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(A) Step $1$: Find the point of intersection of the lines $2x - 3y = -5$ and $3x + 4y = 0$. Multiplying the first equation by $4$ and the second by $3

Vedclass · Accepted Answer

$\frac{130}{17\sqrt{29}}$

Vedclass · Answer

(A) Step $1$: Find the point of intersection of the lines $2x - 3y = -5$ and $3x + 4y = 0$. Multiplying the first equation by $4$ and the second by $3$: $8x - 12y = -20$ $9x + 12y = 0$ Adding these,we get $17x = -20$,so $x = -\frac{20}{17}$. Substituting $x$ into $3x + 4y = 0$: $3(-\frac{20}{17}) + 4y = 0 \implies 4y = \frac{60}{17} \implies y = \frac{15}{17}$. The point of intersection is $(-\frac{20}{17}, \frac{15}{17})$. Step $2$: Calculate the perpendicular distance from the point $(x_1, y_1) = (-\frac{20}{17}, \frac{15}{17})$ to the line $Ax + By + C = 0$ where $5x - 2y = 0$. The formula is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. $d = \frac{|5(-\frac{20}{17}) - 2(\frac{15}{17})|}{\sqrt{5^2 + (-2)^2}} = \frac{|-\frac{100}{17} - \frac{30}{17}|}{\sqrt{25 + 4}} = \frac{|-\frac{130}{17}|}{\sqrt{29}} = \frac{130}{17\sqrt{29}}$.

The distance of the point of intersection of the lines $2x - 3y + 5 = 0$ and $3x + 4y = 0$ from the line $5x - 2y = 0$ is

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