Find the equation of line through the point of intersection of the line x+3y−11=0 and 5x−4y+2=0 and perpendicular to 4x+2y+9=0.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 109160 by john santu last updated on 21/Aug/20

$F i n d t h e e q u a t i o n o f l i n e t h r o u g h$ $t h e p o i n t o f i n t e r s e c t i o n o f t h e$ $l i n e x + 3 y - 11 = 0 a n d 5 x - 4 y + 2 = 0$ $a n d p e r p e n d i c u l a r t o 4 x + 2 y + 9 = 0 .$
	Answered by bemath last updated on 21/Aug/20

	Answered by Dwaipayan Shikari last updated on 21/Aug/20

	$4 x + 2 y + 9 = 0 \Rightarrow y = - 2 x - \frac{9}{2}$ $s l o p e (m) = - 2$ ${m m}_{0} = - 1$ $m_{0} = \frac{1}{2}$ $x + 3 y - 11 + Ψ (5 x - 4 y + 2) = 0$ $(1 + 5 Ψ) x + (3 - 4 Ψ) y + 2 Ψ - 11 = 0$ $y = \frac{1 + 5 Ψ}{4 Ψ - 3} x + 11 - 2 Ψ$ $m_{0} = \frac{1 + 5 Ψ}{4 Ψ - 3} = \frac{1}{2} \Rightarrow Ψ = - \frac{5}{6}$ $(1 - \frac{25}{6}) x + (3 + \frac{20}{6}) y - \frac{5}{3} - 11 = 0$ $x - 2 y + 4 = 0$ $y = \frac{x}{2} + 2$
	Answered by Aziztisffola last updated on 21/Aug/20

	$x + 3 y - 11 = 0 \Rightarrow y = \frac{11 - x}{3}$ $5 x - 4 y + 2 = 0 \Rightarrow y = \frac{5 x + 2}{4}$ $\Rightarrow \frac{11 - x}{3} = \frac{5 x + 2}{4} \Rightarrow x = 2 \Rightarrow y = 3$ $(x + 3 y - 11 = 0) \cap (5 x - 4 y + 2 = 0) = (2; 3)$ $4 x + 2 y + 9 = 0 \Rightarrow \vec{n} (\begin{matrix} 4 \\ 2 \end{matrix}) vector normal$ $\Rightarrow a x + b y + c = 0$ $\Rightarrow 2 x - 4 y + c = 0$ $2 \times 2 - 4 \times 3 + c = 0 \Rightarrow c = 8$ $\Rightarrow 2 x - 4 y + 8 = 0 .$ $\Rightarrow x - 2 y + 4 = 0 .$
	Answered by 1549442205PVT last updated on 22/Aug/20
	$Let the point of intersection of the line x+3y−11=0 and 5x−4y+2=0 denoted by A.Then the coordintes of A are root of the system { ((5x−4y+2=0)),((x+3y−11=0)) :}⇔ { ((5x−4y+2=0 (1))),((5x+15y−55=0 (2))) :} Subtracting (1) from (2) we get 19y−57=0⇒y=3,x=2⇒A(2;3) The line d go through A and perpendicular to the line 4x+2y+9=0 must have the direction vector p^(→) //n^(→) (4;2) ⇒p^(→) =(2;1).Therefore ,the equation of d is:((x−2)/2)=((y−3)/1)⇔x−2y+4=0 Thus,the line we need find has the equation is (d):x−2y+4=0$
	$Let t h e p o i n t o f i n t e r s e c t i o n o f t h e$ $l i n e x + 3 y - 11 = 0 a n d 5 x - 4 y + 2 = 0$ $denoted by A . Then the coordintes$ $of A are root of the system$ ${\begin{cases} 5 x - 4 y + 2 = 0 \\ x + 3 y - 11 = 0 \end{cases} \Leftrightarrow {\begin{cases} 5 x - 4 y + 2 = 0 (1) \\ 5 x + 15 y - 55 = 0 (2) \end{cases}$ $Subtracting (1) from (2) we get$ $19 y - 57 = 0 \Rightarrow y = 3, x = 2 \Rightarrow A (2; 3)$ $The line d go through A and perpendicular$ $to the line 4 x + 2 y + 9 = 0 must have$ $the direction vector \vec{p} / / \vec{n} (4; 2)$ $\Rightarrow \vec{p} = (2; 1) . Therefore, the equation of$ $d is : \frac{x - 2}{2} = \frac{y - 3}{1} \Leftrightarrow x - 2 y + 4 = 0$ $Thus, the line we need find has the$ $equation is (d) : x - 2 y + 4 = 0$
	Commented by peter frank last updated on 22/Aug/20

	$thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum