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

y=mx
line must pass through prependicular bisector of PQ
mid point = 4,9
thus slope = 9/4
IMO B

aatech wrote: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

Since the line is equidistant from P = (1, 11) and Q = (7, 7), it must pass through the midpoint between P = (1, 11) and Q = (7, 7). We can use the midpoint formula:

Midpoint = ((x1 + x2)/2, (y1 + y2)/2)

Midpoint = ((1 + 7)/2, (11 + 7)/2)

Midpoint = (4, 9)

Since the line also passes through the origin, (0, 0), the slope is:

Slope = change in y/change in x

(9 - 0)/(4 - 0) = 9/4 = 2.25

Answer: B

aatech wrote: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

I understood the dubts raised by some members.
So let me clarify that all the lines passing through the midpoint
of two different points X and Y shall be equidistant from these points X and Y.
From all such lines we have to select only that line which passes through the origin also and find its slope.
Since midpoint is [(1+7)/2,(11+7)/2] i.e. (4,9) ,
we have to find slope of the lone passing through (4,9) and origin that is (0,0)
the slope is =9/4= 2.25
Hence option B is correct.
Hope this clarifies all the doubts raised by some members.