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 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 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