# Linear equations in two variables

• May 23rd 2008, 06:29 PM
rowdy3
Linear equations in two variables
Find the x and y intercepts and slope (if they exist) for the line given.
-2x + y = 14
x intercept Would it be (-2,0)
y intercept Would it be ( 0,4)
Slope Would it be 2

The slope of a line is 1/3
What is the slope of a line parallel to the given line? Would it be 1/3
What is the slope of a line perpendicular to the given line? Would it be -1/3

What if the slope (if it exists) of a line given by the equation x = -5? Would it be undefined?
Thanks
• May 23rd 2008, 07:41 PM
topsquark
Quote:

Originally Posted by rowdy3
Find the x and y intercepts and slope (if they exist) for the line given.
-2x + y = 14
x intercept Would it be (-2,0)
y intercept Would it be ( 0,4)
Slope Would it be 2

These points aren't even on the line! How can they be intercepts??

You got the slope right, though.
$y = 2x + 14$
The x-intercept is where y = 0. Solve this for x.
The y-intercept is where x = 0. Solve this for y.

Quote:

Originally Posted by rowdy3
The slope of a line is 1/3
What is the slope of a line parallel to the given line? Would it be 1/3
What is the slope of a line perpendicular to the given line? Would it be -1/3

Yes to the first, no to the second. Two perpendicular lines have slopes that have the following property:
$m_1m_2 = -1$

So you know that $m_1 = 1/3$.
$\frac{1}{3} \cdot m_2 = -1$
Solve for $m_2$.

Quote:

Originally Posted by rowdy3
What if the slope (if it exists) of a line given by the equation x = -5? Would it be undefined?
Thanks

Yes. The slope of a line is rise/run. Consider two points on this line, say (-5, 1) and (-5, 6).
$\text{slope} = \frac{1 - 6}{-5 - (-5)} \to \frac{-5}{0}$
which is undefined.

-Dan
• May 24th 2008, 07:24 PM
rowdy3
This is what I did.
y= 2x + 14
x- intercept 0= 2x + 14 14/2
x- intercept (7,0)
y- intercept y= 2(0) + 14
y- intercept (0,14)

The slope of a line is 1/3
1/3 * - 3/1 =-1
What is the slope of a line perpendicular to the given line? Is it -3/1
• May 24th 2008, 09:48 PM
Reckoner
Quote:

Originally Posted by rowdy3
This is what I did.
y= 2x + 14
x- intercept 0= 2x + 14 14/2
x- intercept (7,0)
y- intercept y= 2(0) + 14
y- intercept (0,14)

The slope of a line is 1/3
1/3 * - 3/1 =-1
What is the slope of a line perpendicular to the given line? Is it -3/1

This is correct except for your x-intercept. Again, $\left(7,\;0\right)$ isn't even on the line (since $0\neq2\cdot7 + 14$). Your method is fine, but you made a slight mistake when solving for $x$.
• May 25th 2008, 09:30 AM
rowdy3
Write an equation of the line passing through the point (12, -10) having slope 4/3. Write your final answer in slope-intercept form.

Write an equation of the line passing through the point (0, 2) and parallel to the line x-y = 7. Write your final answer in slope-intercept form.

1. y+10= 4/3 ·(x- 12) My answer is y= 4/3·x -26

2. P(0,2) slope= 1--> y-2= 1(x-0) My answer is y= x + 2
• May 25th 2008, 10:58 AM
earboth
Quote:

Originally Posted by rowdy3
Write an equation of the line passing through the point (12, -10) having slope 4/3. Write your final answer in slope-intercept form.

Write an equation of the line passing through the point (0, 2) and parallel to the line x-y = 7. Write your final answer in slope-intercept form.

1. y+10= 4/3 ·(x- 12) My answer is y= 4/3·x -26

2. P(0,2) slope= 1--> y-2= 1(x-0) My answer is y= x + 2

Everything is OK (Clapping)

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