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Enter the equation of the line in any format that we have learned that passes the two points: (2, -5) and (0, 1).

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mhanifa

Given points:

  • (2, -5) and (0, 1)

Find the slope:

  • m = (1 - (-5))/(0 - 2) = 6/(-2) = -3

Point - slope form:

  • y - (-5) = -3(x - 2) ⇒ y + 5 = -3(x - 2)

Convert into slope- intercept form:

  • y = -3x + 6 - 5 ⇒ y = -3x + 1

Convert into standard form:

  • 3x + y = 1

DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

let's find slope (m) :

  • [tex] \dfrac{y2 - y1}{x2 - x1} [/tex]

  • [tex] \dfrac{1 - ( - 5)}{0 - 2} [/tex]

  • [tex] \dfrac{6}{ - 2} [/tex]

  • [tex] - 3[/tex]

Plugging value of x and y from point (0 , 1) and slope (m) = -3, in general equation of line we will get the value of y - intercept (c)

that is :

  • [tex]y = mx + c[/tex]

  • [tex]1 = (0 \times - 3) + c[/tex]

  • [tex]c = 1[/tex]

y intercept (c) = 1, now let's plug the value of slope(m) and y - intercept (c) in general equation of line to get the required equation of line which passes from given two points.

  • [tex]y = - 3x + 1[/tex]

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