intercept form

\(y-3=-\frac{1}{4}x + 3\) (distribute \(-\frac{1}{4}\) to \(x\) and \(- 12\))
\(y=-\frac{1}{4}x+3 + 3\)
\(y=-\frac{1}{4}x + 6\)

Answer:

\(y =-\frac{1}{4}x+6\)