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

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

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$5: 8$

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(B) Let the line $AB$ pass through $A(2,-1)$ and $B(6,5)$. The slope of $AB$ 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 line $PD$ is perpendicular to $AB$ and passes through $P(4,1)$. The slope of $PD$ is $-\frac{1}{m} = -\frac{2}{3}$.The equation of $PD$ 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 $\left(\frac{6k+2}{k+1}, \frac{5k-1}{k+1}\right)$.Since $D$ lies on $PD$ $(2x + 3y - 11 = 0)$,we have $2\left(\frac{6k+2}{k+1}\right) + 3\left(\frac{5k-1}{k+1}\right) - 11 = 0$.$12k + 4 + 15k - 3 - 11(k + 1) = 0$.$27k + 1 - 11k - 11 = 0$.$16k - 10 = 0 \Rightarrow k = \frac{10}{16} = \frac{5}{8}$.Thus,the ratio is $5:8$.

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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