A line,with a slope greater than 1,passes thr | vedclass

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(C) Let the point $B$ be $(x_1, y_1)$. Since $B$ lies on $x - y - 2 = 0$,we have $y_1 = x_1 - 2$,so $B = (x_1, x_1 - 2)$.The distance $AB = \sqrt{(x_1

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$x + 2y = 6$

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(C) Let the point $B$ be $(x_1, y_1)$. Since $B$ lies on $x - y - 2 = 0$,we have $y_1 = x_1 - 2$,so $B = (x_1, x_1 - 2)$.The distance $AB = \sqrt{(x_1 - 4)^2 + (x_1 - 2 - 3)^2} = \frac{\sqrt{29}}{3}$.Squaring both sides: $(x_1 - 4)^2 + (x_1 - 5)^2 = \frac{29}{9}$.$x_1^2 - 8x_1 + 16 + x_1^2 - 10x_1 + 25 = \frac{29}{9} \implies 2x_1^2 - 18x_1 + 41 = \frac{29}{9}$.$18x_1^2 - 162x_1 + 369 = 29 \implies 18x_1^2 - 162x_1 + 340 = 0 \implies 9x_1^2 - 81x_1 + 170 = 0$.Solving the quadratic equation: $(3x_1 - 10)(3x_1 - 17) = 0$,so $x_1 = \frac{10}{3}$ or $x_1 = \frac{17}{3}$.If $x_1 = \frac{10}{3}$,then $y_1 = \frac{10}{3} - 2 = \frac{4}{3}$. The slope $m = \frac{4/3 - 3}{10/3 - 4} = \frac{-5/3}{-2/3} = 2.5 > 1$.If $x_1 = \frac{17}{3}$,then $y_1 = \frac{17}{3} - 2 = \frac{11}{3}$. The slope $m = \frac{11/3 - 3}{17/3 - 4} = \frac{2/3}{5/3} = 0.4 Since the slope must be greater than $1$,we take $B = (\frac{10}{3}, \frac{4}{3})$.Checking the options for $(\frac{10}{3}, \frac{4}{3})$: $x + 2y = \frac{10}{3} + 2(\frac{4}{3}) = \frac{10+8}{3} = 6$. Thus,$B$ lies on $x + 2y = 6$.

$A$ line,with a slope greater than $1$,passes through the point $A(4, 3)$ and intersects the line $x - y - 2 = 0$ at the point $B$. If the length of the line segment $AB$ is $\frac{\sqrt{29}}{3}$,then $B$ also lies on the line:

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