
If A and B are two fixed points, then find the nature of the locus of the point P, such that
\[P{A^2} + P{B^2} = A{B^2}\].
A. Circle with AB as a diameter
B. Right angle with \[\angle P = {90^ \circ }\]
C. Semi-circle with AB as a diameter
D. Circle with AB as a diameter, excluding point A and B.
Answer
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Hint: First apply the converse of Pythagorean theorem in th given relation. Then assume the coordinate of A, B and P. Then find the slope of line PA and PB. Using the the slope we will find the equation of locus of the point P.
Formula used:
Converse of Pythagorean theorem: If square of side is equal to the sum of the square of other sides, then the triangle must be right angle triangle.
The slope of a line passes through the points \[\left(x_1,y_1\right)\] and \[\left(x_2,y_2\right)\] is \[\dfrac{y_2-y_1}{x_2-x_1}\].
The product of slope of two perpendicular lines is -1.
The equation of circle having \[\left(x_1,y_1\right)\] and \[\left(x_2,y_2\right)\] as endpoint of the diameter is \[(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\]
Complete step by step solution:
Given that, \[P{A^2} + P{B^2} = A{B^2}\]
According to converse of Pythgorean theorem,
PAB is a right angled triangle and \[\angle APB = \dfrac{\pi}{2}\].
Thus PA and PB are perpendicular to each other.
Assume that \[\left(x_1,y_1\right)\], \[\left(x_2,y_2\right)\], and \[\left(h,k\right)\] be coordinates of A,B, and P respectively.
The slope of the line PA is \[\dfrac{k-y_1}{h-x_1}\].
The slope of the line PB is \[\dfrac{k-y_2}{h-x_2}\].
Since PA and PB are perpendicular to each other, so the product of the slope is -1.
\[\dfrac{k-y_1}{h-x_1}\times \dfrac{k-y_2}{h-x_2} =-1\]
\[\dfrac{(k-y_1)(k-y_2)}{(h-x_1)(h-x_2)}=-1\]
Cross multiply:
\[{(k-y_1)(k-y_2)}=-{(h-x_1)(h-x_2)}\]
\[{{(h-x_1)(h-x_2)+(k-y_1)(k-y_2)}}=0\]
Now replace h with x and k with y:
\[{{(x-x_1)(x-x_2)+(y-y_1)(y-y_2)}}=0\]
Which is an equation of circle having AB as diameter.
But in the locus A and B are not included. Because A,B,P are colinear, then PAB is not a right angled triangle.
Hence, the. option D is correct.
Additional information:
A locus is a curve or other shape created in mathematics from all the points that meet a specific equation describing the relationship between the coordinates, or from a point, line, or moving surface. The locus defines all shapes as a set of points, including circles, ellipses, parabolas, and hyperbolas.
Note: As it is given that A and B are two fixed points students often take them as (1,0) and (0,1) and calculate as per the given condition and ended up with an equation of a circle, though the calculation is correct but we need to proceed for our given options, so we need to calculate accordingly.
Formula used:
Converse of Pythagorean theorem: If square of side is equal to the sum of the square of other sides, then the triangle must be right angle triangle.
The slope of a line passes through the points \[\left(x_1,y_1\right)\] and \[\left(x_2,y_2\right)\] is \[\dfrac{y_2-y_1}{x_2-x_1}\].
The product of slope of two perpendicular lines is -1.
The equation of circle having \[\left(x_1,y_1\right)\] and \[\left(x_2,y_2\right)\] as endpoint of the diameter is \[(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\]
Complete step by step solution:
Given that, \[P{A^2} + P{B^2} = A{B^2}\]
According to converse of Pythgorean theorem,
PAB is a right angled triangle and \[\angle APB = \dfrac{\pi}{2}\].
Thus PA and PB are perpendicular to each other.
Assume that \[\left(x_1,y_1\right)\], \[\left(x_2,y_2\right)\], and \[\left(h,k\right)\] be coordinates of A,B, and P respectively.
The slope of the line PA is \[\dfrac{k-y_1}{h-x_1}\].
The slope of the line PB is \[\dfrac{k-y_2}{h-x_2}\].
Since PA and PB are perpendicular to each other, so the product of the slope is -1.
\[\dfrac{k-y_1}{h-x_1}\times \dfrac{k-y_2}{h-x_2} =-1\]
\[\dfrac{(k-y_1)(k-y_2)}{(h-x_1)(h-x_2)}=-1\]
Cross multiply:
\[{(k-y_1)(k-y_2)}=-{(h-x_1)(h-x_2)}\]
\[{{(h-x_1)(h-x_2)+(k-y_1)(k-y_2)}}=0\]
Now replace h with x and k with y:
\[{{(x-x_1)(x-x_2)+(y-y_1)(y-y_2)}}=0\]
Which is an equation of circle having AB as diameter.
But in the locus A and B are not included. Because A,B,P are colinear, then PAB is not a right angled triangle.
Hence, the. option D is correct.
Additional information:
A locus is a curve or other shape created in mathematics from all the points that meet a specific equation describing the relationship between the coordinates, or from a point, line, or moving surface. The locus defines all shapes as a set of points, including circles, ellipses, parabolas, and hyperbolas.
Note: As it is given that A and B are two fixed points students often take them as (1,0) and (0,1) and calculate as per the given condition and ended up with an equation of a circle, though the calculation is correct but we need to proceed for our given options, so we need to calculate accordingly.
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