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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
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Video solution from Quant Reasoning:
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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
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Bunuel wrote:
In the xy-coordinate plane, if the point (0,2) lies on the graph of the line 2x + ky = 4, what is the value of the constant k ?

A. 2

B. 1

C. 0

D. –1

E. –2


Substituting 0 for x and 2 for y into the equation, we have:

2(0) + k(2) = 4

0 + 2k = 4

k = 2

Answer: A
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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
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Solution:

Step 1: If the point (0, 2) lies on the graph of the line 2x + ky = 4, then, the coordinate x = 0, y = 2 are a possible solution of the equation 2x + ky = 4.

Step 2: Following from Step 1 above, when x = 0 and y = 2, the equation 2x + ky = 4 becomes 2(0) + k(2) = 4. Hence, 0 + 2K = 4.

Therefore, k = 4/2 = 2. The answer is
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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
Does this equation not need to be converted to the standard y = mx + b structure?
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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
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Bunuel wrote:
In the xy-coordinate plane, if the point (0,2) lies on the graph of the line 2x + ky = 4, what is the value of the constant k ?

A. 2

B. 1

C. 0

D. –1

E. –2

(PS02382)


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Answer: Option A

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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
To find the value of the constant k in the equation 2x + ky = 4, we can substitute the given point (0, 2) into the equation and solve for k.

Substituting x = 0 and y = 2 into the equation, we get:

2(0) + k(2) = 4
0 + 2k = 4
2k = 4
k = 4/2 = 2

Hence, the correct answer is A. 2.
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Re: In the xy-coordinate plane, if the point (0,2) lies on the graph of th [#permalink]
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