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the point on the line 3x 4y 5 which is equidistant
Question:
The point on the line $3x+4y=5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is
Show Hint
The locus of points equidistant from two points is the perpendicular bisector of the segment joining them.
MET - 2009
MET
Updated On:
Apr 10, 2026
$(7, -4)$
$(15, -10)$
$(\frac{1}{7}, \frac{8}{7})$
$(0, \frac{5}{4})$
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The Correct Option is
B
Solution and Explanation
Step 1: Equidistance Equation
Let the point be $(x_1, y_1)$. Distance from $(1, 2)$ equals distance from $(3, 4)$:
$(x_1 - 1)^2 + (y_1 - 2)^2 = (x_1 - 3)^2 + (y_1 - 4)^2$.
Step 2: Simplify
$x_1^2 - 2x_1 + 1 + y_1^2 - 4y_1 + 4 = x_1^2 - 6x_1 + 9 + y_1^2 - 8y_1 + 16$.
$4x_1 + 4y_1 = 20 \Rightarrow x_1 + y_1 = 5$.
Step 3: Linear System
The point is also on $3x + 4y = 5$.
Solve the system:
1) $x_1 + y_1 = 5 \Rightarrow y_1 = 5 - x_1$.
2) $3x_1 + 4(5 - x_1) = 5$.
Step 4: Solve
$3x_1 + 20 - 4x_1 = 5 \Rightarrow -x_1 = -15 \Rightarrow x_1 = 15$.
$y_1 = 5 - 15 = -10$.
Final Answer:
(b)
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