Select ALL of the points that are on the graph of the line y = -1/2x + 5

Group of answer choices

(0, 5)

(0, 10)

(1, 2)

(1, 4)

(5, 0)

(10, 0)

We conclude that only the points (0, 5) and (10, 0) are on the graph of the line.

Step-by-step explanation:

Given the line

Any points which will satisfy the equation would be the points on the graph of the line.

Let us substitute and check all the points one by one.

FOR (0, 5)

Putting (0, 5) in the equation

y = -1/2x + 5

(5) = -1/2(0)+5

5 = 0+5

5 = 5

L.H.S = R.H.S, so (0, 5)  satistfies the equation.

Hence,

(0, 5) is on the graph of the line.

FOR (0, 10)

Putting (0, 10) in the equation

y = -1/2x + 5

(10) = -1/2(0)+5

10 = 0+5

10 = 5

L.H.S ≠ R.H.S, so (0, 10) does not satisfy the equation.

Hence,

(0, 10) is not on the graph of the line.

FOR (1, 2)

Putting (1, 2) in the equation

y = -1/2x + 5

(2) = -1/2(1)+5

2 = 9/2

L.H.S ≠ R.H.S, so (1, 2) does not satisfy the equation.

Hence,

(1, 2) is not on the graph of the line.

FOR (1, 4)

Putting (1, 4) in the equation

y = -1/2x + 5

(4) = -1/2(1)+5

4 = 9/2

L.H.S ≠ R.H.S, so (1, 4) does not satisfy the equation.

Hence,

(1, 4) is not on the graph of the line.

FOR (5, 0)

Putting (5, 0) in the equation

y = -1/2x + 5

(0) = -1/2(5)+5

0 = 5/2

L.H.S ≠ R.H.S, so (5, 0) does not satisfy the equation.

Hence,

(5, 0) is not on the graph of the line.

FOR (10, 0)

Putting (10, 0) in the equation

y = -1/2x + 5

(0) = -1/2(10)+5

0 = -5+5

0 = 0

L.H.S = R.H.S, so (10, 0) satisfies the equation.

Hence,

(10, 0) is on the graph of the line.

Thus, we conclude that only the points (0, 5) and (10, 0) are on the graph of the line.

89

step-by-step explanation:

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