The equation of line perpendicular to 3 x -5 y -19 and passing through 3-2 isA- 5 x-3 y-9-0B- 5 x-3 y-9-0- 5 x -3 y -9-0D- 5 x -3 y -9-0

The correct option is A 5x 3y 9 = 0Given line nbsp;3x+5y= 19 Line perpendicular to this 5x 3y+λ = 0 This line passing through (3, 2), so 15 6+λ = 0 ⇒λ = 9 ...

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Standard XII

Mathematics

Perpendicular Form of a Straight Line

The equation ...

Question

The equation of line perpendicular to $3 x + 5 y = 19$ and passing through $(3, 2)$ is

5 x + 3 y + 9 = 0

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5 x - 3 y - 9 = 0

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5 x - 3 y + 9 = 0

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5 x + 3 y - 9 = 0

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Solution

The correct option is B $5 x - 3 y - 9 = 0$
Given line $3 x + 5 y = 19$
Line perpendicular to this
$5 x - 3 y + λ = 0$
This line passing through $(3, 2)$ , so
$15 - 6 + λ = 0 \Rightarrow λ = - 9$
Now, the required line is
$5 x - 3 y - 9 = 0$

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Similar questions

The equation of the line passing through $(1, 5)$ and perpendicular to the line $3 x - 5 y + 7 = 0$ is

Q. The equation of the line passing through (1, 5) and perpendicular to the line 3x − 5y + 7 = 0 is
(a) 5x + 3y − 20 = 0
(b) 3x − 5y + 7 = 0
(c) 3x − 5y + 6 = 0
(d) 5x + 3y + 7 = 0

Q. If in a parallelogram

A B D C

, the coordinates of

A, B

and

C

are respectively

(1, 2), (3, 4)

and

(2, 5)

, then the equation of the diagonal

A D

is :

\frac{3}{x} - \frac{1}{y} + 9 = 0, \frac{2}{x} + \frac{3}{y} = 5 (x \neq 0, y \neq 0 .)

Q. Find the equation of line through the intersection of

5 x - 3 y = 1

and

2 x + 3 y = 23

and perpendicular to the line

5 x - 3 y - 1 = 0

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Perpendicular Form of a Straight Line

Standard XII Mathematics

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The equation of line perpendicular to 3x+5y=19 and passing through (3,2) is

The correct option is B 5x−3y−9=0Given line 3x+5y=19 Line perpendicular to this 5x−3y+λ=0 This line passing through (3,2), so 15−6+λ=0⇒λ=−9 Now, the required line is 5x−3y−9=0

The equation of line perpendicular to $3 x + 5 y = 19$ and passing through $(3, 2)$ is

The correct option is B $5 x - 3 y - 9 = 0$
Given line $3 x + 5 y = 19$
Line perpendicular to this
$5 x - 3 y + λ = 0$
This line passing through $(3, 2)$ , so
$15 - 6 + λ = 0 \Rightarrow λ = - 9$
Now, the required line is
$5 x - 3 y - 9 = 0$