The distance of the point (1, 2) from the lin | vedclass

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(C) Let the equation of the line parallel to $3x - y = 7$ be $3x - y = \lambda$. Since it passes through $(1, 2)$,this point must satisfy the equation

Vedclass · Accepted Answer

$\sqrt{40}$

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(C) Let the equation of the line parallel to $3x - y = 7$ be $3x - y = \lambda$. Since it passes through $(1, 2)$,this point must satisfy the equation $3x - y = \lambda$. $	herefore 3(1) - 2 = \lambda \Rightarrow \lambda = 1$. So,the line is $3x - y = 1$. Now,we find the intersection of $x + y + 5 = 0$ and $3x - y = 1$. Adding the two equations: $(x + y + 5) + (3x - y) = 0 + 1$ $\Rightarrow 4x + 5 = 1$ $\Rightarrow 4x = -4$ $\Rightarrow x = -1$. Substituting $x = -1$ into $x + y + 5 = 0$: $-1 + y + 5 = 0 \Rightarrow y = -4$. So,the point of intersection is $(-1, -4)$. The required distance is the distance between $(1, 2)$ and $(-1, -4)$. Using the distance formula: $d = \sqrt{(1 - (-1))^2 + (2 - (-4))^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$.

The distance of the point $(1, 2)$ from the line $x + y + 5 = 0$ measured along the line parallel to $3x - y = 7$ is equal to

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