What is the equation of the line passing through #(3,7)# and #(-8,12)#?

1 Answer
Nov 20, 2015

It is important that you be more specific in your question. I am assuming you mean a 'strait line graph'.

#color(green)(y= -5/11x+92/11)#
You did not request the y and x intercept so not calculated

Explanation:

#color(green)("The standard (std) form of this type of plot is")#

#color(green)(y=mx+c)#

#color(blue)(~~~~~~~~~~~~ "Pre-amble"~~~~~~~~~~~~~~~~~)#

#color(brown)(y " is the dependent variable as it is the outcome of and thus")#
#color(brown)("controlled by what is on the right of the =")#

#color(brown)(x" is the independent variable as it can take on any value you")#
#color(brown)("chose.")#

#color(brown)(m color(white)(x) "is the gradient of the 'curve'. Yes it is mathematically correct")#
#color(brown)("to call a strait line plot a curve. People do not tend to")#
#color(brown)("though!")#

#color(blue)(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~)#

#color(green)("We start by determining the gradient. We then")#
#color(green)("substitute a pair of the given co-ordinates (Ordered pairs)")#
#color(green)("to find the value of the constant.")#

#color(blue)("Find the gradient")#

#m = ("change in up or down")/("change in along") -> (y_2-y_1)/(x_2-x_1)#

Let #(x_1,y_1) = ( 3,7)#
the left most pair chosen to be so as you listed them first!

Let #(x_2,y_2)=(-8,12)#

Then #m = (y_2-y_1)/(x_2-x_1) = (12-7)/((-8)-3) = 5/(-11)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Find the constant")#

Using ; #y=mx+c#

then for#color(white)(xx) (x_1,y_1) -> y_1=mx_1+c#

#=> 7=(-5/11)3+c#

#c= 7+15/11 = 8 4/11 " or " 92/11#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together!")#

Thus, we now have what we need: the gradient and the constant

The equation is : #y= -5/11x+92/11#

#color(green)("Fractions are precise:::: Decimals are less so!!! Use fractions in")#
#color(green)("preference unless specifically instructed otherwise!!!!!!!")#

TonyB