The point on the line 4x - y - 2 = 0 which is | vedclass

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(B) Let the point on the line $4x - y - 2 = 0$ be $P(x, y)$.Let $A = (-5, 6)$ and $B = (3, 2)$.Since $P$ is equidistant from $A$ and $B$,we have $AP =

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$(4, 14)$

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(B) Let the point on the line $4x - y - 2 = 0$ be $P(x, y)$.Let $A = (-5, 6)$ and $B = (3, 2)$.Since $P$ is equidistant from $A$ and $B$,we have $AP = PB$,which implies $AP^2 = PB^2$.Using the distance formula: $(x + 5)^2 + (y - 6)^2 = (x - 3)^2 + (y - 2)^2$.Expanding both sides: $x^2 + 10x + 25 + y^2 - 12y + 36 = x^2 - 6x + 9 + y^2 - 4y + 4$.Simplifying: $16x - 8y + 48 = 0$,which reduces to $2x - y + 6 = 0$ (Equation $2$).We are given the line $4x - y - 2 = 0$ (Equation $1$).Subtracting Equation $2$ from Equation $1$: $(4x - y - 2) - (2x - y + 6) = 0$.$2x - 8 = 0 \Rightarrow x = 4$.Substituting $x = 4$ into $4x - y - 2 = 0$: $4(4) - y - 2 = 0$ $\Rightarrow 16 - y - 2 = 0$ $\Rightarrow y = 14$.Thus,the required point is $(4, 14)$.

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is:

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