A line passes through the point (-1, 1) and m | vedclass

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(B) Given the angle $	heta = \sin^{-1}(\frac{3}{5})$,we have $	an 	heta = \frac{3}{4}$. The slope of the line is $m = \frac{3}{4}$.The equation of

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$\frac{5}{4}$ unit

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(B) Given the angle $	heta = \sin^{-1}(\frac{3}{5})$,we have $	an 	heta = \frac{3}{4}$. The slope of the line is $m = \frac{3}{4}$.The equation of the line passing through $(-1, 1)$ with slope $m = \frac{3}{4}$ is $y - 1 = \frac{3}{4}(x + 1)$,which simplifies to $4y - 4 = 3x + 3$,or $4y = 3x + 7$.Substitute $4y = 3x + 7$ into the curve equation $x^2 = 4y - 9$:$x^2 = (3x + 7) - 9$$x^2 - 3x + 2 = 0$$(x - 1)(x - 2) = 0$So,$x = 1$ and $x = 2$.For $x = 1$,$4y = 3(1) + 7 = 10 \Rightarrow y = \frac{5}{2}$. Point $A = (1, \frac{5}{2})$.For $x = 2$,$4y = 3(2) + 7 = 13 \Rightarrow y = \frac{13}{4}$. Point $B = (2, \frac{13}{4})$.The distance $|AB| = \sqrt{(2 - 1)^2 + (\frac{13}{4} - \frac{5}{2})^2} = \sqrt{1^2 + (\frac{13-10}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$ units.

$A$ line passes through the point $(-1, 1)$ and makes an angle $\sin^{-1}(\frac{3}{5})$ in the positive direction of the $x$-axis. If this line meets the curve $x^2 = 4y - 9$ at points $A$ and $B$,then the length $|AB|$ is equal to

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