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Question

Answers

A. a, b, c

B. c, b, a

C. b, c, a

D. a, c, b

Answer

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Hint: Let us find the distance of the point from the two given points using distance formula and then equate that to get the correct values of a, b and c.

Now as we know that, in geometry, a locus is a set of all points whose location satisfies or is determined by one or more specified conditions.

Complete step-by-step answer:

Now here we are given the locus of an unknown point whose distance from two given points is the same.

So, let that point be (x, y)

So, according to the question, the distance between (x, y) and (-2, 3) is equal to the distance between (x, y) and (6, -5).

So, now we will find the distance between the points.

As we know that if \[\left( {{{\text{x}}_{\text{1}}}{\text{, }}{{\text{y}}_{\text{1}}}} \right)\] and \[\left( {{{\text{x}}_{\text{2}}}{\text{, }}{{\text{y}}_{\text{2}}}} \right)\] are the two points. Then according to distance formula distance between them will be \[\sqrt {{{\left( {{{\text{x}}_{\text{1}}}{\text{ - }}{{\text{x}}_{\text{2}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{y}}_{\text{1}}}{\text{ - }}{{\text{y}}_{\text{2}}}} \right)}^{\text{2}}}} \]

So, now distance between (x, y) and (-2, 3) will be \[\sqrt {{{\left( {{\text{x - ( - 2)}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - 3}}} \right)}^{\text{2}}}} \]

And the distance between (x, y) and (6, -5) will be \[\sqrt {{{\left( {{\text{x - 6}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - ( - 5)}}} \right)}^{\text{2}}}} \]

Now according to the question,

$\Rightarrow$ \[\sqrt {{{\left( {{\text{x - ( - 2)}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - 3}}} \right)}^{\text{2}}}} \] = \[\sqrt {{{\left( {{\text{x - 6}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - ( - 5)}}} \right)}^{\text{2}}}} \] (1)

So, now we had to solve equation 1 to find the locus of point (x, y).

So, squaring both sides of equation 1. We get,

$\Rightarrow$ \[{\left( {{\text{x + 2}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y - 3}}} \right)^{\text{2}}}\] = \[{\left( {{\text{x - 6}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y + 5}}} \right)^{\text{2}}}\]

Now opening the squares of the above equation. We get,

$\Rightarrow$ \[\left( {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4}}} \right){\text{ + }}\left( {{{\text{y}}^{\text{2}}}{\text{ - 6y + 9}}} \right){\text{ = }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 12x + 36}}} \right){\text{ + }}\left( {{{\text{y}}^{\text{2}}}{\text{ + 10y + 25}}} \right)\]

Now subtracting RHS from the LHS of the above equation. We get,

$\Rightarrow$ 16x – 16y – 48 = x – y – 3 = 0

So, now as we can see that the locus of point (x, y) such that it is equidistant from the points (-2, 3), (6, -5) is x – y – 3 = 0 .

And in the question locus given is ax + by + c = 0, where a > 0.

So, x – y – 3 = ax + by + c

So, on comparing LHS and RHS of the above equation. We get,

a = 1, b = -1 and c = -3

Now, as we know that in ascending order the smallest number comes first and largest number comes at last position.

So, the ascending order of a, b and c will be c, b, a.

Hence, the correct option will be B.

Note: Whenever we are given with the equation of locus and asked to find the variables in equation or place it in ascending or descending order then first we had to find the equation of locus of that point using given conditions (here equating the distance between points) and then compare that equation with the given equation of locus of that point to get the required value of the variables.

Now as we know that, in geometry, a locus is a set of all points whose location satisfies or is determined by one or more specified conditions.

Complete step-by-step answer:

Now here we are given the locus of an unknown point whose distance from two given points is the same.

So, let that point be (x, y)

So, according to the question, the distance between (x, y) and (-2, 3) is equal to the distance between (x, y) and (6, -5).

So, now we will find the distance between the points.

As we know that if \[\left( {{{\text{x}}_{\text{1}}}{\text{, }}{{\text{y}}_{\text{1}}}} \right)\] and \[\left( {{{\text{x}}_{\text{2}}}{\text{, }}{{\text{y}}_{\text{2}}}} \right)\] are the two points. Then according to distance formula distance between them will be \[\sqrt {{{\left( {{{\text{x}}_{\text{1}}}{\text{ - }}{{\text{x}}_{\text{2}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{\text{y}}_{\text{1}}}{\text{ - }}{{\text{y}}_{\text{2}}}} \right)}^{\text{2}}}} \]

So, now distance between (x, y) and (-2, 3) will be \[\sqrt {{{\left( {{\text{x - ( - 2)}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - 3}}} \right)}^{\text{2}}}} \]

And the distance between (x, y) and (6, -5) will be \[\sqrt {{{\left( {{\text{x - 6}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - ( - 5)}}} \right)}^{\text{2}}}} \]

Now according to the question,

$\Rightarrow$ \[\sqrt {{{\left( {{\text{x - ( - 2)}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - 3}}} \right)}^{\text{2}}}} \] = \[\sqrt {{{\left( {{\text{x - 6}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - ( - 5)}}} \right)}^{\text{2}}}} \] (1)

So, now we had to solve equation 1 to find the locus of point (x, y).

So, squaring both sides of equation 1. We get,

$\Rightarrow$ \[{\left( {{\text{x + 2}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y - 3}}} \right)^{\text{2}}}\] = \[{\left( {{\text{x - 6}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y + 5}}} \right)^{\text{2}}}\]

Now opening the squares of the above equation. We get,

$\Rightarrow$ \[\left( {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4}}} \right){\text{ + }}\left( {{{\text{y}}^{\text{2}}}{\text{ - 6y + 9}}} \right){\text{ = }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 12x + 36}}} \right){\text{ + }}\left( {{{\text{y}}^{\text{2}}}{\text{ + 10y + 25}}} \right)\]

Now subtracting RHS from the LHS of the above equation. We get,

$\Rightarrow$ 16x – 16y – 48 = x – y – 3 = 0

So, now as we can see that the locus of point (x, y) such that it is equidistant from the points (-2, 3), (6, -5) is x – y – 3 = 0 .

And in the question locus given is ax + by + c = 0, where a > 0.

So, x – y – 3 = ax + by + c

So, on comparing LHS and RHS of the above equation. We get,

a = 1, b = -1 and c = -3

Now, as we know that in ascending order the smallest number comes first and largest number comes at last position.

So, the ascending order of a, b and c will be c, b, a.

Hence, the correct option will be B.

Note: Whenever we are given with the equation of locus and asked to find the variables in equation or place it in ascending or descending order then first we had to find the equation of locus of that point using given conditions (here equating the distance between points) and then compare that equation with the given equation of locus of that point to get the required value of the variables.