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2 years ago

What is the equation of the line perpendicular to 2x - 3y = 13 that passes through the point (-6, 5)?

Mathematics

1 answer:

Arte-miy333 [17]2 years ago

3 0

Answer: 2y + 3x = -8.

Step-by-step explanation:

Given our first equation 2x-3y =13, we rewrite in the form of equation of a line, i.e y = mx + c

Where m is the slope of the line and c is the intersection on y axis.

By rewriting, we have...

3y= 2x-13

y= 2x/3 - 13/3.

m= 2/3, c= -13/3.

For two perpendicular lines, the product of their gradients is -1

i.e

m¹ * m² = -1

Hence m² = -1 * 3/2

m² = -3/2.

Hence, to get the equation of the line passing through the point (-6,5) we first find the intersection when x=0 given as

[Y-c]/[x-0] = -3/2

2y - 2c = -3x

2(5) - 2c =- 3(-6)

10-18= 2c

C = -8/2

C= -4

Then the equation is given as

y = mx + c

y = [-3/2]x - 4

2y= -3x - 8

2y + 3x = -8.

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Step-by-step explanation:

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Step-by-step explanation:

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VikaD [51]

Not sure why such an old question is showing up on my feed...

Anyway, let $x=\tan^{-1}\dfrac43$ and $y=\sin^{-1}\dfrac35$ . Then we want to find the exact value of $\cos(x-y)$ .

Use the angle difference identity:

$\cos(x-y)=\cos x\cos y+\sin x\sin y$

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Step-by-step explanation:

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