Find-two-possible-values-of-p-if-the-lines-px-y-0-and-3x-y-1-0-intersect-at-45-

Question Number 151513 by pete last updated on 21/Aug/21

Find two possible values of p if the lines

p x - y = 0 and 3 x + y + 1 = 0 intersect at 45 °

Answered by Olaf_Thorendsen last updated on 21/Aug/21

Δ_{1} : p x - y = 0

Δ_{2} : 3 x + y + 1 = 0

{\vec{u}}_{Δ_{1}} = (\begin{matrix} 1 \\ p \end{matrix})

{\vec{u}}_{Δ_{2}} = (\begin{matrix} 1 \\ - 3 \end{matrix})

{\vec{u}}_{Δ_{1}} ∙ {\vec{u}}_{Δ_{2}} = ∣∣ {\vec{u}}_{Δ_{1}} ∣∣ \times ∣∣ {\vec{u}}_{Δ_{2}} ∣∣ \times \cos (\hat{{\vec{u}}_{Δ_{1}}, {\vec{u}}_{Δ_{2}}})

(1) (1) + p (- 3) = \sqrt{1 + p^{2}} \times \sqrt{10} \times \frac{1}{\sqrt{2}}

\sqrt{1 + p^{2}} = \frac{1}{\sqrt{5}} (1 - 3 p)

\Rightarrow 1 + p^{2} = \frac{1}{5} (9 p^{2} - 6 p + 1)

2 p^{2} - 3 p - 2 = 0

2 (p - 2) (p + \frac{1}{2}) = 0

p = - \frac{1}{2} or p = 2

Commented by pete last updated on 21/Aug/21

thank you sir

Answered by nadovic last updated on 21/Aug/21

l_{1} : p x - y = 0

g r a d i e n t, m_{1} = p

l_{2} : 3 x + y + 1 = 0

g r a d i e n t, m_{2} = - 3

\Rightarrow ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ = \tan 45 °

\frac{p - (- 3)}{1 + p (- 3)} = \pm 1

p + 3 = \pm 1 \mp 3 p

4 p = - 2

p = - \frac{1}{2}

OR

- 2 p = - 4

p = 2

Commented by bramlexs22 last updated on 21/Aug/21

it should be \tan 45 ° =∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣

it should be \tan 45 ° =∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣

Commented by nadovic last updated on 21/Aug/21

A l r i g h t S i r . T h a n k y o u!

A l r i g h t S i r . T h a n k y o u!

Commented by pete last updated on 21/Aug/21

thanks very much sir

thanks very much sir

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