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In the xy-coordinate plane, Line J passes through points (−2,0) and (0

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In the xy-coordinate plane, Line J passes through points (−2,0) and (0  [#permalink]

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New post 07 Feb 2018, 20:50
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A
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E

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81% (01:02) correct 19% (04:27) wrong based on 80 sessions

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In the xy-coordinate plane, Line J passes through points (−2,0) and (0,2). Line K passes through points (4,0) and (0,4). If point (j,k) is the point at which Line J and Line K intersect, which of the following is true of point (j,k)?

A. j>0 and k>0
B. j>0 and k<0
C. j<0 and k<0
D. j<0 and k>0
E. The product jk=0

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Re: In the xy-coordinate plane, Line J passes through points (−2,0) and (0  [#permalink]

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New post 07 Feb 2018, 20:57
1
As we draw lines in the xy plane, the only quadrant the lines will meet is 1 St quadrant . So j and k should be >o ..A is correct

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Re: In the xy-coordinate plane, Line J passes through points (−2,0) and (0  [#permalink]

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New post 07 Feb 2018, 22:40
Since, line j lies on x-axis, the intersection will also lie on the x-axis and as such, the y-coordinate of the intersection point will be zero.

So, the product of JK will be '0'. Answer is E.

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Re: In the xy-coordinate plane, Line J passes through points (−2,0) and (0  [#permalink]

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New post 08 Feb 2018, 21:06
When we draw the line J and K using the points given, the line J and K would intersect when both J and K is greater than 0.

Ans: A
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In the xy-coordinate plane, Line J passes through points (−2,0) and (0  [#permalink]

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New post 09 Feb 2018, 09:37
Bunuel wrote:
In the xy-coordinate plane, Line J passes through points (−2,0) and (0,2). Line K passes through points (4,0) and (0,4). If point (j,k) is the point at which Line J and Line K intersect, which of the following is true of point (j,k)?

A. j>0 and k>0
B. j>0 and k<0
C. j<0 and k<0
D. j<0 and k>0
E. The product jk=0

I usually graph, but finding the equations for these lines is quick.

To find the intersection point for two lines: set the equations for y equal; solve for x (x-coordinate = j); then solve for y (y-coordinate = k)

Write the equation of the lines in slope-intercept form.
y = mx + b
m = slope = \(\frac{rise}{run}=\frac{(y_2-y_1)}{(x_2-x_1)}\)
b = y-intercept

Line J has slope \(\frac{(0-2)}{(-2-0)}=\frac{-2}{-2} =
1\)

From the prompt, when x=0, b = 2 (y-intercept)
Line J: y = (1)x + 2

Line K's slope: \(\frac{(4-0)}{(0-4)}=\frac{4}{-4} =
-1\)

When x=0, b = 4 (y-intercept)
Line K: y = (-1)x + 4

Set equations for y equal, solve for x
x + 2 = -x + 4
2x = 2
x = 1 (= the x-coordinate of the intersection point, = j)
Plug x in to find y-coordinate:
y = 1 + 2
y = 3 (= y-coordinate of intersection point, k)

Intersection point is (j,k) = (1,3)

Answer A
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In the xy-coordinate plane, Line J passes through points (−2,0) and (0 &nbs [#permalink] 09 Feb 2018, 09:37
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