Question

The locus of the point P(x, y) satisfying the relation
$\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}+\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}=6\text{ is }$
(a) Straight line
(b) Pair of straight lines
(c) Circle
(d) Ellipse

Answer 1

Hint: The curve we get by solving the given condition is nothing but the locus of point P. Just solve it normally as an algebraic expression. By sending one term to the right hand side and then applying a square on both sides which will in turn increase the algebra we need to solve but it is the only way possible to solve this question. At last try to write the second degree equation you get in the form of multiplication of 2 lines. This way you can say that the obtained equation is a pair of the lines which are in the multiplication we wrote.

Complete step-by-step answer:
Locus is a set of all points, whose location satisfies or is determined by one or more specified conditions. In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property.
By above definition we can see that the coordinates (x, y) of the point P are satisfying the given condition.
So the curve we get by solving the given condition is nothing but the locus of the point P.
$\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}+\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}=6$
By subtracting the second term on left side from both sides, we get:
$\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}+\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}-\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}=6-\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}$
By simplifying, we get:
$\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}=6-\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}$
By squaring on both sides, we get:
${{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( 6-\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}} \right)}^{2}}$
By solving more, we get:
${{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=36+{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}-12\left( \sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}} \right)$
By expanding every square in the equation, we get:
${{x}^{2}}+9-6x+{{y}^{2}}+1-2y=36+{{x}^{2}}+9+6x+{{y}^{2}}+1-2y-12\left( \sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}} \right)$
By cancelling the common terms, we get:
$-6x=36+6x-12\left( \sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}} \right)$
By simplifying more, we get:
$12\left( \sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}} \right)=36+12x$
By dividing 12 on both sides, we get:
$\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}=x+3$
By squaring on both sides, we get:
${{\left( x+3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( x+3 \right)}^{2}}$
By cancelling the common terms, we get:
${{\left( y-1 \right)}^{2}}=0$
We can write the above equation in this form:
$\left( y-1 \right).\left( y-1 \right)=0$
So we get two equations from the above condition. These two are the same lines. But we say that they are 2 same lines, also called as coincident lines. So the locus is called a pair of coincident lines. According to options we can select as a pair of straight lines. So option (b) is correct.

Note: While calculating locus the calculation looks long, but you must proceed with confidence at last many terms will get canceled and you will get your required result. The idea of writing the final equation as multiplication of 2 lines is to find the name of the locus; it is done from checking options.