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Line k passes through the points (6,2) and P and has a slope of − 3/5

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Line k passes through the points (6,2) and P and has a slope of − 3/5  [#permalink]

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New post 22 Jul 2016, 12:16
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Line k passes through the points (6,2) and P and has a slope of \(\frac{−3}{5}\) . If the line that passes through the origin and point P has a slope of –2, which of the following are the xy-coordinates for point P ?

A. (\(\frac{-40}{7}\),\(\frac{80}{7}\))
B. (-4,8)
C. (-3,6)
D. (\(\frac{11}{5}\),\(\frac{-22}{5}\))
E. (\(\frac{28}{13}\),\(\frac{-56}{13}\))

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Line k passes through the points (6,2) and P and has a slope of − 3/5  [#permalink]

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New post 22 Jul 2016, 13:17
AbdurRakib wrote:
Line k passes through the points (6,2) and P and has a slope of \(\frac{−3}{5}\) . If the line that passes through the origin and point P has a slope of –2, which of the following are the xy-coordinates for point P ?

A. (\(\frac{-40}{7}\),\(\frac{80}{7}\))
B. (-4,8)
C. (-3,6)
D. (\(\frac{11}{5}\),\(\frac{-22}{5}\))
E. (\(\frac{28}{13}\),\(\frac{-56}{13}\))



This is how I tried.

Given Line k passes through the points (6,2) and P and has a slope of \(\frac{−3}{5}\).

Let P be (x2,y2).

slope between (6,2) and (x2,y2) is -3/5.

=> \(\frac{y2-2}{x2-6}\) = \(\frac{-3}{5}\). (Here used slope formula )
=> 3x2 + 5y2 -28 = 0 ---eq 1.

Then given line that passes through the origin and point P has a slope of –2.

Again same point P and line passes through origin and slope is -2.

\(\frac{y2-0}{x2-0}\) = -2.

y2 = -2x2 -- eq 2.

Now sub eq 2 in eq 1 we get 3x2 - 10x2 = 28
=> x2 = -4 and y2 = 8.

(-4,8)...option B is correct answer.
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Re: Line k passes through the points (6,2) and P and has a slope of − 3/5  [#permalink]

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New post 21 Feb 2019, 18:00
AbdurRakib wrote:
Line k passes through the points (6,2) and P and has a slope of \(\frac{−3}{5}\) . If the line that passes through the origin and point P has a slope of –2, which of the following are the xy-coordinates for point P ?

A. (\(\frac{-40}{7}\),\(\frac{80}{7}\))
B. (-4,8)
C. (-3,6)
D. (\(\frac{11}{5}\),\(\frac{-22}{5}\))
E. (\(\frac{28}{13}\),\(\frac{-56}{13}\))


We can let the point P be (x,y).

Since we know the slope of both lines, we have:

Slope of k:

(y - 2)/(x - 6) = -3/5

The second line passes through the origin, so its slope is:

(y - 0)/(x - 0) = -2

y/x = -2

y = -2x

Substituting, we have:

(-2x - 2)/(x - 6) = -3/5

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

-10x - 10 = -3x + 18

-7x = 28

x = -4; thus, y = 8

Answer: B.
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Re: Line k passes through the points (6,2) and P and has a slope of − 3/5   [#permalink] 21 Feb 2019, 18:00
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