Question

gent to the the line 4x-3y=6 at (3,2) and passing through (2,-1)

Answer 1

Answer on Question #63800 – Math – Analytic Geometry

Question

Tangent to the line $4x - 3y = 6$ at (3,2) and passing through (2,-1).

Solution

The direction vector from (3,2) to (2,-1) is $\vec{a} = (3 - 2,2 - (-1)) = (1,3)$.

The vector $(4, -3)$ is orthogonal to the line $4x - 3y = 6$.

The vector $\vec{b} = (3,4)$ is orthogonal to the vector $(4, -3)$.

Consequently, $\vec{b} = (3,4)$ is the direction vector of the line $4x - 3y = 6$.

Let $\varphi$ be the acute angle between the lines with the direction vectors $\vec{a}$ and $\vec{b}$.

Then $0 \leq \varphi \leq \frac{\pi}{2}$ and

$$\cos \varphi = \frac{|(\vec{a}, \vec{b})|}{\sqrt{(\vec{a}, \vec{a})} \sqrt{(\vec{b}, \vec{b})}} = \frac{|1 \cdot 3 + 3 \cdot 4|}{\sqrt{1 \cdot 1 + 3 \cdot 3} \sqrt{3 \cdot 3 + 4 \cdot 4}} = \frac{3}{\sqrt{10}}.$$

Hence the required tangent is

$$\tan \varphi = \sqrt{\frac{1}{(\cos \varphi)^2} - 1} = \sqrt{\frac{10}{9} - 1} = \frac{1}{3}.$$

Answer: $\frac{1}{3}$.

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