# Thread: Graph Involving Linear Equation on degrees

1. ## Graph Involving Linear Equation on degrees

Ok here is the Problem.

Weather Temperature. The National Oceanic and Atmospheric ADministration (NO) has an online cconversion chart that relates degrees Fahrenheit to degrees Celsius. 77deg F is equevalent to 25deg C and 68deg F is equivalent to 20deg C. Assuming relationship is linear, write the equation relating degrees Celsius C to degrees Fahrenheit F. What temperature is the same in both degress Celsius and degress Fahrenheit?

Here steps I did:

77deg F = 25deg C
68deg F = 20deg C

Let x = Fahrenheit and y = Celsius
Linear Equation: y=mx + b or C = mF + b

Two Points of this Equation is (77,25), (68,20)

Solving the equation:

Calculate the change of rate:
$\frac {Y2 + Y1} {X2-X1}$

$m= \frac {25-20} {68-77} = \frac {-5} {-9}$

$m= \frac {-5} {-9}$

Subtitute the slope into the linear equation.

$m= \frac {-5} {-9} x + b$

finding b:

I use either point to satisfy the equation. I use (77,25)
Let x = 77
$
C = \frac {-5}{-9}F + b$

$
25= \frac {-5} {-9} (77) + b$

$
25= \frac {385} {9} + b$

$
b = \frac {160} {9}$

$Equation is : C = \frac {-5} {-9}F + \frac {160} {9}$

I don't know if i'm this right. I still haven't answer the other question: What temperature are both the same degrees.

2. you were correct throughout, but you made the slope negative in the denominator and numerator, making it positive . using what you said at first for the variables you would get the points (77,25) and (68,20).

using slope formula you get:
$
\frac{20-25}{68-77}
$

giving you a slope of $\frac{5}{9}$

next i would use point slope form at first then rearrange,so..

$
y-y_{1}=m(x-x_{1})
$

using the point (68,20)

$
y-(20)=\frac{5}{9}(x-68)
$

$
y=(\frac{5}{9}x-\frac{340}{9})+20
$

$
y=\frac{5}{9}x-\frac{160}{9}
$

now to find when they are the same.to do this set y=x

$
x=\frac{5}{9}x-\frac{160}{9}
$

$
x-\frac{5}{9}x=-\frac{160}{9}
$

$
\frac{4x}{9}=-\frac{160}{9}
$

$
4x=-160
$

$
x=-40
$

so at -40 degrees they are the same

3. Would you not just use the $y-y1=m(x-x1)$ formula?
You have a point (77,25) so substituting that into the equation:
$y-25=m(x-77)$

And using the rule for calculating a gradient:
$m=\frac{y2-y1}{x2-x1}$
Gives:
$m=\frac{25-20}{77-68}$
$=\frac{5}{9}$

$y-25=\frac{5}{9}(x-77)$
$9C-225=5(F-77)$
$9C-225=5F-385$
$9C=5F-160$
Checking this by substituting the values back into it suggests that it is correct.

For the second part, you are trying to find when C=F. So just substitute this into the equation:
9F=5F-160
4F=-160
F=-40