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2 years ago

Find the equation of the line which is parallel to y = 4x + 3 and goes through the point (4, 5)

Mathematics

1 answer:

skad [1K]2 years ago

7 0

Answer:

y = 4x - 11

Step-by-step explanation:

The given line has a slope m = 4.

We know parallel lines have equal slopes.

<u>Use point - slope equation and substitute the coordinates of the given point to get:</u>

y - 5 = 4(x - 4)
y - 5 = 4x - 16
y = 4x - 16 + 5
y = 4x - 11

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See Below.

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We want to verify the equation:

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To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

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Hence verified.

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2 years ago

Find the equation of the line which is parallel to y = 4x + 3 and goes through the point (4, 5)​

Find the equation of the line which is parallel to y = 4x + 3 and goes through the point (4, 5)