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Re: In the [m]xy[/m] coordinate plane, does the point (3,4) lie [#permalink]

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15 Apr 2013, 05:09

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y=mx+q the question is: 4=m3+q?

1) The line \(5y-45=-x\) is perpendicular to the line \(t\). \(y=-\frac{1}{5}x+9\) those are perpendicular so m=5. is \(3m+q=3*5+q=4\)? it depends on q. Not sufficient

2) The line with the equation \(y= \frac{3}{4}x - 11\) intersects the line t when \(y=-11\). \(-11=\frac{3}{4}x-11\) x=0 y=-11 is where those lines intersect, but this point defines the intersenction of t with the y-axis => defines q. Is \(3m-11=3\)? it depends on m. Not sufficient

1+2) We have m and q. Sufficient to say if 4=5*3-11. So the answer is YES _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: In the xy coordinate plane, does the point (3,4) lie on line [#permalink]

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15 Apr 2013, 05:21

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In the xy coordinate plane, does the point (3,4) lie on line t?

Minimum calculations are needed to answer this question. Even the calculations below are not necessary and are just to illustrate.

(1) The line 5y-45=-x is perpendicular to the line t. Given that line y=9-x/5 (5y-45=-x) is perpendicular to the line t means that the slope of line t is 5 (negative reciprocal of the slope of line y=9-x/5, which is -1/5). Now, line with the slope of 5 may or may not pass through point (3,4). For example, y=5x-11 (YES) and y=5x (NO). Not sufficient.

(2) The line with the equation \(y= \frac{3}{4}x - 11\) intersects the line t when y=-11 --> \(-11= \frac{3}{4}x - 11\) --> \(x=0\). So, we have that line t passes through point (0, -11). Obviously this line may also pass through (3,4) but may as well not. Not sufficient.

(1)+(2) We know the slope of line t and point (0, -11) it passes, which is enough to get its equation. Therefore we can determine whether it passes point (3,4). Sufficient.

Re: In the xy coordinate plane, does the point (3,4) lie on line [#permalink]

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02 Jul 2015, 03:42

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