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. (9, 4), parallel to 2x - y = 6 (a) write the equation of the line in slope - intercept form. -2x + y = -14 (simplify your answer. use integers or fractions for any numbers in the equation.)

Explanation:

Step1: Find the slope of the given line

First, rewrite the given line \(2x - y = 6\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
We can solve \(2x - y=6\) for \(y\):
\(-y=-2x + 6\), multiply both sides by \(- 1\) to get \(y = 2x-6\). So the slope \(m\) of the line \(2x - y = 6\) is \(2\).

Step2: Determine the slope of the parallel line

Since parallel lines have the same slope, the line we want to find also has a slope \(m = 2\).

Step3: Use the point - slope form to find the equation of the line

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(9,4)\) and \(m = 2\).
Substitute \(x_1 = 9\), \(y_1 = 4\) and \(m = 2\) into the point - slope form:
\(y-4=2(x - 9)\)

Step4: Simplify to slope - intercept form

Expand the right - hand side: \(y-4=2x-18\)
Add \(4\) to both sides: \(y=2x-18 + 4\), so \(y=2x-14\)

Step5: Convert to standard form (\(Ax+By = C\), where \(A\), \(B\), and \(C\) are integers and \(A\geq0\))

Starting from \(y = 2x-14\), subtract \(2x\) from both sides: \(-2x + y=-14\) (or we can multiply both sides by \(-1\) to get \(2x-y = 14\), but the given answer in the box is \(-2x + y=-14\), and we can check:
If \(y=2x - 14\), then moving \(2x\) to the left side gives \(-2x + y=-14\))

Answer:

(a) The equation of the line in slope - intercept form is \(y = 2x-14\)
(b) The equation of the line in standard form is \(-2x + y=-14\) (or \(2x - y=14\))