Question

Find the value of k so that the line y+5 = k(x-3) is numerically 3 unit distant from the origin.

Answer 1

Answer on Question #53015 – Math – Analytic Geometry

Find the value of $k$ so that the line $y + 5 = k(x - 3)$ is numerically 3 unit distant from the origin.

Solution

$$y = k(x - 3) - 5$$

Let's write the equation of a line in the normal form:

$$\begin{array}{l} Ax + By = C, \\ Ax + By - C = 0 \\ x - y - 3k - 5 = 0 \\ A = k, B = -1, C = 3k + 5 \\ \end{array}$$

Let's normalize the equation of the line dividing it by the square root of $\sqrt{A^2 + B^2}$ or multiplying by $\frac{1}{\sqrt{A^2 + B^2}}$.

$$\frac{A}{\sqrt{A^2 + B^2}} x + \frac{B}{\sqrt{A^2 + B^2}} y - \frac{C}{\sqrt{A^2 + B^2}} = 0$$ $$\sqrt{A^2 + B^2} = \sqrt{k^2 + 1}$$ $$\frac{k}{\sqrt{k^2 + 1}} x + \frac{-1}{\sqrt{k^2 + 1}} y - \frac{3k + 5}{\sqrt{k^2 + 1}} = 0$$

This normalized equation contains very important information: the coordinates of the unit normal vector $\vec{n} = \left(\frac{k}{\sqrt{k^2 + 1}}, \frac{-1}{\sqrt{k^2 + 1}}\right)$ and the distance from the origin $d = \left|\frac{3k + 5}{\sqrt{k^2 + 1}}\right|$.

According to the statement of the problem: $\left|\frac{3k + 5}{\sqrt{k^2 + 1}}\right| = 3$.

Let's solve the equation and find the value (or values) of $k$.

$$\begin{array}{l} \frac{3k + 5}{\sqrt{k^2 + 1}} = \pm 3 \\ \frac{(3k + 5)^2}{k^2 + 1} = 9 \Rightarrow \frac{9k^2 + 30k + 25 - 9k^2 - 9}{k^2 + 1} = 0 \\ \frac{30k + 16}{k^2 + 1} = 0 \Rightarrow 30k + 16 = 0 \Rightarrow k = -\frac{8}{15}. \\ \end{array}$$

So, the equation of the line $y + 5 = k(x - 3)$ which is numerically 3 unit distant from the origin is

$$y + 5 = -\frac{8}{15}(x - 3).$$

Now, let's check the answer: $d = \frac{\left|3\cdot\left(-\frac{8}{15}\right) + 5\right|}{\sqrt{\frac{64}{225} + 1}} = \frac{-\frac{8}{5} + 5}{\sqrt{\frac{289}{225}}} = \frac{\left|\frac{17}{5}\right|}{\left|\frac{17}{15}\right|} = 3.$

Answer: $k = -\frac{8}{15}$ .

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