the line passes through the points s(6,6) and T(0,-2) determine the gradient of the line

the equation of the line
the midpoint of the line segment,TS
the length of the line segment,TS

Ohh dear!

midpoint is a "point" so it must look like ( ? , ? )
your method should be
the x of the midpoint is (6+0)/2= 3
the y of the midpoint is (6-2)/2 = 2
so the midpoint is (3,2)
(notice that in effect we are taking the average of the x's and the average of the y's)

length
= √(6-0)^2 + (6-(-2))^2)
= √(36 +64)
= √100
= 10

gradient

= slope
= (6-(-2))/(6-0)
= 8/6 = 4/3

since (0,-2) is the y-intercept or the b of the y = mx + b notation, we can simply write

y = (4/3) - 2

surely you have 2 little formulas in your text or your notebook to find the midpoint and the length.
Let me know what you get.

To determine the gradient of a line passing through two points, we can use the formula:

Gradient (m) = (y₂ - y₁) / (x₂ - x₁)

In this case, the points are S(6,6) and T(0,-2). Using the formula, we can substitute the coordinates into the equation:

m = (-2 - 6) / (0 - 6)
m = (-8) / (-6)
m = 4/3

Therefore, the gradient of the line is 4/3.

To find the equation of the line, we can use the point-slope form:

y - y₁ = m(x - x₁)

Using one of the points, let's use S(6,6), and substitute the values into the equation:

y - 6 = (4/3)(x - 6)

Expanding and simplifying, we get the equation:

y = (4/3)x - 4

Therefore, the equation of the line is y = (4/3)x - 4.

To find the midpoint (M) of the line segment TS, we can use the midpoint formula:

M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

Using the coordinates of S(6,6) and T(0,-2), we substitute the values into the formula:

M = ((6 + 0) / 2, (6 + (-2)) / 2)
M = (6 / 2, 4 / 2)
M = (3, 2)

Therefore, the midpoint of the line segment TS is M(3,2).

To find the length of the line segment TS, we can use the distance formula:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Using the coordinates of S(6,6) and T(0,-2), we substitute the values into the formula:

Distance = √[(0 - 6)² + (-2 - 6)²]
Distance = √[(-6)² + (-8)²]
Distance = √[36 + 64]
Distance = √100
Distance = 10

Therefore, the length of the line segment TS is 10 units.

mid point

y2+y1/2
-2+6/2
4/2
=2

x2-x1/2
0-6/2
-6/2
=-3