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Afina-wow [57]
4 years ago
10

Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR.

Mathematics
1 answer:
Vinvika [58]4 years ago
5 0

Answer:

2x + y - 1 = 0

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

x - 2y + 4 = 0 ( subtract x + 4 from both sides )

- 2y = - x - 4 ( divide all terms by - 2 )

y = \frac{1}{2} x + 2 ← in slope- intercept form

with slope m =  \frac{1}{2}

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{\frac{1}{2} } = - 2 , thus

y = - 2x + c ← is the partial equation of the perpendicular line

To find c substitute (1, - 1) into the partial equation

- 1 = - 2 + c ⇒ c = - 1 + 2 = 1

y = - 2x + 1 ← in slope intercept form

Subtract - 2x + 1 from both sides

2x + y - 1 = 0 ← in general form

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y=\frac{x^2-4x+3}{x^2-9}

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Firstly, we will factor numerator and denominator

we get

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We can see that (x-3) is common in both numerator and denominator

so, we will only set x+3 to 0

and then we can find vertical asymptote

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Horizontal asymptote:

We can see that degree of numerator is 2

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for finding horizontal asymptote, we find ratio of leading coefficients of numerator and denominator

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now, we can draw graph

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Answer:

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