Coordinates

Step 1: Find the gradient.

The formula of gradient is

$m=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

Let $C=({ x }_{ 2 },{ y }_{ 2 })$ and $D=({ x }_{ 1 },{ y }_{ 1 })$

m=\frac { 5-2 }{ 2-(-1) } =1

Step 2: Form an equation

An equation of a line is y=mx+c where c is an interception of the line and the y-axis. (Basically where the line and the y-axis meet up)

How to find C? Get x and y first. To get x and y, choose either the coordinate of point C or D. I choose C.

$5=(1)(2)+c\\ 5-2=c\\ 3=c$

So the equation of the line is $y=x+3$ .

Step 3: Find coordinate E.

Just test the four coordinates given. The correct one is (1,4).

gradient= y2 - y1 / x2 - x1 so, (5-2)/(2-(-1))=3/3=1 all the point lie on line thus, (2-y)/(-1-x) =1 so the value of x=1 and y=4

The formula for the slope is: m=y2-y1 / x2-x1

Substituting, we get:

5-2   /   2-(-1)

=3 / 3 =1

Slope=1

Using the formula;

and using (2,5)

M(x-X)=y-Y

1(x-2)=y-5

x-2=y-5

y=x+3

We substitute each of the choices to x and y and see what point fits in the given equation.

gradient of line = 1 C(2,5) , Y(1,4) gradient = 1

Points are said to be in same line If their slopes are equal .... Slope of two points (x1,y1) , (x2, y2) = y2- y1/ x2-x1 So slopes of CD & DE are equal .... Verify the options

let C is point 1 and D is point 2..then by using..

 ( y-y1 ) =( x-x1)*(y2-y1)/(x1-x2)

we will get equation x-y = -3... so acc.. to options given.(1,4) satisfy it..

Sorry to everyone but... I'm confused with gradients.

I did it so i found a sequence, and got the answer from there. So: (-1,2) , (0,3) , (1,4) , (2,5) ...