MHB Finding the Equation of a Line Given Two Points

  • Thread starter Thread starter mathdad
  • Start date Start date
AI Thread Summary
To find the equation of the line passing through the points (6, -3) and with a y-intercept of 8, start with the slope-intercept formula y = mx + b, where b = 8. The y-intercept can be represented as the point (0, 8). The slope m is calculated using the formula m = (y2 - y1) / (x2 - x1), substituting the known values. After determining the slope, plug m and b into the slope-intercept equation to find the complete equation of the line. This method effectively utilizes both points and the y-intercept to derive the line's equation.
mathdad
Messages
1,280
Reaction score
0
Find an equation of the line that passes through (6, -3) and has y-intercept 8.

I know y = mx + b is the slope-intercept formula. In the formula, b represents the y-intercept. I also see that 8 is given to be b in this case.

The y-intercept can be written as (0, 8).

Do I now find the slope m?
Afterward, use one of the points and m to plug into the point-slope formula. Finally, I must isolate y.

Is this right?
 
Mathematics news on Phys.org
Like you, I would begin with the slope-intercept form of a line:

$$y=mx+b$$

We are given $b=8$, and we know two points on the line, so we can compute the slope $m$:

$$m=\frac{8-(-3)}{0-6}=$$?

Then, just plug in the values for $m$ and $b$. :)
 
MarkFL said:
Like you, I would begin with the slope-intercept form of a line:

$$y=mx+b$$

We are given $b=8$, and we know two points on the line, so we can compute the slope $m$:

$$m=\frac{8-(-3)}{0-6}=$$?

Then, just plug in the values for $m$ and $b$. :)

I can take it from here. Thanks.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top