The equation of line passing through 3-4 and parallel to 5 x -9 y -12-0 isA- 5 x-9 y-51-0B- 5 x-9 y-51-0C- 5 x-9 y-51-0D- 5 x -9 y -51-0

The correct option is D 5x+9y 51= 0Given line nbsp;5x+9y +12 = 0 Line parallel to this 5x+9y+λ = 0 It passes through (3, 4), we get nbsp; 15+36 +λ = 0 ⇒λ ...

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Standard XII

Mathematics

Intercept of a Straight Line

The equation ...

Question

The equation of line passing through $(3, 4)$ and parallel to $5 x + 9 y + 12 = 0$ is

5 x - 9 y + 51 = 0

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5 x - 9 y - 51 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 x + 9 y + 51 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 x + 9 y - 51 = 0

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Solution

The correct option is D $5 x + 9 y - 51 = 0$
Given line $5 x + 9 y + 12 = 0$
Line parallel to this
$5 x + 9 y + λ = 0$
It passes through $(3, 4)$ , we get
$15 + 36 + λ = 0 \Rightarrow λ = - 51$
Therefore the required equation of line is
$5 x + 9 y - 51 = 0$

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Similar questions

Q. The equation of line passing through

(3, 4)

and parallel to

5 x + 9 y + 12 = 0

Simplify:

8 (5 x + 9 y) + 12 (5 x + 9 y)

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and

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Q. Let PS be the median of the triangle with vertices P(2, 2), Q(6, -1) and R(7, 3). The equation of the line passing through (1, -1) and parallel to PS is

Q. Let

P S

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The equation of line passing through (3,4) and parallel to 5x+9y+12=0 is

The correct option is D 5x+9y−51=0Given line 5x+9y+12=0 Line parallel to this 5x+9y+λ=0 It passes through (3,4), we get 15+36+λ=0⇒λ=−51 Therefore the required equation of line is 5x+9y−51=0

The equation of line passing through $(3, 4)$ and parallel to $5 x + 9 y + 12 = 0$ is

The correct option is D $5 x + 9 y - 51 = 0$
Given line $5 x + 9 y + 12 = 0$
Line parallel to this
$5 x + 9 y + λ = 0$
It passes through $(3, 4)$ , we get
$15 + 36 + λ = 0 \Rightarrow λ = - 51$
Therefore the required equation of line is
$5 x + 9 y - 51 = 0$