Question

Find the equation of a line that passes through points (9,3) and (-6,5)?

Answer 1

Answer:

y-3 = -2/15 (x-9)

Step-by-step explanation:

First, we should find the gradient, m, which is $\frac{rise (change.in.the.y.value)}{run (change.in.the.x.value)}$

This is $\frac{3-5}{9-(-6)} =-\frac{2}{15}$

A linear (straight) line has a general equation y=mx+c

y = $-\frac{2}{15\\}$x+c

We can rewrite the equation into the general form:

y-y1 = m(x-x1)

where y1 is the value of a y coordinate and x1 is the x value of the coordinate that the y1 value was taken from.

This stems from the idea that 0 = 0.

To make that equation true, we can make both sides equal to 0.

To do that, y - y1 has to be - and x-x1 has to be zero. m can be ignored as anything multiplied by 0 is 0.

Up until here, y1 and x1 are arbitrary constants. They are values that we have to find. However, we don't actually. We have coordinates that the x and y will equal to. (9,3)

Because 9 - 9 = 0 and 3 - 3 = 0, we can state that the value of y1 is 3 and x1 is 9.

Hence, the equation is y-3 = -2/15 (x-9)

You can rearrange the equation to make it into the form y = mx + c, but this is usually not necessary.

I hope this helped.  

Answer 2

Answer:

Y= X - 31.5

Step-by-step explanation:

find the gradient then the equation