The points (2,3) and (-4,-4/3) lie on the opp | vedclass

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(D) Let $f(x, y) = 5x - 6y + k$. Since $(2, 3)$ and $(-4, -4/3)$ lie on opposite sides,$f(2, 3) \cdot f(-4, -4/3) < 0$. $f(2, 3) = 5(2) - 6(3) + k = 1

Vedclass · Accepted Answer

$\frac{11}{\sqrt{61}}$

Vedclass · Answer

(D) Let $f(x, y) = 5x - 6y + k$. Since $(2, 3)$ and $(-4, -4/3)$ lie on opposite sides,$f(2, 3) \cdot f(-4, -4/3) $f(2, 3) = 5(2) - 6(3) + k = 10 - 18 + k = k - 8$. $f(-4, -4/3) = 5(-4) - 6(-4/3) + k = -20 + 8 + k = k - 12$. So,$(k - 8)(k - 12) Since $k$ is an integer,$k \in \{9, 10, 11\}$. Now,$(1, 2)$ and $(4, 5)$ lie on the same side,so $f(1, 2) \cdot f(4, 5) > 0$. $f(1, 2) = 5(1) - 6(2) + k = k - 7$. $f(4, 5) = 5(4) - 6(5) + k = k - 10$. So,$(k - 7)(k - 10) > 0$. For $k = 9$: $(9 - 7)(9 - 10) = 2(-1) = -2 For $k = 10$: $f(4, 5) = 0$ (Points lie on the line,not same side). For $k = 11$: $(11 - 7)(11 - 10) = 4(1) = 4 > 0$ (True). Thus,$k = 11$. The line is $5x - 6y + 11 = 0$. The perpendicular distance from origin $(0, 0)$ is $d = \frac{|5(0) - 6(0) + 11|}{\sqrt{5^2 + (-6)^2}} = \frac{11}{\sqrt{25 + 36}} = \frac{11}{\sqrt{61}}$.

The points $(2,3)$ and $(-4,-4/3)$ lie on the opposite sides of the line $L \equiv 5x - 6y + k = 0$,where $k$ is an integer. If the points $(1,2)$ and $(4,5)$ lie on the same side of the line $L = 0$,then the perpendicular distance from the origin to the line $L = 0$ is:

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