The equation of the line joining the point (3 | vedclass

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(A) First,find the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$. Multiplying the first equation by $3$,we get $12x + 3y

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(A) First,find the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$. Multiplying the first equation by $3$,we get $12x + 3y - 3 = 0$. Adding this to $7x - 3y - 35 = 0$,we get $19x - 38 = 0$,which implies $x = 2$. Substituting $x = 2$ into $4x + y - 1 = 0$,we get $4(2) + y - 1 = 0$,so $y = -7$. The point of intersection is $(2, -7)$. The line joining $(3, 5)$ and $(2, -7)$ has the slope $m = \frac{-7 - 5}{2 - 3} = \frac{-12}{-1} = 12$. The equation of the line is $y - 5 = 12(x - 3)$,which simplifies to $12x - y - 31 = 0$. The distance of this line from $(0, 0)$ is $d_1 = \frac{|12(0) - 0 - 31|}{\sqrt{12^2 + (-1)^2}} = \frac{31}{\sqrt{145}}$. The distance of this line from $(8, 34)$ is $d_2 = \frac{|12(8) - 34 - 31|}{\sqrt{12^2 + (-1)^2}} = \frac{|96 - 65|}{\sqrt{145}} = \frac{31}{\sqrt{145}}$. Since $d_1 = d_2$,the statement is True.

The equation of the line joining the point $(3, 5)$ to the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34)$.

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