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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]  24 May 2019, 08:04
Expert's post
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 [ 18.98 KiB | Viewed 1584 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 $$7y+6x=30$$

Math Review
Question: 17
Page: 245
Difficulty: medium

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Re: In the coordinate system below, ﬁnd the following. (a) Coor [#permalink]  25 May 2019, 09:20
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Expert's post
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 $$7y+6x=30$$

(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 .

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

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

(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 $$y = -\frac{6}{7}x+\frac{30}{7}$$

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

Great.

regards
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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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