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work out the gradient of the straight line that passes through (2,6) and (6,12)

Sagot :

Answer:

The gradient of the straight line that passes through (2, 6) and (6, 12) is [tex]m = \frac{3}{2}[/tex].

Step-by-step explanation:

Mathematically speaking, lines are represented by following first-order polynomials of the form:

[tex]y = b + m\cdot x[/tex] (1)

Where:

[tex]x[/tex] - Independent variable.

[tex]y[/tex] - Dependent variable.

[tex]m[/tex] - Slope.

[tex]b[/tex] - Intercept.

The gradient of the function is represented by the first derivative of the function:

[tex]y' = m[/tex]

Then, we conclude that the gradient of the staight line is the slope. According to Euclidean Geometry, a line can be form after knowing the locations of two distinct points on plane. By definition of secant line, we calculate the slope:

[tex]m = \frac{y_{B}-y_{A}}{x_{B}-x_{A}}[/tex] (2)

Where:

[tex]x_{A}[/tex], [tex]y_{A}[/tex] - Coordinates of point A.

[tex]x_{B}[/tex], [tex]y_{B}[/tex] - Coordinates of point B.

If we know that [tex]A(x,y) = (2,6)[/tex] and [tex]B(x,y) = (6,12)[/tex], then the gradient of the straight line is:

[tex]m = \frac{12-6}{6-2}[/tex]

[tex]m = \frac{6}{4}[/tex]

[tex]m = \frac{3}{2}[/tex]

The gradient of the straight line that passes through (2, 6) and (6, 12) is [tex]m = \frac{3}{2}[/tex].