SOLUTION: This is on page 542 number 9
The question is Find an equation of the line containing the point(5,4) and perpendicular to the line 2x - 4y = 12
I tried to solve this and this is
Question 184354This question is from textbook introductory and intermediate algebra
: This is on page 542 number 9
The question is Find an equation of the line containing the point(5,4) and perpendicular to the line 2x - 4y = 12
I tried to solve this and this is how I did it
2x - 4y = 12
-2x -2x
-4y = -2x + 12
then i divided both side by -4
y = 4/2x + -3
so then I plugged that in to the next equation
y - 4 = 4/2(x - 5)
y - 4 = 4/2x -20/5
then I added 4 to both sides
y = 4/2x - 0
I got this wrong and I do not understand how I did it wrong can you please try to help me? This question is from textbook introductory and intermediate algebra
First, when you divided by -4 when deriving the slope-intercept form of the given equation, you should have had a result like this:
Rather than
Second you used the same slope number for your derived equation that you developed when you put your given equation in slope-intercept form. What you developed was an equation for a line parallel to your incorrect slope-intercept version of the given equation rather than perpendicular to the given line.
You forgot the rule that perpendicular lines have slopes that are negative reciprocals of each other, or:
In your case, , therefore
And your derived equation should look like:
Third, when you went from:
y - 4 = 4/2(x - 5)
to
y - 4 = 4/2x -20/5
You multiplied the -5 by 4 but then forgot to divide by 2. Your result should have been
y - 4 = 4/2x - 10/5
Having said all of that, you really have the right idea about all of this. You just made a couple of easy-to-make mistakes. Keep up the good work.
