The length of the chord joining the points in | vedclass

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(B) The given line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.The circle is ${x^2} + {y^2} = \frac{169}{25}$,so the

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$2$

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(B) The given line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.The circle is ${x^2} + {y^2} = \frac{169}{25}$,so the radius $r = \sqrt{\frac{169}{25}} = \frac{13}{5}$.The perpendicular distance $d$ from the center $(0, 0)$ to the line $4x + 3y - 12 = 0$ is given by $d = \frac{|4(0) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = \frac{12}{5}$.The length of the chord is $2\sqrt{r^2 - d^2} = 2\sqrt{(\frac{13}{5})^2 - (\frac{12}{5})^2} = 2\sqrt{\frac{169 - 144}{25}} = 2\sqrt{\frac{25}{25}} = 2(1) = 2$.

The length of the chord joining the points in which the straight line $\frac{x}{3} + \frac{y}{4} = 1$ cuts the circle ${x^2} + {y^2} = \frac{169}{25}$ is

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