If L_1 is a line passing through the point P( | vedclass

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(A) The slope of the line $3x - 4y + k = 0$ is $m = \frac{3}{4}$.Since $L_1$ is perpendicular to this line,its slope is $m_1 = -\frac{4}{3}$.The equat

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(A) The slope of the line $3x - 4y + k = 0$ is $m = \frac{3}{4}$.Since $L_1$ is perpendicular to this line,its slope is $m_1 = -\frac{4}{3}$.The equation of line $L_1$ passing through $P(4, -3)$ is $y - (-3) = -\frac{4}{3}(x - 4)$,which simplifies to $3y + 9 = -4x + 16$,or $4x + 3y = 7$.We need the distance of $P(4, -3)$ from the line $L: 5x - 3y - 2 = 0$ measured along $L_1$. This is the distance between $P(4, -3)$ and the intersection point of $L_1$ and $L$.Solving the system:$4x + 3y = 7$$5x - 3y = 2$Adding the equations: $9x = 9 \Rightarrow x = 1$.Substituting $x = 1$ into $4x + 3y = 7$: $4(1) + 3y = 7$ $\Rightarrow 3y = 3$ $\Rightarrow y = 1$.The intersection point is $(1, 1)$.The distance $d$ between $P(4, -3)$ and $(1, 1)$ is $\sqrt{(4 - 1)^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

If $L_1$ is a line passing through the point $P(4, -3)$ and perpendicular to the line $3x - 4y + k = 0$,then the distance of $P$ from the line $5x - 3y - 2 = 0$ measured along the line $L_1$ is

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