MA005 Calculus 1 Coaching Session 1.2.1 # 11

ma005

#1

Hi this is Swapnila! I am a Student Success Coach and will be explaining some practice problems from the MA005 Calculus course. Today we will review an assigned problem from Unit 1. In the “Homework Assessment” you were asked to solve this problem:

Let P = (2 ,3) and Q = (8 ,11) . Verify that if 0 ≤ a ≤ 1, then the point
R = (x,y) with x = 2a + 8(1 – a) and y = 3a + 11(1 – a) is on the line from P to Q and image

We will be approaching this problem as follows:

  1. Find the slope of the line passing through P and Q.

  2. Find equation of a line PQ.

  3. Check whether the given coordinates x and y satisfy the equation of the line PQ found in step 2.

  4. Use distance formula to verify the distance relation.

Step 1: Find the slope of the line passing through P and Q:
image
image
image
By simplifying the numerator and denominator,
image
Therefore,
image

Step 2: Find the equation of line PQ:
Consider the point slope form of the equation of a line: image whereimage be any point on the line.
Substitute image = (2,3) and slope image in the above equation:
image

To get rid of fraction, multiply both sides by 6:
6(y - 3) = 8 (x - 2)
By using distributive property, it gives
6y - 18 = 8x - 16
Subtract 8x from both sides; it gives
6y - 18 - 8x = 8x -16 - 8x
Simplifying,
6y - 18 - 8x = -16
Add 18 on both sides; it becomes
6y - 18 - 8x +18 = -16 +18
Simplifying,
6y - 8x = 2
Therefore, the equation of the line passing through points P = (2, 3) and Q = (8, 11) is
image
Or 6y - 8x = 2

Step 3: Check whether the given coordinates x and y satisfy the equation of the line PQ found in step 2.
Now consider the point R = (x, y) where x = 2a + 8(1 – a) and y = 3a + 11(1 – a)
To check if this point lies on line PQ, substitute the values of x and y in the equation of the line PQ found in step 2.
6y - 8x = 2 …………equation of the line
Replace x by 2a + 8(1 – a) and y by 3a + 11(1 – a) in this equation:
6 (3a + 11(1 – a)) - 8 (2a + 8(1 – a)) = 2
By using distributive property on the left side of the equation,
6(3a + 11 - 11a) - 8(2a + 8 - 8a) = 2
Simplifying the left side of the equation,
6(11 - 8a) - 8(8 - 6a) = 2
Apply distributive property on the left side of the equation:
66 - 48a - 64 + 48a = 2
Simplifying,
66 - 64 = 2
2 = 2 ………for every value of a
It means the point R(x, y) satisfies the equation of the line, and therefore, it lies in line PQ.

Step 4: Use distance formula to verify the distance relation.
Formula to find distance between two points image is image
Find distance from P to R by using distance formula: image
Substitute image and image in the distance formula:
image
Simplifying,
image
image
image
image
Factor out the greatest common factor; it becomes
image
Simplifying,
image
image
By taking square root,
D= 10|1 - a| ………. since the distance can never be negative
D= 10(1 - a) ………since 0 ≤ a ≤ 1,|1 - a| = (1 - a)
Thus, the Dist(PR) = 10(1 - a) ………………… (i)
Similarly, find distance from P to Q by using the distance formula:
image
Substitute image and image in the distance formula; it becomes
image
Simplifying,
image
image
image

By taking square root,
D = 10 ……………………………. since the distance can never be negative
Thus, the Dist(PQ) = 10 ………………… (ii)
From (i) and (ii),
The distance from P to R equals the distance from P to Q times (1 - a), that is, image

Conclusion:
If 0 ≤ a ≤ 1, then the point R = (x,y) with x = 2a + 8(1 – a) and y =3a + 11(1 – a) is on the line 6y-8x=2 from P = (2 ,3) to Q = (8 ,11) and image

In addition to this, I am also attaching the pdf document of the practice problem for your ready reference.
Question 11.pdf (282.0 KB)

Please let me know if you have any question on this problem, or on this topic generally. I will be here in the forum for the next hour.