Question

what do the following two equations represent?

• $y + 6 = -\frac{1}{3}(x - 15)$
• $2x + 6y = 24$

choose 1 answer:
a the same line
b distinct parallel lines
c perpendicular lines
d intersecting, but not perpendicular lines

Explanation:

Step1: Find slope of first line

The first equation is in point - slope form \(y - y_1=m(x - x_1)\), where \(m\) is the slope. For \(y + 6=-\frac{1}{3}(x - 15)\), the slope \(m_1 =-\frac{1}{3}\).

Step2: Find slope of second line

Rewrite the second equation \(2x + 6y = 24\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Subtract \(2x\) from both sides: \(6y=-2x + 24\).
Divide each term by 6: \(y=-\frac{2}{6}x+\frac{24}{6}\), simplify to \(y =-\frac{1}{3}x + 4\). The slope \(m_2=-\frac{1}{3}\).

Step3: Check if lines are parallel or same

Since \(m_1 = m_2=-\frac{1}{3}\), the lines are parallel. Now check if they are the same line.
Take a point from the first line. Let's find the y - intercept of the first line. From \(y + 6=-\frac{1}{3}(x - 15)\), when \(x = 0\), \(y+6=-\frac{1}{3}(- 15)=5\), so \(y=5 - 6=-1\). The first line has a point \((0,-1)\)? Wait, no, let's use the slope - intercept form of the first line.
From \(y + 6=-\frac{1}{3}(x - 15)\), expand: \(y+6=-\frac{1}{3}x + 5\), then \(y=-\frac{1}{3}x+5 - 6=-\frac{1}{3}x - 1\).
The second line is \(y =-\frac{1}{3}x + 4\). Since the y - intercepts are different (\(-1\) vs \(4\)), the lines are distinct parallel lines.

Answer:

B. Distinct parallel lines