Find two possible values of p if the lines px−y=0 and 3x+y+1=0 intersect at 45°


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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

	$it should be \tan 45 ° =∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣$
	Commented by nadovic last updated on 21/Aug/21

	$A l r i g h t S i r . T h a n k y o u!$
	Commented by pete last updated on 21/Aug/21

	$thanks very much sir$
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