A line segment is bisected by a line with the equation 4y3x=2. If one end of the line segment is at (7,5), where is the other end?

2 Answers
Jul 7, 2016

The other point is on the line 4y3x=5
(any point on this line will satisfy the given requirement)

Explanation:

Given
XXXThe line 4y3x=2
and a point (7,5)

Consider the vertical line through (7,5)
This vertical line will intersect 4y3x=2 at (x=7,y)
where y can be determined by solving
XXX4y3×7=2

XXXy=234

The distance from (7,5) to (7,234)
is 2345=34

So a point (7,5+2×34)=(7,132) will be twice as far away from (7,5) as the point (7,234) on the same vertical line

Note that any point on a line parallel to 4y3x=2 through (7,132) will also be twice as far away from (7,5) as the point from (7,5) on 4y3x=2 to that point.
enter image source here

If L1 parallel to L2
then ABCADE
|AB|:|AD|=|AC|:|AE|

Since 4y3x=2 has a slope of 34

Our required line will also have a slope of 34
and since it passes through ((7,132))
using the slope-point form, we have
XXXy132=34(x7)
or
XXX4y3x=5
enter image source here

Jul 7, 2016

The other end-pt. lies on the line given by the eqn. 4y3x=5.

Explanation:

Suppose that the other end-pt. is P(X,Y).

Let the given end-pt. be Q(7,5), and the given line be L:4y3x=2.

If M is the mid-pt. of the segment PQ, then co-ords. of M as obtained by using Section Formula for Mid-pt. are =M(7+X2,5+Y2)

Now, PQ is bisected by L at M, so, ML.

Therefore, co-ords. of M must satisfy the eqn. of L.

Hence, 4{5+Y2}3{7+X2}=220+4Y213X=4, i.e., 4Y3X=5, as Sir Alan P. has readily derived!

This shows that :
(i) the co-ords. of other end-pt. can not be uniquely derived under the given conds.
(ii) What we can say about it (the other end-pt.) is that it lies on :4y3x=5. This eqn. represents a line to L.