Respuesta :

Nayefx

Answer:

[tex]\rm\displaystyle \theta = \frac{\pi}{4} [/tex]

Step-by-step explanation:

we want to find the acute angle θ (as θ is between (0,π/2)) formed by the line y=-2x+4 and y=3x-3 to do so we can consider the following formula:

[tex] \displaystyle\tan( \theta) = \bigg| \frac{ m_{2} - m_{1} }{1 + m_{1} m_{2} } \bigg | [/tex]

[tex] \rm \displaystyle \implies\theta = \arctan \left( \bigg | \frac{ m_{2} - m_{1} }{1 + m_{1} m_{2} } \bigg | \right)[/tex]

From the first equation we obtain that [tex]m_1[/tex] is -2 and from the second that [tex]m_2[/tex] is 3 therefore substitute:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 3 - ( - 2) }{1 + ( - 2) (3)} \bigg | \right)[/tex]

simplify multiplication:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 3 - ( - 2) }{1 + ( - 6)} \bigg | \right)[/tex]

simplify Parentheses:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 3 + 2 }{1 + ( - 6)} \bigg | \right)[/tex]

simplify addition:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 5 }{ - 5} \bigg | \right)[/tex]

simplify division:

[tex] \rm\displaystyle \theta = \arctan \left( | - 1| \right)[/tex]

calculate the absolute of -1:

[tex]\rm\displaystyle \theta = \arctan \left( 1\right)[/tex]

calculate the inverse function:

[tex]\rm\displaystyle \theta = \frac{\pi}{4} [/tex]

hence,

the angle θ which is formed by the line y = -2x+4 and y = 3x-3 is π/4

(for more info about the formula refer the attachment thank you!)

Ver imagen Nayefx
LammettHash

Consider two vector-valued functions,

[tex]\vec r(t) = \left\langle t, -2t+4\right\rangle \text{ and } \vec s(t) = \left\langle t, 3t-3\right\rangle[/tex]

Differentiate both to get the corresponding tangent/direction vectors:

[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt} = \left\langle1,-2\right\rangle \text{ and } \dfrac{\mathrm d\vec s(t)}{\mathrm dt} = \left\langle1,3\right\rangle[/tex]

Recall the dot product identity: for two vectors [tex]\vec a[/tex] and [tex]\vec b[/tex], we have

[tex]\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)[/tex]

where [tex]\theta[/tex] is the angle between them.

We have

[tex]\langle1,-2\rangle \cdot \langle1,3\rangle = \|\langle1,-2\rangle\| \|\langle1,3\rangle\| \cos(\theta) \\\\ 1\times1 + (-2)\times3 = \sqrt{1^2 + (-2)^2} \times \sqrt{1^2+3^2} \cos(\theta) \\\\ \cos(\theta) = \dfrac{-5}{\sqrt5\times\sqrt{10}} = -\dfrac1{\sqrt2} \\\\ \implies \theta = \cos^{-1}\left(-\dfrac1{\sqrt2}\right) = \dfrac{3\pi}4[/tex]

Then the acute angle between the lines is π/4.

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