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ArbitrLikvidat [17]
3 years ago
11

A line goes through the points (7, 3) and (–1, 5). Write the equation of the line in slope-intercept form. Show your work for fu

ll credit. (How did you go from those two points to having the equation in slope-intercept form? Or, how did you go form having the equation in point-slope form to slope-intercept form?)
Mathematics
1 answer:
Sergio [31]3 years ago
4 0

Answer:

The equation in slope-intercept form is;

y = -\dfrac{1}{4} \cdot x + 4\frac{3}{4}

Step-by-step explanation:

The equation of a straight line in slope and intercept form is given as follows;

y = m·x + c

Where;

m = The slope of the line

c = The y-intercept of the line

The points through which the line passes are;

(7, 3) and (-1, 5), therefore, the slope of the line, m, is given as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Where (x₁, y₁) = (7, 3) and (x₂, y₂) = (-1, 5), we have;

Slope, \, m =\dfrac{5-3}{(-1)-7} = -\dfrac{1}{4}

Therefore;

\dfrac{y-3}{x-7} =  -\dfrac{1}{4}

The equation in point slope form is therefore;

y - 3 = -\dfrac{1}{4} \times (x  - 7)

From which we have;

y - 3 = -1/4×(x - 7) = -x/4 + 7/4

y = -x/4 + 7/4 + 3 = -x/4 + 19/4

The equation in slope-intercept form is therefore;

y = -\dfrac{1}{4} \cdot x + 4\frac{3}{4}

Which can be expressed as follows

y = -0.25·x + 4.75.

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