A line passes through point A (14,21). A second point on the line has an x-value that is 125% of the x-value of point A and a y-value
that is 75% of the y-value of point A. Use point A to write an equation of the line in point-slope form.

?:(

Answer:

The equation of the line in point-slope form is [tex]y-21 = - \frac{3}{2}\cdot (x-14)[/tex].

Step-by-step explanation:

According to the statement, let [tex]A(x,y) = (14,21)[/tex] and [tex]B(x,y) = (1.25\cdot x_{A},0.75\cdot y_{A})[/tex]. The equation of the line in point-slope form is defined by the following formula:

[tex]y-y_{A} = m\cdot (x-x_{A})[/tex] (1)

Where:

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

[tex]m[/tex] - Slope, dimensionless.

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

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

In addition, the slope of the line is defined by:

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

If we know that [tex]x_{A} = 14[/tex] and [tex]y_{A} = 21[/tex], then the equation of the line in point-slope form is:

[tex]x_{B} = 1.25\cdot (14)[/tex]

[tex]x_{B} = 17.5[/tex]

[tex]y_{B} = 0.75\cdot (21)[/tex]

[tex]y_{B} = 15.75[/tex]

From (2):

[tex]m = \frac{15.75-21}{17.5-14}[/tex]

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

By (1):

[tex]y-21 = - \frac{3}{2}\cdot (x-14)[/tex]

The equation of the line in point-slope form is [tex]y-21 = - \frac{3}{2}\cdot (x-14)[/tex].

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