Use slope to determine if lines PQ and RS are parallel, perpendicular, or neither. P(-4,17),Q (1,-3), R(-9,3),S(-5,4)

Mathematics

1 answer:

vitfil [10]3 years ago

3 0

Answer:

Both lines are perpendicular

Step-by-step explanation:

We want to determine the relationship between the two lines.

The best thing to do is to calculate the slope of the two lines;

Mathematically;

Slope = y2-y1/x2-x1

For PQ; slope will be;

(-3-17)/(1-(-4) =-20/5 = -4

For RS

slope = (4-3)/(-5-(-9)) = 1/4

Now we have -4 and 1/4 as slopes

The product of both is 1/4 * -4 = -1

When two lines are perpendicular, the product of their slopes is -1

We can conclude that both lines are perpendicular to one another

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2 years ago

I can't figure this out

andriy [413]

Suppose that equation of parabola is
y =ax² + bx + c

Since parabola passes through the point (2,−15) then
−15 = 4a + 2b + c

Since parabola passes through the point (-5,-29), then
−29 = 25a − 5b + c

Since parabola passes through the point (−3,−5), then
−5 = 9a − 3b + c

Thus, we obtained following system:
4a + 2b + c = −15
25a − 5b + c = −29
9a − 3b + c = −5

Solving it we get that
a = −2, b = −4, c = 1

Thus, equation of parabola is
y = −2x²− 4x + 1

____________________

Rewriting in the form of
(x - h)² = 4p(y - k)

i) -2x² - 4x + 1 = y

ii) -3x² - 7x = y - 11
(-3x² and -7x are isolated)

iii) -3x² - 7x - 49/36 = y - 1 - 49/36
(Adding -49/36 to both sides to get perfect square on LHS)

iv) -3(x² + 7/3x + 49/36) = y - 3
(Taking out -3 common from LHS)

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vi) (x + 7/6)² = -⅓(y - 445/36)
(Shifting -⅓ to RHS)

vii) (x + 1)² = 4(-1/12)(y - 445/36)
(Rewriting in the form of 4(-1/12) ; This is 4p)

So, after rewriting the equation would be -

(x + 7/6)² = 4(-⅛)(y - 445/36)

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I hope this is what you wanted.

Regards,
Divyanka♪
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