What is the equation of the line passing through #(34,5)# and #(4,-31)#?

2 Answers
Nov 11, 2015

#y = (6x-179)/5#.

Explanation:

We will set up the co-ordinates as:
#(34, 5)#
#(4, -31)#.

Now we do subtraction of the #x#s and the #y#s.

#34 - 4 = 30#,
#5 -(-31) = 36#.

We now divide the difference in #y# over that in #x#.

#36/30 = 6/5#.
So #m# (gradient) #= 6/5#.

Equation of a straight line:
#y = mx +c#. So, let's find #c#. We substitute values of any of the coordinates and of #m#:
#5 = 6/5 * 34 + c#,
#5 = 204/5 + c#,
#c = 5 - 204/5#,
#c = -179/5#. So,

#y = (6x-179)/5#.

Nov 11, 2015

#color(blue)(y= 6/5x-35.8)#

Explanation:

Standard form equation is:

#color(blue)(y=mx+c............................(1))#

Where m is the slope (gradient) and c is the point where the plot crosses the y-axis in this context.

The gradient is the amount of up (or down) of y for the amount of along for the x-axis. #color(blue)("Always considered from left to right .")#

So #m -> (y_2-y_1)/(x_2-x_1) = ((-31)-5)/(4-34)#

As #(34,5)# is listed first you assume this is the left most point of the two.

#m= (-36)/(-30)# dividing negative into negative gives positive

#color(blue)(m=(36)/(30) = 6/5 .........................(2))#

Substitute (2) into (1) giving:

#color(blue)(y=6/5x+c............................(3))#

Now all we need to do is substitute known values for x and y to obtain that for c

Let #(x,y) -> (34,5)#

Then #y=6/5x+c" "# becomes:

#color(brown)(5=(6/5 times 34) +c)# #color(white)(xxx)#brackets used for grouping only

Subtract #color(green)((6/5 times 34))# from both sides giving

#color(brown)(5) -color(green)((6/5 times 34)) color(white)(xx) = color(white)(xx)color(brown)( ( 6/5 times 34)) -color(green)((6/5 times 34))color(brown)( +c)#

#c=5-(6/5times 34)#

#color(blue)(c = -35.8....................................(4))#

Substitute (4) into (3) giving:

#color(blue)(y= 6/5x-35.8)#