The coordinates of the points on the line 2x | vedclass

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(A) Let the point be $P(x_1, y_1)$. Since $P$ lies on $2x - y = 5$,we have $y_1 = 2x_1 - 5$. The distance of $P(x_1, y_1)$ from the line $3x + 4y - 5

Vedclass · Accepted Answer

$\left(\frac{30}{11}, \frac{5}{11}\right), \left(\frac{20}{11}, \frac{-15}{11}\right)$

Vedclass · Answer

(A) Let the point be $P(x_1, y_1)$. Since $P$ lies on $2x - y = 5$,we have $y_1 = 2x_1 - 5$. The distance of $P(x_1, y_1)$ from the line $3x + 4y - 5 = 0$ is given by $d = \left|\frac{3x_1 + 4y_1 - 5}{\sqrt{3^2 + 4^2}}\right| = 1$. Substituting $y_1 = 2x_1 - 5$ into the distance formula: $\left|\frac{3x_1 + 4(2x_1 - 5) - 5}{5}\right| = 1$ $\left|\frac{3x_1 + 8x_1 - 20 - 5}{5}\right| = 1$ $\left|\frac{11x_1 - 25}{5}\right| = 1$ $11x_1 - 25 = 5$ or $11x_1 - 25 = -5$ Case $1$: $11x_1 = 30 \implies x_1 = \frac{30}{11}$. Then $y_1 = 2(\frac{30}{11}) - 5 = \frac{60-55}{11} = \frac{5}{11}$. Case $2$: $11x_1 = 20 \implies x_1 = \frac{20}{11}$. Then $y_1 = 2(\frac{20}{11}) - 5 = \frac{40-55}{11} = \frac{-15}{11}$. Thus,the points are $\left(\frac{30}{11}, \frac{5}{11}\right)$ and $\left(\frac{20}{11}, \frac{-15}{11}\right)$.

The coordinates of the points on the line $2x - y = 5$ which are at a distance of $1$ unit from the line $3x + 4y = 5$ are:

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