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Find the equation of a locus of a point which is equidistance from the points (4,2) and (3,5).​

1 Answer

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Answer:

3y = x + 7.

Explanation:

The locus will be a line perpendicular to and passing through the centre of the line joining the 2 points.

The midpoint of (4, 2) and (3, 5) is

[(4+3)/2, (2+5 / 2)=

= (3.5 , 3.5).

The slope of the line joining the 2 points is

(5-2)(3-4) = -3.

So, the slope of the perpendicular line = -1 / -3 = 1/3.

Now we find the required locus by using the point-slope form of the straight line and the above results:

y - y1 = m(x - x1)

Here m = 1/3, x1 = 3.3 and y1 = 3.5 so:

y - 3.5 = 1/3(x - 3.5)

y = 1/3x - 3.5/3 + 3.5

Multiplying through by 3:

3y = x - 3.5 + 10.5

3y = x + 7.

answered
User Svens
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