The staright line L passes through the points A(-1,4) and B(5,8).

a. Calculate the gradient of L
b. Find the equation of L
c. The Line L also passes through the point P(8,y). Find the value of y.

a. The gradient of a line can be found using the formula: m = (y2 - y1)/(x2 - x1)

Therefore, the gradient of line L can be calculated as follows:

m = (8 - 4)/(5 - (-1))
= 4/6
= 2/3

b. To find the equation of line L, we need to use the formula: y - y1 = m(x - x1)
Using point A(-1,4), we have:

y - 4 = (2/3)(x - (-1))
y - 4 = (2/3)(x + 1)
Multiplying through by 3 to remove the fraction:

3(y - 4) = 2(x + 1)
3y - 12 = 2x + 2
Rearranging the equation:

2x - 3y = -14

c. Since point P(8,y) lies on line L, we can substitute these coordinates into the equation of line L and solve for y:

2(8) - 3y = -14
16 - 3y = -14
Subtracting 16 from both sides:

-3y = -30
Dividing by -3:

y = 10

Therefore, the value of y when line L passes through point P(8,y) is y = 10.

a. To calculate the gradient of the line passing through points A(-1,4) and B(5,8), we use the formula:

Gradient (m) = (change in y)/(change in x)

Let's assign the coordinates of points A and B as follows:

A: (x₁, y₁) = (-1, 4)
B: (x₂, y₂) = (5, 8)

Substituting the values into the formula:

m = (y₂ - y₁)/(x₂ - x₁)
= (8 - 4)/(5 - (-1))
= 4/6
= 2/3

Therefore, the gradient of line L is 2/3.

b. To find the equation of the line passing through points A(-1,4) and B(5,8), we can use the point-slope form of a linear equation:

y - y₁ = m(x - x₁)

Substituting the values:

y - 4 = (2/3)(x - (-1))
y - 4 = (2/3)(x + 1)
y - 4 = (2/3)x + 2/3
y = (2/3)x + 2/3 + 4
y = (2/3)x + 14/3

Therefore, the equation of line L is y = (2/3)x + 14/3.

c. We are given that line L passes through the point P(8, y). To find the value of y, we substitute the x and y coordinates of point P into the equation of line L:

y = (2/3)x + 14/3
y = (2/3)(8) + 14/3
y = 16/3 + 14/3
y = 30/3
y = 10

Therefore, when line L passes through the point P(8, y), the value of y is 10.

To calculate the gradient of line L, you can use the formula:

Gradient (m) = (change in y-coordinates) / (change in x-coordinates)

a. Calculate the gradient of L:
The coordinates of point A are (-1,4) and the coordinates of point B are (5,8).
So, the change in y-coordinates = 8 - 4 = 4
And the change in x-coordinates = 5 - (-1) = 6

Therefore, the gradient of line L is:
m = (change in y-coordinates) / (change in x-coordinates) = 4 / 6 = 2/3

b. Find the equation of L:
To find the equation of a straight line, we can use the point-slope form: y - y₁ = m(x - x₁), where (x₁, y₁) represent the coordinates of any point on the line, and m is the gradient.

Let's use point A as (x₁, y₁) = (-1, 4) and substitute the values into the equation:
y - 4 = (2/3)(x - (-1))

Simplifying the equation:
y - 4 = (2/3)(x + 1)
y - 4 = (2/3)x + 2/3

To convert it into the standard form (Ax + By = C), we multiply through by 3 to eliminate fractions:
3y - 12 = 2x + 2
2x - 3y = -14

c. The line L passes through point P(8, y). To find the value of y, we can substitute the x-coordinate and solve for y in the equation of the line.

Using the equation of line L found in part b:
2x - 3y = -14

Substituting x = 8:
2(8) - 3y = -14
16 - 3y = -14
-3y = -14 - 16
-3y = -30
y = -30 / -3
y = 10

Therefore, the value of y is 10.

the gradient is, of course, (8-4)/(5+1) = 2/3

so the equation is
y-4 = 2/3 (x+1)
Now plug in the point and solve
y-4 = 2/3 (8+1)