Identify the point on the line 2x + 3y + 7 = 0, which is at a distance of + 3 units from (1, − 3).
Identify the point on the line 2x + 3y + 7 = 0, which is at a distance of + 3 units from (1, − 3).
- $\left(\frac{\sqrt{13}+9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)$
- $\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}-6}{\sqrt{13}}\right)$
- $\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)$
- $\left(\frac{\sqrt{13}+9}{\sqrt{13}}, \frac{3 \sqrt{13}-6}{\sqrt{13}}\right)$
Solution
Let $P(\alpha, \beta)$ be the point on the line
2x + 3y + 7 = 0
$\begin{array}{rlrl} & \Rightarrow & 2 \alpha+3 \beta+7 & =0 \\ & \beta & =\left(\frac{-7-2 \alpha}{3}\right) \\ \therefore & & P & =\left(\alpha, \frac{-7-2 \alpha}{3}\right)\end{array}$
Given, point A = (1, − 3)
$\begin{aligned} & \because \quad A P=3 \\ & \quad \sqrt{(\alpha-1)^2+\left(\frac{-7-2 \alpha}{3}+3\right)^2}=3 \\ & \alpha^2+1-2 \alpha+\frac{49+4 \alpha^2+28 \alpha}{9}+9-14-4 \alpha=9 \\ & 9 \alpha^2+9-18 \alpha+49+4 \alpha^2+28 \alpha-126-36 \alpha=0\end{aligned}$
$\begin{aligned} & 13 \alpha^2-26 \alpha-68=0 \\ & 13 \alpha^2-26 \alpha-68=0 \\ & \alpha=\frac{26 \pm \sqrt{676+3536}}{26} \\ & \alpha=1 \pm \frac{9}{\sqrt{13}}=\frac{\sqrt{13} \pm 9}{\sqrt{13}}\end{aligned}$
When, $\alpha=\frac{\sqrt{13}-9}{\sqrt{13}}, \beta=\frac{-7-2 \alpha}{3}$
$\begin{aligned} & =\frac{-3 \sqrt{13}+6}{\sqrt{13}} \\ \therefore \quad P & =\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3 \sqrt{13}+6}{\sqrt{13}}\right)\end{aligned}$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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