Find the point on the line such that...

[leoduong]

Find the point on the line 4x - 2y + 3 = 0 that is equidistant from (3,3) and (7,-3)

 

[stapel]

Express y in terms of x, so the point on the line, (x, y), is expressed only in terms of x. Then plug the three points into the Distance Formula, set the two distances equal to each other, and solve for x. Back-solve for y. :wink:

Eliz.

 

[Deleted member 4993]

Since the distances are expressed as square-roots - you may choose to equate "squares" of distances.

 

[soroban]

Hello, leoduong!

Here's another approach . . .

\(\displaystyle \text{Find the point on the line }\,L_1\!\!:\,4x - 2y + 3 \:= \:0\,\text{ that is equidistant from }P(3,3)\text{ and }Q(7,-3)\)

\(\displaystyle \text{Find the equation of }L_2\text{, the perpendicular bisector of }PQ.\)

. . \(\displaystyle \text{(You will need the midpoint of }PQ\text{ and the slope of }PQ.)\)


\(\displaystyle \text{Then find the intersection of }L_1\text{ and }L_2.\)