
If $P(1,0),Q( - 1,0),R(2,0)$ are three given points, then find the locus of S satisfying the relation ${\left( {SQ} \right)^2} + {\left( {SR} \right)^2} = 2{\left( {SP} \right)^2}$ .
A. A straight line parallel to the x-axis.
B. A parabola whose vertex at origin
C. A straight line parallel to y-axis
D. A circle centered at (0,0).
Answer
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Hint: First suppose the coordinate of the point S. Then use the distance formula to obtain the results SQ, SR and SP. Then substitute the values in the given relation and simplify to obtain the required result.
Formula Used:
The distance between the points $(a,b)$ and $(c,d)$ is
$\sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $ .
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$
Complete step by step solution:
The given points are $P(1,0),Q( - 1,0),R(2,0)$.
Suppose the coordinate of S is (x, y).
Hence,
$SQ = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{(0 - y)}^2}} $
So, $S{Q^2} = {\left( {1 + x} \right)^2} + {y^2}$
Now,
$SR = \sqrt {{{\left( {2 - x} \right)}^2} + {{(0 - y)}^2}} $
So, $S{R^2} = {\left( {2 - x} \right)^2} + {y^2}$
And,
$SP = \sqrt {{{\left( {1 - x} \right)}^2} + {{(0 - y)}^2}} $
So, $S{P^2} = {\left( {1 - x} \right)^2} + {y^2}$
Substitute the obtained values of $S{P^2},S{Q^2},S{R^2}$ in the given equation ${\left( {SQ} \right)^2} + {\left( {SR} \right)^2} = 2{\left( {SP} \right)^2}$ to obtain the required result.
${\left( {1 + x} \right)^2} + {y^2} + {\left( {2 - x} \right)^2} + {y^2} = 2\left[ {{{\left( {1 - x} \right)}^2} + {y^2}} \right]$
$1 + {x^2} + 2x + {y^2} + 4 - 2x + {x^2} + {y^2} = 2\left[ {1 - 2x + {x^2} + {y^2}} \right]$
$2{x^2} + 2{y^2} + 5 = 2{x^2} + 2{y^2} - 4x + 2$
$ - 4x = 3$
$x = - \dfrac{3}{4}$
Now, x=constant line is always parallel to the y axis.
Hence the line $x = - \dfrac{3}{4}$is parallel to the y axis.
Option ‘C’ 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: Students sometimes get confused with the fact that x= constant line is parallel to x axis or the y axis, so for those students can take the help of a graph and draw a line through that point in roughwork to conclude the result.
Formula Used:
The distance between the points $(a,b)$ and $(c,d)$ is
$\sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $ .
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$
Complete step by step solution:
The given points are $P(1,0),Q( - 1,0),R(2,0)$.
Suppose the coordinate of S is (x, y).
Hence,
$SQ = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{(0 - y)}^2}} $
So, $S{Q^2} = {\left( {1 + x} \right)^2} + {y^2}$
Now,
$SR = \sqrt {{{\left( {2 - x} \right)}^2} + {{(0 - y)}^2}} $
So, $S{R^2} = {\left( {2 - x} \right)^2} + {y^2}$
And,
$SP = \sqrt {{{\left( {1 - x} \right)}^2} + {{(0 - y)}^2}} $
So, $S{P^2} = {\left( {1 - x} \right)^2} + {y^2}$
Substitute the obtained values of $S{P^2},S{Q^2},S{R^2}$ in the given equation ${\left( {SQ} \right)^2} + {\left( {SR} \right)^2} = 2{\left( {SP} \right)^2}$ to obtain the required result.
${\left( {1 + x} \right)^2} + {y^2} + {\left( {2 - x} \right)^2} + {y^2} = 2\left[ {{{\left( {1 - x} \right)}^2} + {y^2}} \right]$
$1 + {x^2} + 2x + {y^2} + 4 - 2x + {x^2} + {y^2} = 2\left[ {1 - 2x + {x^2} + {y^2}} \right]$
$2{x^2} + 2{y^2} + 5 = 2{x^2} + 2{y^2} - 4x + 2$
$ - 4x = 3$
$x = - \dfrac{3}{4}$
Now, x=constant line is always parallel to the y axis.
Hence the line $x = - \dfrac{3}{4}$is parallel to the y axis.
Option ‘C’ 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: Students sometimes get confused with the fact that x= constant line is parallel to x axis or the y axis, so for those students can take the help of a graph and draw a line through that point in roughwork to conclude the result.
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