MGMAT - slope of line

[jayhawk2001]

Interesting question on slope of a line. OA after a few folks reply.

In the xy-coordinate system, what is the slope of the line that goes through the origin and is equidistant from the two points P = (1, 11) and Q = (7, 7)?
A. 2
B. 2.25
C. 2.50
D. 2.75
E. 3

[800GMAT]

Since the line passes through the origin, one set of coordinates is (0,0)
we just another set of coordinates to find the slope

since the line is equidistant from both P and Q, the line passes through the midpoint of P and Q

Midpoint of PQ =(x,y) = (1+7)/2 , (11+7)/2
=(4,9)

Using (0,0) and (4,9), the slope is 2.25

Hence B

[abkhan]

let equation of line be = > y=mx+c

Since it passes origin (0,0) on replacing (x,y)
we have

=>y=mx
Mid point is ( (1+7)/2,(11+9)/2)= (4,9)

replacing in above eq. we have m= 2.25
so slope is 2.25

[f2001290]

If a line is equi-distant from two points doesn't mean that it is passing through the middle.

for example, if the two points lie on the same side of the line and are equidistant, then mid-point formula doesn't help .

[abkhan]

If a line is equi-distant from two points doesn't mean that it is passing through the middle.

for example, if the two points lie on the same side of the line and are equidistant, then mid-point formula doesn't help .

Your right, i tried that approach but the data seemed insufficient if we take that apporoach. calculating (x,y) with distance theorem the equation we get is
2y=3m+8
need another equation to find (x,y).

[takami]

Why isn't the slope -2/3 as in y=(-2/3)x

if u connect pt P and Q and create a line, its slope will be

(y2-y1)/(x2-x1)=(7-11)/(7-1) =-4/6 =-2/3

Now it is possible to set any slope to pass thru the origin. Subbing (0,0) into y-y1=m(x-x1) , you will get y=-(2/3)x.

This line will cross the orgin and the distance from this line with pt P and Q will be the shortest equal distance, since line PQ and this line is parallel.

I must have done something wrong, and my GMAT is in a few days... I'm screwed

[jayhawk2001]

Yup. OA is B

[800GMAT]

Thanks jayhawk..... can u also pls post the official explanation for this question......thnks

[jayhawk2001]

Thanks jayhawk..... can u also pls post the official explanation for this question......thnks

Here's the OE from MGMAT -- pretty much along the same lines
of discussion here

The question asks us to find the slope of the line that goes through the origin and is equidistant from the two points P=(1, 11) and Q=(7, 7). It's given that the origin is one point on the requested line, so if we can find another point known to be on the line we can calculate its slope. Incredibly the midpoint of the line segment between P and Q is also on the requested line, so all we have to do is calculate the midpoint between P and Q! (This proof is given below).

Let's call R the midpoint of the line segment between P and Q. R's coordinates will just be the respective average of P's and Q's coordinates. Therefore R's x-coordinate equals 4 , the average of 1 and 7. Its y-coordinate equals 9, the average of 11 and 7. So R=(4, 9).

Finally, the slope from the (0, 0) to (4, 9) equals 9/4, which equals 2.25 in decimal form.

Proof

To show that the midpoint R is on the line through the origin that's equidistant from two points P and Q, draw a line segment from P to Q and mark R at its midpoint. Since R is the midpoint then PR = RQ.

Now draw a line L that goes through the origin and R. Finally draw a perpendicular from each of P and Q to the line L. The two triangles so formed are congruent, since they have three equal angles and PR equals RQ. Since the triangles are congruent their perpendicular distances to the line are equal, so line L is equidistant from P and Q.

[Cybermusings]

Equidistant from the two points P and Q will have co-ordinates of (1+7)/2 and (11+7)/2 = 4,9
So slope of this line = 9-0/4-0 = 9/4 = 2.25
Should be C