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
We will be approaching this problem as follows:

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

Find equation of a line PQ.

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

Use distance formula to verify the distance relation.
Step 1: Find the slope of the line passing through P and Q:
By simplifying the numerator and denominator,
Therefore,
Step 2: Find the equation of line PQ:
Consider the point slope form of the equation of a line: where be any point on the line.
Substitute = (2,3) and slope in the above equation:
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
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 is
Find distance from P to R by using distance formula:
Substitute and in the distance formula:
Simplifying,
Factor out the greatest common factor; it becomes
Simplifying,
By taking square root,
D= 101  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:
Substitute and in the distance formula; it becomes
Simplifying,
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,
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 6y8x=2 from P = (2 ,3) to Q = (8 ,11) and
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.