Question

If any point \[P\] is at the equal distances from points \[A\left( {a + b,a - b} \right)\] and \[B\left( {a - b,a + b} \right)\], then what is the locus of point \[P\]?
A. \[x - y = 0\]
B. \[ax + by = 0\]
C. \[bx + ay = 0\]
D. \[x + y = 0\]

Answer 1

Hint: Let at any instant, the coordinates of the point \[P\] be \[\left( {h,k} \right)\]. Then find the distances of \[P\] from the two given points \[A\left( {a + b,a - b} \right)\] and \[B\left( {a - b,a + b} \right)\]. It is said in the question that these two distances are equal. So, make an equation using it and simplify the equation. After simplification, the locus of the point will be obtained.

Formula Used:
Distance between two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is \[\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \] units.
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]

Complete step-by-step answer:
Let at any instant, the coordinates of the point \[P\] be \[\left( {h,k} \right)\]
Then distance of \[P\]\[\left( {h,k} \right)\] from \[A\left( {a + b,a - b} \right)\] is \[PA = \sqrt {{{\left\{ {\left( {a + b} \right) - h} \right\}}^2} + {{\left\{ {\left( {a - b} \right) - k} \right\}}^2}} \]
And distance of \[P\]\[\left( {h,k} \right)\] from \[B\left( {a - b,a + b} \right)\] is \[PB = \sqrt {{{\left\{ {\left( {a - b} \right) - h} \right\}}^2} + {{\left\{ {\left( {a + b} \right) - k} \right\}}^2}} \]
It is given that these two distances are equal.
So, \[\sqrt {{{\left\{ {\left( {a + b} \right) - h} \right\}}^2} + {{\left\{ {\left( {a - b} \right) - k} \right\}}^2}} = \sqrt {{{\left\{ {\left( {a - b} \right) - h} \right\}}^2} + {{\left\{ {\left( {a + b} \right) - k} \right\}}^2}} \]
Squaring both sides, we get
\[{\left\{ {\left( {a + b} \right) - h} \right\}^2} + {\left\{ {\left( {a - b} \right) - k} \right\}^2} = {\left\{ {\left( {a - b} \right) - h} \right\}^2} + {\left\{ {\left( {a + b} \right) - k} \right\}^2}\]
Use the identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {\left( {a + b} \right)^2} - 2h\left( {a + b} \right) + {h^2} + {\left( {a - b} \right)^2} - 2k\left( {a - b} \right) + {k^2} = {\left( {a - b} \right)^2} - 2h\left( {a - b} \right) + {h^2} + {\left( {a + b} \right)^2} - 2k\left( {a + b} \right) + {k^2}\]
Cancel the terms \[{\left( {a + b} \right)^2},{\left( {a - b} \right)^2},{h^2},{k^2}\] from both sides.
\[ \Rightarrow - 2h\left( {a + b} \right) - 2k\left( {a - b} \right) = - 2h\left( {a - b} \right) - 2k\left( {a + b} \right)\]
Take \[\left( { - 2} \right)\] common from both sides and cancel.
\[ \Rightarrow h\left( {a + b} \right) + k\left( {a - b} \right) = h\left( {a - b} \right) + k\left( {a + b} \right)\]
\[ \Rightarrow ah + bh + ak - bk = ah - bh + ak + bk\]
Cancel the terms \[ah\] and \[ak\] from both sides.
\[ \Rightarrow bh - bk = - bh + bk\]
\[ \Rightarrow bh + bh = bk + bk\]
\[ \Rightarrow 2bh = 2bk\]
Cancel \[\left( {2b} \right)\] from both sides.
\[ \Rightarrow h = k\]
\[ \Rightarrow h - k = 0\]
Replacing \[h\] by \[x\] and \[k\] by \[y\], we get
\[x - y = 0\]
This is the equation of the locus of the point \[P\].
Hence option A is correct.

Note: Locus of a point is a path created by the point satisfying some conditions. To find a locus of a point just use the given condition(s) and make an equation. After simplification, you’ll get the locus. In this problem, it is said that the two distances of the points \[A\left( {a + b,a - b} \right)\] and \[B\left( {a - b,a + b} \right)\] from the moving point \[P\] are equal. So, the distances have been calculated and equated. After simplification of the equation the desired answer has been obtained.