# Find an equation of the line that satisfies the given conditions

• February 17th 2009, 10:28 PM
rckstr4lfe
Find an equation of the line that satisfies the given conditions
I have three "find the equation of the line" problems I cannot complete correctly.

1.
Through points (-1, -6) and (2, -4)

2.
Through the origin; parallel to the line 3x + 15y = 22

--- on this one, I was able to find the slope. m= -3/15 but I got stuck after that.

3.
Through (-1, -2); perpendicular to the line 2x + 5y + 8 = 0

---I also found the slope for this one: m= 5/2, but again, I got stuck after.

If anyone could help me with these, I would be grateful.

Thanks!
• February 17th 2009, 11:01 PM
earboth
Quote:

Originally Posted by rckstr4lfe
I have three "find the equation of the line" problems I cannot complete correctly.

1.
Through points (-1, -6) and (2, -4)

2.
Through the origin; parallel to the line 3x + 15y = 22

--- on this one, I was able to find the slope. m= -3/15 but I got stuck after that.

3.
Through (-1, -2); perpendicular to the line 2x + 5y + 8 = 0

---I also found the slope for this one: m= 5/2, but again, I got stuck after.

If anyone could help me with these, I would be grateful.

Thanks!

to #1:

If you have 2 points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ then this is the equation of the line

$P_1P_2: \dfrac{y-y_1}{x-x_1} = \underbrace{\dfrac{y_2-y_1}{x_2-x_1}}_{slope\ of\ line}$
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to #2:

A parallel to $3x + 15y = 22$ has the equation $3x + 15y = c$ . Plug in the coordinates of the origin to calculate the value of c.
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to #3:
If you have a points $P_1(x_1, y_1)$ and the slope of the line then this is the equation of the line

$P_1m: \dfrac{y-y_1}{x-x_1} = m~\implies~y-y_1=m(x-x_1)$

Plug in the values you already know.