Question:
On the locus of the point \(P(x,y)\) equidistant from \((3,0)\) and \((0,4)\), if \(A\) and \(B\) are two points that satisfy \(4x=3y\) and \(x=y\) respectively, then the distance between \(A\) and \(B\) is:
On the locus of the point \(P(x,y)\) equidistant from \((3,0)\) and \((0,4)\), if \(A\) and \(B\) are two points that satisfy \(4x=3y\) and \(x=y\) respectively, then the distance between \(A\) and \(B\) is:
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The locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining those points.
Updated On: Jun 26, 2026
- \(\dfrac{5}{2}\)
- \(5\)
- \(\dfrac{25}{4}\)
- \(25\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the locus of points equidistant from \((3,0)\) and \((0,4)\).
Given, \[ \sqrt{(x-3)^2+y^2} = \sqrt{x^2+(y-4)^2} \] Squaring both sides, \[ (x-3)^2+y^2 = x^2+(y-4)^2 \] Expanding, \[ x^2-6x+9+y^2 = x^2+y^2-8y+16 \] \[ -6x+9=-8y+16 \] \[ 6x-8y+7=0 \] Thus, the required locus is \[ 6x-8y+7=0 \]
Step 2: Find point \(A\).
Point \(A\) lies on the locus and satisfies \[ 4x=3y \] Hence, \[ y=\frac{4x}{3} \] Substituting into the locus equation, \[ 6x-8\left(\frac{4x}{3}\right)+7=0 \] \[ 18x-32x+21=0 \] \[ -14x+21=0 \] \[ x=\frac{3}{2} \] \[ y=2 \] Therefore, \[ A\left(\frac{3}{2},2\right) \]
Step 3: Find point \(B\).
Point \(B\) lies on the locus and satisfies \[ x=y \] Substituting into the locus equation, \[ 6x-8x+7=0 \] \[ -2x+7=0 \] \[ x=\frac{7}{2} \] \[ y=\frac{7}{2} \] Therefore, \[ B\left(\frac{7}{2},\frac{7}{2}\right) \]
Step 4: Find the distance \(AB\).
Using the distance formula, \[ AB= \sqrt{ \left(\frac{7}{2}-\frac{3}{2}\right)^2 + \left(\frac{7}{2}-2\right)^2 } \] \[ = \sqrt{ 2^2+\left(\frac{3}{2}\right)^2 } \] \[ = \sqrt{ 4+\frac{9}{4} } \] \[ = \sqrt{\frac{25}{4}} \] \[ =\frac{5}{2} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{5}{2}} \]
Given, \[ \sqrt{(x-3)^2+y^2} = \sqrt{x^2+(y-4)^2} \] Squaring both sides, \[ (x-3)^2+y^2 = x^2+(y-4)^2 \] Expanding, \[ x^2-6x+9+y^2 = x^2+y^2-8y+16 \] \[ -6x+9=-8y+16 \] \[ 6x-8y+7=0 \] Thus, the required locus is \[ 6x-8y+7=0 \]
Step 2: Find point \(A\).
Point \(A\) lies on the locus and satisfies \[ 4x=3y \] Hence, \[ y=\frac{4x}{3} \] Substituting into the locus equation, \[ 6x-8\left(\frac{4x}{3}\right)+7=0 \] \[ 18x-32x+21=0 \] \[ -14x+21=0 \] \[ x=\frac{3}{2} \] \[ y=2 \] Therefore, \[ A\left(\frac{3}{2},2\right) \]
Step 3: Find point \(B\).
Point \(B\) lies on the locus and satisfies \[ x=y \] Substituting into the locus equation, \[ 6x-8x+7=0 \] \[ -2x+7=0 \] \[ x=\frac{7}{2} \] \[ y=\frac{7}{2} \] Therefore, \[ B\left(\frac{7}{2},\frac{7}{2}\right) \]
Step 4: Find the distance \(AB\).
Using the distance formula, \[ AB= \sqrt{ \left(\frac{7}{2}-\frac{3}{2}\right)^2 + \left(\frac{7}{2}-2\right)^2 } \] \[ = \sqrt{ 2^2+\left(\frac{3}{2}\right)^2 } \] \[ = \sqrt{ 4+\frac{9}{4} } \] \[ = \sqrt{\frac{25}{4}} \] \[ =\frac{5}{2} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{5}{2}} \]
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