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In the coordinate system below, find the following. (a) Coor

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In the coordinate system below, find the following. (a) Coor [#permalink] New post 24 May 2019, 08:04
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In the coordinate system below, find the following.

(a) Coordinates of point \(Q\)

(b) Lengths of \(PQ, QR,\) and \(PR\)

(c) Perimeter of \(\triangle PQR\)

(d) Area of \(\triangle PQR\)

(e) Slope, y-intercept, and equation of the line passing through points \(P\) and \(R\)

Attachment:
#GREexcercises  In the coordinate system below, find the following. (a) Coordinate.jpg
#GREexcercises In the coordinate system below, find the following. (a) Coordinate.jpg [ 18.98 KiB | Viewed 172 times ]


[Reveal] Spoiler: OA
(a) (−2,0) (b) PQ=6, QR=7, PR=\(\sqrt{85}\) (c) \(13+\sqrt{85}\) (d) 21 (e) Slope: -\(\frac{6}{7}\); y-intercept: \frac{30}{7} ; equation of line: \(y=- \frac{6}{7}x+\frac{30}{7}\), or \(7y+6x=30\)



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Question: 17
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Difficulty: medium

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Re: In the coordinate system below, find the following. (a) Coor [#permalink] New post 25 May 2019, 09:20
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Carcass wrote:
In the coordinate system below, find the following.

(a) Coordinates of point \(Q\)

(b) Lengths of \(PQ, QR,\) and \(PR\)

(c) Perimeter of \(\triangle PQR\)

(d) Area of \(\triangle PQR\)

(e) Slope, y-intercept, and equation of the line passing through points \(P\) and \(R\)

Attachment:
#GREexcercises In the coordinate system below, find the following. (a) Coordinate.jpg


[Reveal] Spoiler: OA
(a) (−2,0) (b) PQ=6, QR=7, PR=\(\sqrt{85}\) (c) \(13+\sqrt{85}\) (d) 21 (e) Slope: -\(\frac{6}{7}\); y-intercept: \frac{30}{7} ; equation of line: \(y=- \frac{6}{7}x+\frac{30}{7}\), or \(7y+6x=30\)


Image


(a) Coordinates of point \(Q\)
Since point Q is on the x-axis, it's y-coordinate must be 0.

Also, since PQ is PARALLEL to the y-axis, we know that points P and Q are the SAME distance from the y-axis.
This means the two points will have the SAME x-coordinate .
Image
So, the coordinates of point Q are (-2, 0)
----------------------------------

(b) Lengths of \(PQ, QR,\) and \(PR\)

PQ is the line segment from (-2, 0) and (-2, 6).
Length of PQ = 6-0 = 6

QR is the line segment from (-2, 0) and (5, 0).
Length of QR = 5-(-2) = 7

Finally, PR is the hypotenuse of the right triangle.
Let x = the length of PR
Apply the Pythagorean Theorem to get: 6² + 7² = x²
Evaluate: 36 + 49 = x²
So, 85 = x², which means x = √85

Image
----------------------------------

(c) Perimeter of \(\triangle PQR\)
Perimeter = sum of all 3 sides
= 6 + 7 + √85 = 13 + √85
----------------------------------

(d) Area of \(\triangle PQR\)
area = (base)(height)/2
= (7)(6)/2
= 21

----------------------------------

(e) Slope, y-intercept, and equation of the line passing through points \(P\) and \(R\)
Slope = rise/run = -6/7

Let's write the equation of the line in slope y-intercept form: y = mx + b, where m = slope and b = y-intercept
We get: \(y = -\frac{6}{7}x+b\) (we don't know the value of b yet, but we soon will)

Since the point (5, 0) lies ON the line, we know that x = 5 and y = 0 is a solution to the equation of the line.

So, plug in x = 5 and y = 0 to get: \(0 = -\frac{6}{7}(5)+b\)

Simplify: \(0 = -\frac{30}{7}+b\)

Solve: \(b = \frac{30}{7}\)

This means the y-intercept is 30/7

Finally, the equation of the line (in slope y-intercept form) is \(y = -\frac{6}{7}x+\frac{30}{7}\)
Image

Cheers,
Brent
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Re: In the coordinate system below, find the following. (a) Coor [#permalink] New post 26 May 2019, 10:11
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360° explanation.

Great.

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Re: In the coordinate system below, find the following. (a) Coor   [#permalink] 26 May 2019, 10:11
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