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ollegr [7]
2 years ago
7

Write all three forms (slope intercept, point slope and standard) of a line containing the points (4, -11/3) and (1/2, -29/6).

Mathematics
1 answer:
Zanzabum2 years ago
3 0

Answer:

See explanations below

Step-by-step explanation:

Slope intercept form of the equation is expressed as y = mx+b

m is the slope

b is the intercept

Given the coordinates (4, -11/3) and (1/2, -29/6)

Slope m = -29/6-(-11/3)÷(1/2-4)

m = (-29/6+11/3)÷(-7/2)

m = -29+22/6÷(-7/2)

m = -7/6÷(-7/2)

m = -7/6×2/-7

m = 2/6

m = 1/3

Get the intercept b

Substitute m= 1/3 and (4, -11/3) into y = MX+b

-11/3 = 1/3(4)+b

-11/3 = 4/3+b

b= -11/3-4/3

b= -15/3

b =-5

Get the required equation

y = 1/3 x -5 (slope-intercept form)

Point slope form of the equation is expressed as y-y0 = m(x-x0)

Substituting m = 1/3 and the point (4,-11/3)

x0 = 4

y0 = -11/3

y - (-11/3) = 1/3(x-4)

y+11/3 = 1/3(x-4) (point-slope form)

For standard format

Recall that y = 1/3x - 5

Multiply through by 3

3y = 1/3x(3)-5(3)

3y = x-15

3y-x = -15(standard form)

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The value of x is 27°.

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Given \triangle \mathrm{ACB} \cong \Delta \mathrm{DCE}.

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