eqn of line equidistant from (4, 2) and (-4, 3)

[NEHA]

Write the equation of the line that is equidistant from the points (4, 2) and (-4, 3).

distance (x, y) to (-4, 3) = distance (x, y) to (4, 2)

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sqrt( [(x) - (-4)]^2 + [(3) - (y)]^2 ) = sqrt( [(4) - (x)]^2 + [(2) - (y)]^2 )

(x + 4)^2 + (3 - y)^2 = (4 - x)^2 + (2 - y)^2

x^2 + 16x + 16 + 9 - 6y + y^2 = 16 - 8x + x^2 + 4 - 4y + y^2

16x + 16 + 9 - 6y = 16 - 8x + 4 - 4y

24x + 10y - 5 = 0

y = - 12/5x + 5/10

 

[stapel]

(x + 4)^2 + (3 - y)^2 = (4 - x)^2 + (2 - y)^2

x^2 + 16x + 16 + 9 - 6y + y^2 = 16 - 8x + x^2 + 4 - 4y + y^2

The line passing between these points will be perpendicular to the line through these points. The slope of the line through the points is:

. . . . .\(\displaystyle \L m\, =\, \frac{3\, -\, 2}{-4\, -\, 4}\, =\, \frac{1}{-8}\, =\, -\frac{1}{8}\)

Then the line you seek must have a slope of 8.

You might want to check your squaring:

. . . . .(x + 4)² = (x + 4)(x + 4) = x² + 4x + 4x + 16 = x² + 8x + 16

See if that helps.

Note: You can always check your answer by graphing.

Eliz.

 

[Mrspi]

The line equidistant from two points is the perpendicular bisector of the segment joining the two points.

1) So, you're looking for the line which goes through the midpoint of the segment joining (4, 2) and (-4, 3). Find the coordinates of the midpoint.

2) And, the line you're must be perpendicular to the line through (4, 2) and (-4, 3). So, your line will have as its slope the opposite reciprocal of the slope through the given points.

Once you know the slope (from step 2) and the coordinates of a point on the line (that would be the midpoint you found in step 1), you can use the point-slope form of the equation of a line:

y - y<SUB>1</SUB> = m(x - x<SUB>1</SUB>)

 

[NEHA]

ok so what did i do....that isn't right than? in what problem can i use that way that i uused in this problem

 

[stapel]

Is my earlier reply (correcting not your method but an addition error) not displying for you...?

Thank you.

Eliz.