Question

write an equation (a) in slope - intercept form and (b) in standard form for the line passing through (1,9) and perpendicular to 2x + 5y = 1.

Explanation:

Step1: Find the slope of the given line

Rewrite \(2x + 5y=1\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Solve for \(y\):
\(5y=-2x + 1\)
\(y=-\frac{2}{5}x+\frac{1}{5}\)
The slope of the line \(2x + 5y = 1\) is \(m_1=-\frac{2}{5}\).

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the perpendicular line be \(m_2\). Then \(m_1\times m_2=-1\).
Substitute \(m_1 = -\frac{2}{5}\) into the equation:
\(-\frac{2}{5}\times m_2=-1\)
Solve for \(m_2\):
\(m_2=\frac{-1}{-\frac{2}{5}}=\frac{5}{2}\)

Step3: Write the equation in slope - intercept form

We use the point - slope form \(y - y_1=m(x - x_1)\) where \((x_1,y_1)=(1,9)\) and \(m = \frac{5}{2}\).
\(y - 9=\frac{5}{2}(x - 1)\)
Expand the right - hand side:
\(y-9=\frac{5}{2}x-\frac{5}{2}\)
Add 9 to both sides:
\(y=\frac{5}{2}x-\frac{5}{2}+9\)
\(y=\frac{5}{2}x+\frac{- 5 + 18}{2}\)
\(y=\frac{5}{2}x+\frac{13}{2}\) (This is the slope - intercept form \(y=mx + b\) with \(m = \frac{5}{2}\) and \(b=\frac{13}{2}\))

Step4: Write the equation in standard form

The standard form of a line is \(Ax+By = C\) where \(A\), \(B\), and \(C\) are integers and \(A\geq0\).
Start with \(y=\frac{5}{2}x+\frac{13}{2}\)
Multiply both sides by 2 to eliminate the fractions:
\(2y = 5x+13\)
Rearrange to get:
\(5x-2y=-13\)

Answer:

(a) Slope - intercept form: \(y=\frac{5}{2}x+\frac{13}{2}\)
(b) Standard form: \(5x - 2y=-13\)