Identify the point on the line 2x + 3y + 7 = | vedclass

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Question

(C) Let $P(\alpha, \beta)$ be the point on the line $2x + 3y + 7 = 0$.Then $2\alpha + 3\beta + 7 = 0$,so $\beta = \frac{-7-2\alpha}{3}$.The point $P$

Vedclass · Accepted Answer

$\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)$

Vedclass · Answer

(C) Let $P(\alpha, \beta)$ be the point on the line $2x + 3y + 7 = 0$.Then $2\alpha + 3\beta + 7 = 0$,so $\beta = \frac{-7-2\alpha}{3}$.The point $P$ is $\left(\alpha, \frac{-7-2\alpha}{3}\right)$.The distance between $P(\alpha, \beta)$ and $A(1, -3)$ is $3$.Using the distance formula: $(\alpha - 1)^2 + \left(\frac{-7-2\alpha}{3} + 3\right)^2 = 3^2$.$(\alpha - 1)^2 + \left(\frac{-7-2\alpha+9}{3}\right)^2 = 9$.$(\alpha - 1)^2 + \left(\frac{2-2\alpha}{3}\right)^2 = 9$.$(\alpha - 1)^2 + \frac{4(1-\alpha)^2}{9} = 9$.$(\alpha - 1)^2 \left(1 + \frac{4}{9}\right) = 9$.$(\alpha - 1)^2 \left(\frac{13}{9}\right) = 9$.$(\alpha - 1)^2 = \frac{81}{13}$.$\alpha - 1 = \pm \frac{9}{\sqrt{13}}$.$\alpha = 1 \pm \frac{9}{\sqrt{13}} = \frac{\sqrt{13} \pm 9}{\sqrt{13}}$.For $\alpha = \frac{\sqrt{13}-9}{\sqrt{13}}$,$\beta = \frac{-7 - 2(\frac{\sqrt{13}-9}{\sqrt{13}})}{3} = \frac{-7\sqrt{13} - 2\sqrt{13} + 18}{3\sqrt{13}} = \frac{-9\sqrt{13} + 18}{3\sqrt{13}} = \frac{-3\sqrt{13} + 6}{\sqrt{13}}$.Thus,the point is $\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3\sqrt{13}+6}{\sqrt{13}}\right)$.

Identify the point on the line $2x + 3y + 7 = 0$,which is at a distance of $3$ units from $(1, -3)$.

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