equation of a straight line

Question

Given two points A(2,3) and B(4,-2). Find the equation of a straight line which is parallel

to the line AB and passes through point (4,-6).

Answer 1

Hi Sofea!

First you try to find the slope of line AB since the line we are trying to find would have the same slope.

You can use m=(y₂-y₁)/(x₂-x₁) to find the slope.

m=(-2-3)/(4-2)

m=(-5)/2

Now that we have the slope you can put it into the slope-intercept form.

y=mx+b

y=(-5/2)x+b

-6=(-5/2)(4)+b

-6=(-10)+b

4=b

Finally, with the b and the slope found you can plug it back to the slope-intercept form.

y=mx+b

final answer: y=(-5/2)x+b

I hope that helps!

Answer 2

The equation of a line has the form

y = mx + b

where m is the slope and b is the y-intercept. We start by calculating slope using the points we are given and the equation

m = (y₂-y₁)/(x₂-x₁)

Plug in our values from the given points to get

m = (-2 - 3)/(4 - 2) = -5/2

So our slope is now -5/2. They say we need the equation of a line which is "parallel" to this line. Parallel lines have identical slopes, so we don't have to change anything. Our equation looks like this so far:

y = -5/2x + b

To find the other missing part of our equation, the y-intercept "b", we need to plug in the new point to our equation. It should look like this:

-6 = -5/2(4) + b

From here, we simply solve for b. We should get

4 = b

We now know everything we need to complete our equation. Putting it all together, it looks like

y = -5/2x + 4