Suppose that R is at (8, 6). Let P be a point on the line 8y = 15x and Q be a point on the line 10y = 3x, and suppose that R is the midpoint of PQ. Then the length of PQ can be written as a / b, where a and b have no common factors other than 1. Compute a + b.
Point p a point on the line
8y=15x
Well assuming x=m
8y=15m
y=15m/8
Co-ordinate of p=(m,15m/8)
And Q a point on the line
10y=3x
Y=3n/10
Asumming x=n
Coordinate of Q (n,3n/10)
Mid point through p&q=(8,6)
((M+n)/2,((75m+12n)/80)=(8,6)
(m+n)/2=8
m+n=16.....(1)
75m+12n=80×6
3(25m+3n)=80×3
25m+4n=160......(2)
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m+n=16.....(1)
25m+4n=160......(2)
n=16-m.......(3)
Plug 3 into 2
25m+4(16-m)=160
25m+64-4m=160
21m=96
m=96/21=32/7
N=16-32/7=(112-32)/7=80/7
Point p set it as (32/7,480/7)
And Q set up as (80/7,24/7)
Distance from p to q=√[(x2-x1)²+(y2-y1)²)]
PQ=√[(80/7-32/7)²+(24/7-480/7)²]=√(48)²+(456)²]/7=√(2304+207936)/7=√(210240)/7=(45852)/100
a/b=45852/700
a:b=45652:700=11463:175
a=11463+175=11638
That what I could think off
Correction
at
75m+12n=80×6
3(25m+4n)=80×2
25m+4n=160
typing this wasn't easy
a+b=11463+175=11638
To solve this problem, we need to find the coordinates of points P and Q on the lines 8y = 15x and 10y = 3x, respectively, such that R is the midpoint of PQ. Then we can calculate the length of PQ and express it as a fraction in the form a / b, where a and b have no common factors other than 1. Finally, we will compute the sum a + b.
Step 1: Finding the equation of the line passing through point R:
Since R is at coordinates (8, 6), we can find the slope of the line passing through R by comparing the change in y-coordinates to the change in x-coordinates. The slope is given by:
slope = (change in y-coordinates) / (change in x-coordinates) = (6 - 0) / (8 - 0) = 6/8 = 3/4.
Step 2: Finding the equation of the line perpendicular to the line passing through R:
Since R is the midpoint of PQ, the line passing through R is perpendicular to the lines 8y = 15x and 10y = 3x. The slope of a line that is perpendicular to the line with slope m is given by -1/m. Therefore, the slope of the line perpendicular to the line passing through R is -1/ (3/4) = -4/3.
Step 3: Finding the equation of the line 8y = 15x:
The given line 8y = 15x can be rearranged to slope-intercept form (y = mx + b):
y = (15/8)x.
Step 4: Finding the equation of the line 10y = 3x:
The given line 10y = 3x can be rearranged to slope-intercept form:
y = (3/10)x.
Step 5: Finding the coordinates of point P:
Since point P lies on the line 8y = 15x, we can substitute the value of y from the slope-intercept form into the equation. Let's assume the x-coordinate of point P is x_p. Therefore, we have:
15x_p / 8 = (3/10)x_p.
Cross-multiplying, we get:
15x_p = (3/10)x_p * 8,
15x_p = (3/10) * 8 * x_p,
15x_p = 24 * (3/10) * x_p.
Dividing both sides by x_p gives us:
15 = 24 * (3/10),
15 = 72/10,
15 = 7.2.
This is not possible since we obtain a contradiction. Therefore, there is no point P that satisfies the conditions given in the problem.
Step 6: Conclusion
Since we couldn't find a point P that satisfies the conditions, it means that the problem is not solvable. The length of PQ is undefined, and therefore we cannot compute a + b.