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2 years ago

Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0

Mathematics

1 answer:

wlad13 [49]2 years ago

3 0

Answer:

+6xy+9y

+4x+12y−5=0

Step-by-step explanation:

+6xy+9y

+4x+12y−5=0

Comparing the equation with the general equation of second degree gives

a=1,b=9,h=3,g=2,f=6,c=−5

Angle between a pair of straight lines that is tanθ=

∣

a+b

−ab

∣

tanθ=

∣

1+9

9−1×9

∣

tanθ=0

⇒θ=tan

−

(0)=0

∘

Angle between the pair of straight lines is zero therefore the lines are parallel.

Hence proved

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Answer:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Are AB and CD parallel, perpendicular, or neither?*<br> 1. A(-4, 8), B(4, 10), C(1, 1), D(-2, 13)

ohaa [14]

Answer:

Perpendicular

<em />

Step-by-step explanation:

Given

$A = (-4, 8)$

$B = (4, 10)$

$C = (1, 1)$

$D = (-2, 13)$

Required

Is AB and CD parallel?

First, we need to calculate the slope (m) of AB and CD

$m = \frac{y_2 - y_1}{x_2 - x_1}$

For AB:

$A (x_1,y_1)= (-4, 8)$

$B(x_2,y_2) = (4, 10)$

So:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{10 - 8}{4 - (-4)}$

$m = \frac{10 - 8}{4 +4}$

$m = \frac{2}{8}$

$m = \frac{1}{4}$

For CD:

$C(x_1,y_1) = (1, 1)$

$D(x_2,y_2) = (-2, 13)$

So:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{13 - 1}{-2 - 1}$

$m = \frac{12}{-3}$

$m = -4$

For Lines to be parallel, the slope must be equal:

i.e. $m_1 = m_2$

This condition is not true because:

$\frac{1}{4} \neq -4$

For Lines to be perpendicular, the slope must be:

$m_1 = -\frac{1}{m_2}$

This implies:

$\frac{1}{4} = -\frac{1}{-4}$

$\frac{1}{4} = \frac{-1}{-4}$

$\frac{1}{4} = \frac{1}{4}$

<em>Hence, the lines are perpendicular</em>

6 0

3 years ago

Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0​

Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0