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What is the locus of a point which is equidistant from the point (m + n, n – m) and the point (m – n, n + m)?
- a)mx = ny
- b)nx = – my
- c)nx = my
- d)mx = – ny
Correct answer is option 'C'. Can you explain this answer?
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What is the locus of a point which is equidistant from the point (m + ...
**Explanation:**
To find the locus of a point that is equidistant from two given points, we need to find the equation of the perpendicular bisector of the line segment joining the two points.
Let the two given points be A (m, n) and B (n, m). The midpoint of the line segment AB is the point C, which can be found by taking the average of the x-coordinates and the average of the y-coordinates.
Midpoint C: ((m + n)/2, (n + m)/2)
The slope of the line segment AB can be found using the formula:
Slope of AB = (y2 - y1) / (x2 - x1)
Slope of AB = (m - n) / (n - m) = -1
The negative reciprocal of the slope of AB is the slope of the perpendicular bisector. Therefore, the slope of the perpendicular bisector is 1.
Now, let's find the equation of the perpendicular bisector using the point-slope form of a line:
y - y1 = m(x - x1)
Substituting the values of point C and the slope:
y - (n + m)/2 = 1(x - (m + n)/2)
y - (n + m)/2 = x - (m + n)/2
y = x - (m + n)/2 + (n + m)/2
Simplifying:
y = x
So, the equation of the perpendicular bisector, and hence the locus of points equidistant from A and B, is y = x.
This equation is in the form nx = my, which matches option C.
Therefore, the correct answer is option C) nx = my.
To find the locus of a point that is equidistant from two given points, we need to find the equation of the perpendicular bisector of the line segment joining the two points.
Let the two given points be A (m, n) and B (n, m). The midpoint of the line segment AB is the point C, which can be found by taking the average of the x-coordinates and the average of the y-coordinates.
Midpoint C: ((m + n)/2, (n + m)/2)
The slope of the line segment AB can be found using the formula:
Slope of AB = (y2 - y1) / (x2 - x1)
Slope of AB = (m - n) / (n - m) = -1
The negative reciprocal of the slope of AB is the slope of the perpendicular bisector. Therefore, the slope of the perpendicular bisector is 1.
Now, let's find the equation of the perpendicular bisector using the point-slope form of a line:
y - y1 = m(x - x1)
Substituting the values of point C and the slope:
y - (n + m)/2 = 1(x - (m + n)/2)
y - (n + m)/2 = x - (m + n)/2
y = x - (m + n)/2 + (n + m)/2
Simplifying:
y = x
So, the equation of the perpendicular bisector, and hence the locus of points equidistant from A and B, is y = x.
This equation is in the form nx = my, which matches option C.
Therefore, the correct answer is option C) nx = my.
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What is the locus of a point which is equidistant from the point (m + n, n m) and the point (m n, n + m)?a)mx = nyb)nx = myc)nx = myd)mx = nyCorrect answer is option 'C'. Can you explain this answer?
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