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What is the equation of a line that passes through the points (–3, 4) and (2, 8)?
A. y = 0.8x + 5
B. y = 1.25x + 6.75
C. y = 0.8x + 6.4
D. y = 0.8x – 8

Respuesta :

josshyaby

Answer:

C

Step-by-step explanation:

(–3, 4) and (2, 8)

y2 = 8, y1 = 4, x2 = 2 and x1 = -3

Slope m = {y2-y1}/{x2-x1}

m = {8 - 4}/{2 - (-3)}

m = 4/(2+3)

m = 4/5

And

y - y1 = m(x - x1)

y - 4 = 4/5(x - (-3))

y - 4 = 4/5(x + 3)

Multiply each term by 5

5y - 20 = 4(x + 3)

5y - 20 = 4x + 12

5y = 4x + 20 + 12

5y = 4x + 32

Dividing by 5

y = 4x/5 + 32/5

y = 0.8x + 6.4

The equation of the line is  y = 0.8x + 6.4.

Given that,

The equation of a line that passes through the points (–3, 4) and (2, 8).

We have to determine,

The equation of the line.

According to the question,

The equation of a line that passes through the points (–3, 4) and (2, 8).

The slope of the line is m is,

[tex]\rm m = \dfrac{8-4}{2-(-3)}\\\\m = \dfrac{4}{5}[/tex]

The value of c is at point (-3, 4) is,

[tex]\rm y = mx+c \\\\4 = \dfrac{4}{5} (-3) +c\\\\4 = \dfrac{-12}{5} + c\\\\c =4+\dfrac{12}{5}\\\\c = \dfrac{20+12}{5}\\\\ c = \dfrac{32}{5}[/tex]

Therefore,

The required equation of the line is,

[tex]\rm y = mx +5 \\\\y = \dfrac{4}{5}x+\dfrac{32}{5}\\\\y =0.8x+6.4[/tex]

Hence, The required equation of the line is y = 0.8x + 6.4.

For more details refer to the link given below.

https://brainly.com/question/5245372

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