The Straight Line Question

Ques: Find the equation of the line that passes through the y-intercept of the line 3x+5y+2=0 and is parallel to the line 2x-4y-1=0.

From 3x+5y+2=0 , i found out that y-intercept= -2/5 x-intercept= -2/3

From 2x-4y-1=0 the, i found out that m= 1/2

Since m1=m2, i sub into the formula: y+2/5=1/2(x+2/3)

answer i got is y=1/2x-1/15 but is wrong.

can anybody help? Thanks.

Re: The Straight Line Question

Quote:

Originally Posted by

**freshbox** Ques: Find the equation of the line that passes through the y-intercept of the line 3x+5y+2=0 and is parallel to the line 2x-4y-1=0.

From 3x+5y+2=0 , i found out that y-intercept= -2/5 x-intercept= -2/3 From 2x-4y-1=0 the, i found out that m= 1/2

Why are you doing anything with the x-intercept?

The new line has the same y-intercept, $\displaystyle \left(0,\tfrac{-2}{5}\right)$ as the line $\displaystyle 3x+5y+2=0$

Re: The Straight Line Question

I am trying to find the x-intercept and y-intercept to use it in this formula: y-y1=m(x-x1)

Re: The Straight Line Question

Quote:

Originally Posted by

**freshbox** I am trying to find the x-intercept and y-intercept to use it in this formula: y-y1=m(x-x1)

You can still use that formula: $\displaystyle x_1=\tfrac{-2}{5}~\&~y_1=0$.

Re: The Straight Line Question

How come your x1= -2/5 y1= 0 ?

I thought i found the intercept as y-intercept= -2/5 x-intercept= -2/3

Shouldn't y1=-2/5 and x1=-2/3?

I know i am wrong, please explain it to me if you don't mind, thanks.

Re: The Straight Line Question

Quote:

Originally Posted by

**freshbox** How come your x1= -2/5 y1= 0 ?

I thought i found the intercept as y-intercept= -2/5 x-intercept= -2/3

Shouldn't y1=-2/5 and x1=-2/3? I know i am wrong, please explain it to me if you don't mind, thanks.

If you write the equation of a line through $\displaystyle (p,q)$ with slope $\displaystyle m$ you get $\displaystyle y-q=m(x-p)$ (*point-slope form*).

So line through $\displaystyle \left(0,\tfrac{-2}{5}\right)$ with slope $\displaystyle \tfrac{1}{2}$ is $\displaystyle y+\tfrac{2}{5}\right=\tfrac{1}{2}\left(x-0)$.

Re: The Straight Line Question

Re: The Straight Line Question

(p,q) as (x,y)

(p,q) y-q=m(x-p)

this part i get it

(x,y)

(0,-2/5) y-0=1/2(x+2/5)

this part how come your x and y swap places? i thought the formula is : y1-y=m(x1-x)

Re: The Straight Line Question

You are correct. What we get for copy and paste.

I edited it and it is now correct.

Re: The Straight Line Question

if the question says passes through the y-intercept means x-intercept = 0? Is this always the case?

So the x-intercept i calculated as -2/3 is wrong?

Re: The Straight Line Question

Rewrite in slope-intercept form is what I would do first, personally, for this type of problem.

y=(-3/5)x-2/5

y=(1/2)x-1/4

Since it'll be parallel to equation 2, y=mx+b m=1/2.

Now, with x=0 in the first equation, the y-intercept, y=(-3/5)(0)-2/5=-2/5. So y=(1/2)x+b must equal -2/5 when x=0. -2/5=b.

y=(1/2)x-2/5, or 5x-10y-4=0.

Re: The Straight Line Question

if the question says passes through the y-intercept means x-intercept = 0? Is this always the case?

Re: The Straight Line Question

You aren't supposed to be paying attention to x-intercepts in this question. The y-intercept is the point on the graph where the line intersects the vertical y-axis, which is at x=0, no? Therefore, to find where the graph passes through the y-axis, you must set x=0 and evaluate the equation, which will return the y-intercept.

(Just for reference, the x-intercept is solved by substituting 0 in place of y and solving for x, but that isn't necessary for this problem)

http://www.algebra-class.com/images/intercepts.gif