Find the gradient of the straight line joining (1, 2 + 4p) to (1 – 2p, 2).

The gradient of the straight line joining (1, 2+4p) to (1-2p, 2) is 2 and this can be determined by using the two-point slope formula.

Given :

Points  --  (1,2+4p) and (1-2p,2)

The following steps can be used in order to determine the gradient of the straight line joining (1, 2+4p) to (1-2p, 2):

Step 1 - According to the given data, the points on the straight line are (1, 2+4p) to (1-2p, 2).

Step 2 - The two-point slope formula can be used in order to determine the gradient of the straight line joining (1, 2+4p) to (1-2p, 2).

Step 3 - The two-point slope formula is given below:

Step 4 - Substitute the values of and in the above expression.

Step 5 - Simplify the above expression.

m = 2

For more information, refer to the link given below:

link

2

Step-by-step explanation:

The gradient of the straight line is equal to (y2-y1)/(x2-x1), where

(x1,y1) and (x2,y2) coordinates of the points A and B which belong to the line.

So y2=2 ,x2=1-2p , y1=2+4p, x1=1

k=(2-(2+4p))/((1-2p)-1)= (2-2-4p)/(1-2p-1)= (-4p)/(-2p)=2

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

597

step-by-step explanation: