The point on the line 3x + 4y = 5 which is eq | vedclass

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(B) Let the point be $P(x, y)$. Since $P$ lies on the line $3x + 4y = 5$,we have $3x + 4y = 5$ ... $(i)$.Since $P$ is equidistant from $A(1, 2)$ and $

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$(15, -10)$

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(B) Let the point be $P(x, y)$. Since $P$ lies on the line $3x + 4y = 5$,we have $3x + 4y = 5$ ... $(i)$.Since $P$ is equidistant from $A(1, 2)$ and $B(3, 4)$,we have $PA^2 = PB^2$.$(x - 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 4)^2$$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16$$-2x - 4y + 5 = -6x - 8y + 25$$4x + 4y = 20$$x + y = 5$ ... $(ii)$.From $(ii)$,$y = 5 - x$. Substituting this into $(i)$:$3x + 4(5 - x) = 5$$3x + 20 - 4x = 5$$-x = -15 \Rightarrow x = 15$.Then $y = 5 - 15 = -10$.Thus,the point is $(15, -10)$.

The point on the line $3x + 4y = 5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is

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