If A(2, -1) and B(6, 5) are two points,the ra | vedclass

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(B) Let $A = (2, -1)$ and $B = (6, 5)$. The slope of line $AB$ is $m = \frac{5 - (-1)}{6 - 2} = \frac{6}{4} = \frac{3}{2}$.The equation of line $AB$ i

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$5: 8$

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(B) Let $A = (2, -1)$ and $B = (6, 5)$. The slope of line $AB$ is $m = \frac{5 - (-1)}{6 - 2} = \frac{6}{4} = \frac{3}{2}$.The equation of line $AB$ is $y - (-1) = \frac{3}{2}(x - 2)$,which simplifies to $2y + 2 = 3x - 6$,or $3x - 2y - 8 = 0$.The slope of the perpendicular line $PD$ is $m' = -\frac{1}{m} = -\frac{2}{3}$.The equation of line $PD$ passing through $P(4, 1)$ is $y - 1 = -\frac{2}{3}(x - 4)$,which simplifies to $3y - 3 = -2x + 8$,or $2x + 3y - 11 = 0$.Let $D$ divide $AB$ in the ratio $k:1$. The coordinates of $D$ are $\left(\frac{6k + 2}{k + 1}, \frac{5k - 1}{k + 1}ight)$.Since $D$ lies on $3x - 2y - 8 = 0$,we have $3\left(\frac{6k + 2}{k + 1}ight) - 2\left(\frac{5k - 1}{k + 1}ight) - 8 = 0$.$18k + 6 - 10k + 2 - 8(k + 1) = 0$.$8k + 8 - 8k - 8 = 0$,which is $0=0$. This approach is for finding $D$. Alternatively,for a line $ax + by + c = 0$,the ratio in which the foot of the perpendicular from $(x_1, y_1)$ divides the segment joining $(x_2, y_2)$ and $(x_3, y_3)$ is $-\frac{ax_2 + by_2 + c}{ax_3 + by_3 + c}$.Ratio $= -\frac{3(2) - 2(-1) - 8}{3(6) - 2(5) - 8} = -\frac{6 + 2 - 8}{18 - 10 - 8} = -\frac{0}{0}$.Using the formula for the ratio in which the foot of the perpendicular from $(x_0, y_0)$ divides the line segment $AB$ with $A(x_1, y_1)$ and $B(x_2, y_2)$ is $-\frac{m_1}{m_2}$ where $m_1$ is the slope of $AB$ and $m_2$ is the slope of $PD$. Actually,the ratio is given by $-\frac{(x_1-x_0)(x_2-x_1) + (y_1-y_0)(y_2-y_1)}{(x_2-x_1)^2 + (y_2-y_1)^2}$.Ratio $= -\frac{(2-4)(6-2) + (-1-1)(5-(-1))}{(6-2)^2 + (5-(-1))^2} = -\frac{(-2)(4) + (-2)(6)}{4^2 + 6^2} = -\frac{-8 - 12}{16 + 36} = \frac{20}{52} = \frac{5}{13}$.Re-evaluating the provided solution method: The ratio is $-\frac{L(A)}{L(B)}$ where $L(x,y) = 0$ is the line $AB$. $L(x,y) = 3x - 2y - 8 = 0$.Ratio $= -\frac{3(2) - 2(-1) - 8}{3(6) - 2(5) - 8} = -\frac{0}{0}$. The point $P$ lies on the line $AB$ because $3(4) - 2(1) - 8 = 12 - 2 - 8 = 2 
eq 0$. Wait,$3(4) - 2(1) - 8 = 2$. The ratio is $-\frac{3(2) - 2(-1) - 8}{3(6) - 2(5) - 8} = -\frac{0}{0}$ is incorrect. The ratio is $-\frac{m_{AB}}{m_{PD}} 	imes \dots$ actually,the ratio is $5:8$ as per standard calculation.

If $A(2, -1)$ and $B(6, 5)$ are two points,the ratio in which the foot of the perpendicular from $P(4, 1)$ to $AB$ divides $AB$ is:

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