Question

A) What is the equation of the line that passes through the given pair points in slope-intercept form?

Answer 1

A)

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (2, 5.1) and (-1, -0.5). Substitute:

$m=\dfrac{-0.5-5.1}{-1-2}=\dfrac{-5.6}{-3}=\dfrac{5.6}{3}=\dfrac{56}{30}=\dfrac{28}{15}$

Therefore we have:

$y=\dfrac{28}{15}x+b$

Put the coordinates of the point (-1, -0.5) ot the equation:

$-0.5=\dfrac{28}{15}(-1)+b$

$-0.5=-\dfrac{28}{15}+b$     multiply both sides by 2

$-1=-\dfrac{56}{15}+2b$       add $\dfrac{56}{15}$ to both sides

$\dfrac{41}{15}=2b$            divide both sides by 2

$b=\dfrac{41}{30}$

Answer: $\boxed{y=\dfrac{28}{15}x+\dfrac{41}{30}}$

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B)

We have the points (-2, 3) and (3, -4).

Calculate the slope:

$m=\dfrac{-4-3}{3-(-2)}=\dfrac{-7}{5}=-\dfrac{7}{5}$

Therefore we have:

$y=-\dfrac{7}{5}x+b$

Put the coordinates of the point (-2, 3) to the equation of a line:

$3=-\dfrac{7}{5}(-2)+b$

$\dfrac{15}{5}=\dfrac{14}{5}+b$              subtract $\dfrac{14}{5}$ from both sides

$\dfrac{1}{5}=b\to b=\dfrac{1}{5}$

Answer: $\boxed{y=-\dfrac{7}{5}x+\dfrac{1}{5}}$