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# The line represented by the equation y = 4 – 2x is the

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The line represented by the equation y = 4 – 2x is the  [#permalink]

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18 Jun 2014, 20:52
Hi Bunuel
in your solution you have taken R co-ordinates (4,1) as (x2,y2)
my question is why cannot they be considered (x1,y1)?

Kindly explain
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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18 Jun 2014, 21:08
11yashu wrote:
Hi Bunuel
in your solution you have taken R co-ordinates (4,1) as (x2,y2)
my question is why cannot they be considered (x1,y1)?

Kindly explain

You can consider it either way. The answer will be the same...

$$Slope, m=\frac{1}{2} = \frac{y2-1}{x2- 4}$$

or $$x2-4= 2y2-2$$

Now pick from the options given which satisfies this equation.
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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12 Dec 2014, 08:16
priyalr wrote:
Hello,

This is how I solved the problem. Since the Slope of line is -2, the line perpendicular to it would have SLope as 1/2. So I used R(4,1) and each options to see which one gives Slope as 1/2. Only option D gives me the co ordinates through which the SLope is 1/2. It took me around a minutes time to solve.

Please let me know if I'm correct ?

I solved this way as well ...
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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20 Feb 2015, 07:49
Bunuel wrote:
priyalr wrote:
Hello,

This is how I solved the problem. Since the Slope of line is -2, the line perpendicular to it would have SLope as 1/2. So I used R(4,1) and each options to see which one gives Slope as 1/2. Only option D gives me the co ordinates through which the SLope is 1/2. It took me around a minutes time to solve.

Please let me know if I'm correct ?

That's perfectly valid approach.

Two lines are perpendicular if and only the product of their slopes is -1. The slope of given line is -2, hence the slope of PR must be 1/2 (negative reciprocal of -2): 1/2*(-2)=-1.

Now, the slope of a line (a line segment) passing through two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$m=\frac{y_2-y_1}{x_2-x_1}$$.

So, for our case the slope of PR must be $$m=\frac{1}{2}=\frac{1-y_1}{4-x_1}$$ and you can substitute x and y coordinates of each point from answer choices to see for which one this equation will hold true. Only coordinates of a point from option D fits.

Hi Bunuel ,
I solved i this way :
y = 4 – 2x , set y=0; x=2;
RP will have a slope of 1/2, also the line formed by joining R(4,1) and (2,0) have slope of 1/2 so we know the midpoint is (2,0)
now we can solve P(X,Y) as
(4+X)/2=2
(1+Y)/2=0 to get (0,-1)

Hope this approach is full proof .

thanks
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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18 Jul 2015, 06:42
I've solve this one without involving any formulas for the slope.. Just made a grafik as shown by Bunuel --> then you can see that Y coordinates can be ONLY NEGATIVE and the only answer that has a negative value for Y is (D)
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The line represented by the equation y = 4 – 2x is the  [#permalink]

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03 Apr 2016, 09:46
(2;0) is a point on the line
it is the point of intersection of the two lines
we can create 2 opposed right triangles, of which the point 0;-1 will be the mirror reflection of the point 4;1

alternative way:
slope is -2
slope of the perpendicular line is 1/2
we can build the formula for the line
y=1/2 * 4 +b
1=2+b
b=-1
y=x/2 -1

plug in values, and see which one works.
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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03 Apr 2016, 14:51
here we need to first solve for the intersection of two lines so that we get the coordinates of the midpoint .
and then use the ration theorem => x1+x2/2 ,y1+y2/2 will be the coordinates of the midpoint of RP
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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17 May 2016, 03:57
I solved it by equation method but wanted to learn graphical method..

I got how you plotted the line y = 4 – 2x using Y intercept and X intercept and plotted the point R (4,1) But how do you know, line PR passes through 2 and touches P at (0,-1) Bunuel
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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17 May 2016, 06:31
kapru wrote:
I solved it by equation method but wanted to learn graphical method..

I got how you plotted the line y = 4 – 2x using Y intercept and X intercept and plotted the point R (4,1) But how do you know, line PR passes through 2 and touches P at (0,-1) Bunuel

No other point (blue dots on the image in my solution) can be the mirror reflection of R around the line but (0, -1).
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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29 Aug 2017, 10:24
Just put any values which satisfy the equation y = 4-2x. So, a point is (2,0). Check the slope. Here slope = 1/2 which is inverse and negative of -2. So, (2,0) is the midpoint. Using mid point formula, we find the coordinates of point P.
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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01 Sep 2017, 10:59
enigma123 wrote:
The line represented by the equation y = 4 – 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?

(A) (–4, 1)
(B) (–2, 2)
(C) (0, 1)
(D) (0, –1)
(E) (2, 0)

If the line y = 4 - 2x is the perpendicular bisector of line segment RP, then it’s perpendicular to RP at the midpoint of RP. Recall that two lines are perpendicular if their slopes are negative reciprocals of each other. Since the line has a slope of -2, the line segment should have a slope of ½. Thus, let’s first determine which answer choice can be point P so that the slope of RP is ½. We’ll use the slope formula: m = (y2 - y1)/(x2 - x1):

A) (-4, 1)

m = (1 - 1)/(4 - (-4)) = 0

Point P can’t be (-4, 1).

B) (-2, 2)

m = (1 - 2)/(4 - (-2)) = -1/6

Point P can’t be (-2, 2).

C) (0, 1)

m = (1 - 1)/(4 - 0) = 0

Point P can’t be (0, 1).

D) (0, -1)

m = (1 - (-1))/(4 - 0) = 2/4 = 1/2

Point P could be (0, -1).

E) (2, 0)

m = (1 - 0)/(4 - 2) = 1/2

Point P could be (2, 0).

We see that point P could be either (0, -1) or (2, 0), since either one will make RP’s slope ½. Next let’s determine the midpoint of RP if P is either (0, -1) or (2, 0). We use the midpoint formula: ((x1 + x2)/2 , (y1 + y2)/2)). Recall that R = (4,1).

If P = (0, -1), then the midpoint of RP = ((4 + 0)/2, (1 + (-1))/2) = (2, 0).

If P = (2, 0), then the midpoint of RP = ((4 + 2)/2, (1 + 0)/2) = (3, 1/2).

Recall that the line y = 4 - 2x has to include the midpoint of RP. In other words, the midpoint of RP is a point on the line, and hence its coordinates will satisfy the equation of the line.

If P = (0, -1) and the midpoint of RP is (2, 0), is 0 = 4 - 2(2)? The answer is yes, since 0 = 0. We can see that choice D is the correct choice. However, let’s also show that choice E is not the correct choice:

If P = (2, 0) and the midpoint of RP is (3, 1/2), is 1/2 = 4 - 2(3)? The answer is no, since 1/2 ≠ -2.

Alternate Solution:

Since the line segment RP is perpendicular to y = 4 - 2x, the line containing RP must have a slope of 1/2, since the slopes of perpendicular lines are negative reciprocals of each other. Then, the line containing the line segment RP must be of the form y = (1/2)x + b, for some real number b. Since this line contains the point R, its equation must be satisfied when we substitute x = 4 and y = 1; therefore 1 = (1/2)(4) + b. Then, 1 = 2 + b, and thus b = -1.

We know the line that contains the line segment RP has an equation y = (1/2)x - 1. Let’s find the common point of this line with y = 4 - 2x. We set (1/2)x - 1 = 4 - 2x and solve for x:

(1/2)x -1 = 4 - 2x

(5/2)x = 5

x = 2

So, the x-coordinate of the common point is 2. We can substitute x = 2 in either equation to find the y-coordinate: y = 4 - 2(2) = 0. So, the common point is (2, 0).

Note that this common point is the midpoint of R and P. Therefore, if we let P = (a, b), we must have:

(a + 4)/2 = 2 and (b + 1)/2 = 0. We find that a = 0 and b = -1. Thus, P = (0, -1).

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The line represented by the equation y = 4 – 2x is the  [#permalink]

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Updated on: 06 Nov 2017, 10:17
i have a different approach

y=2x-4
perpendicular line will be having slope 0.5

line rp y= 0.5x+c
4,1 satisfy this line

line rp => y= 0.5x-1
but 2,0 satisfy this line and also the given line
since when we draw x and y intercept to draw on x and y axis we anyhow take x=0 and y=0 so we will come to know both lines cut at 2,0

now 2,0 is mid point

so (x+4)/2 =2 and (y+1)/2 =0
x=0 y=-1
OPTION D IS WINNER
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Originally posted by sahilvijay on 06 Nov 2017, 10:13.
Last edited by sahilvijay on 06 Nov 2017, 10:17, edited 1 time in total.
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The line represented by the equation y = 4 – 2x is the  [#permalink]

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06 Nov 2017, 10:15
another way
take the options a b c d e
with r and p => calculate mid point
that mid point must satisfy y= -2x +4
d is the winner
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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06 Nov 2017, 10:19
Last approach

m1m2=-1
-2 m2 =-1
m2 =1/2
=> Option D satisfy m2=1/2
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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07 Nov 2017, 15:54
One more approach : Advanced Quants

Point (4,1) Reflected point (h,k)
Line 2x+y-4=0

(h-4)/2 = (k-1)/1 = -2(4x2+1x1-4)/(2^2+1^2)

(h-4)/2 = -2 = k-1
h=0 k=-1

Point is (0,-1)

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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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12 Nov 2017, 06:12
Bunuel wrote:
priyalr wrote:
Hello,

This is how I solved the problem. Since the Slope of line is -2, the line perpendicular to it would have SLope as 1/2. So I used R(4,1) and each options to see which one gives Slope as 1/2. Only option D gives me the co ordinates through which the SLope is 1/2. It took me around a minutes time to solve.

Please let me know if I'm correct ?

That's perfectly valid approach.

Two lines are perpendicular if and only the product of their slopes is -1. The slope of given line is -2, hence the slope of PR must be 1/2 (negative reciprocal of -2): 1/2*(-2)=-1.

Now, the slope of a line (a line segment) passing through two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$m=\frac{y_2-y_1}{x_2-x_1}$$.

So, for our case the slope of PR must be $$m=\frac{1}{2}=\frac{1-y_1}{4-x_1}$$ and you can substitute x and y coordinates of each point from answer choices to see for which one this equation will hold true. Only coordinates of a point from option D fits.

I have another approach. Any point on the perpendicular bisector of a line segment is equidistant from the end points of the line segment.
So I tried a random point on the given line and used the options to see if the distance was equal. D satisfies.
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Re: The line represented by the equation y = 4 – 2x is the  [#permalink]

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