Find the equation of a line with a slope of 3/4 that goes through the point (8, 2).

_{1}=m(x-x

_{1})

_{1}= the x coordinate

_{1}= the y coordinate

please help

Find the equation of a line with a slope of 3/4 that goes through the point (8, 2).

Find the equation of a line with a slope of 3/4 that goes through the point (8, 2).

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You may want to use point slope to write the equation of the line, given this information that would be the easiest to use. The form is Y-Y_{1}=m(x-x_{1})

m = slope

x_{1} = the x coordinate

y_{1} = the y coordinate

therefore, replacing that information into the equation would give you

y-2=(3/4)(x-8)

use y = mx + b equation where m is the slope and b is the y-intercept. We are given m which is 3/4. We are also given x and y which are 8 and 2, respectively.

so, we plug in these values to solve for b:

2 = (3/4)8 + b

2 = 6 + b

so, b = -4

therefore, **y = (3/4)x -4**

You want to use the slope-intercept equation for a line. This form is Y =M X + B, or, Y-value = Slope * X-value + Y-Intercept.

What we know is an X value, a Y value and a Slope.

X = 8, Y = 2, M = 3/4

By substitution we can solve for B

2 = 3/4 * 8 + B

or

B = -4

Thus equation for the line would be:** Y = (3/4) * X - 4**

The check (should get 2 for Y if plug in 8 for X):

Y = 3/4 * (8) -4 = 6 - 4 = 2

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