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

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(B) Let the required point be $P(x_1, y_1)$. Since $P$ lies on the line $4x - y - 2 = 0$,we have: $4x_1 - y_1 - 2 = 0$  ... $(i)$ Given that $P$ is eq

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$(4, 14)$

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(B) Let the required point be $P(x_1, y_1)$. Since $P$ lies on the line $4x - y - 2 = 0$,we have: $4x_1 - y_1 - 2 = 0$  ... $(i)$ Given that $P$ is equidistant from $A(-5, 6)$ and $B(3, 2)$,we have $PA^2 = PB^2$: $(x_1 + 5)^2 + (y_1 - 6)^2 = (x_1 - 3)^2 + (y_1 - 2)^2$ $x_1^2 + 10x_1 + 25 + y_1^2 - 12y_1 + 36 = x_1^2 - 6x_1 + 9 + y_1^2 - 4y_1 + 4$ $16x_1 - 8y_1 + 48 = 0$ Dividing by $8$,we get: $2x_1 - y_1 + 6 = 0$  ... (ii) Subtracting (ii) from $(i)$: $(4x_1 - y_1 - 2) - (2x_1 - y_1 + 6) = 0$ $2x_1 - 8 = 0 \implies x_1 = 4$ Substituting $x_1 = 4$ in $(i)$: $4(4) - y_1 - 2 = 0 \implies 16 - y_1 - 2 = 0 \implies y_1 = 14$ Thus,the 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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