# Finding the equation of a line from two given points.

• May 6th 2010, 12:29 PM
daniel323
Finding the equation of a line from two given points.
plz neeed help with this If a line passes through the points (-2, -4) and (3, -1), the equation of the line is y + 4 = 3/5(x + _____).
• May 6th 2010, 12:33 PM
First find the equation of the line as $y=mx+b$. You know how do this right?
Now, $y+4=mx+b+4$ from which you should be able to factor $\frac{3}{5}$.
• May 6th 2010, 12:40 PM
daniel323
i dont know how to do this so 3/5 is the answer
• May 6th 2010, 12:47 PM
undefined
Quote:

Originally Posted by daniel323
i dont know how to do this so 3/5 is the answer

This question is supposed to be testing your understanding of and ability to recognize the point-slope equation.

Do you see it?
• May 6th 2010, 12:47 PM
daniel323
can u help with this problem also Yfrog Image : yfrog.com/j9math14p
• May 6th 2010, 12:48 PM
Ok, first, the equation of the line should be of the form $y=mx+b$ where m is slope and given by $m=\frac{y_2-y_1}{x_2-x_1} = \frac{-1-(-4)}{3-(-2)} = \frac{3}{5}$.
Hence, $y=\frac{3}{5}x+b$. Still need to find b.
the point (-2,-4) belongs to the lines, hence: $-4=\frac{3}{5}\times (-2)+b$, therefore: $b = -4+\frac{6}{4} = -4+\frac{3}{2}$.
Hence, $y=\frac{3}{5}x-4+\frac{3}{2}$.
So, $y+4=\frac{3}{5}x+\frac{3}{2} = \frac{3}{5}(x+\frac{3/2}{3/5}) = \frac{3}{5}(x+\frac{5}{2})$.
$\frac{5}{2}$ is the answer you need.
• May 6th 2010, 12:50 PM
daniel323
i got the first problem already but need second problem
• May 6th 2010, 12:50 PM
For the second question, it is clear from the figure that the two lines are parallel. This means that they have the same slope, hence your answer is $-\frac{4}{3}$.
• May 6th 2010, 12:53 PM
daniel323
thank u so much.
• May 6th 2010, 12:54 PM
masters
Quote:

Originally Posted by daniel323
plz neeed help with this If a line passes through the points (-2, -4) and (3, -1), the equation of the line is y + 4 = 3/5(x + _____).

Hi daniel323,

Are you familiar with the 'point-slope' form of a linear equation:

$y-y_1=m(x-x_1)$

Find your slope, which is already given to you as $\frac{3}{5}$

Use (-2, -4) as $(x_1, y_1)$ and the equation:

$y-y_1=m(x-x_1)$

And fill in the blank: $y-(-4)=\frac{3}{5}(x-(-2))\Longrightarrow \boxed{y+4=\frac{3}{5}(x+{\color{red}2})}$

2 is the answer you need.
• May 6th 2010, 12:56 PM
undefined
Quote:

Ok, first, the equation of the line should be of the form $y=mx+b$ where m is slope and given by $m=\frac{y_2-y_1}{x_2-x_1} = \frac{-1-(-4)}{3-(-2)} = \frac{3}{5}$.
Hence, $y=\frac{3}{5}x+b$. Still need to find b.
the point (-2,-4) belongs to the lines, hence: $-4=\frac{3}{5}\times (-2)+b$, therefore: $b = -4+\frac{6}{4} = -4+\frac{3}{2}$.
Hence, $y=\frac{3}{5}x-4+\frac{3}{2}$.
So, $y+4=\frac{3}{5}x+\frac{3}{2} = \frac{3}{5}(x+\frac{3/2}{3/5}) = \frac{3}{5}(x+\frac{5}{2})$.
$\frac{5}{2}$ is the answer you need.

I know the OP said this problem is already solved, but I'd just like to point out that there is an error here, there was a (6/4) where there should have been a (6/5), and the value for b is wrong.

Using point-slope equation, once you verify that the slope is (3/5), you can immediately see that the answer is 2.

Edit: Ah, masters beat me to it.
• May 6th 2010, 12:59 PM