Find the equation of the line through the point (3, 5) that cuts off the least area from the first quadrant?
y-A=B(x-C)
i know its slope is -3/5 which is B. But what are the values of A and C?
How did you get the slope to be -3/5 ??
let the x-intercept be A(a,0) and the y-intercept be B(0,b)
label point (3,5) as P
By ratios in similar triangles:
(a-3)/5 = 3/(b-5)
ab-5a-3b+5 = 15
b(a-3) = 5a
b = 5a/(a-3)
Area of triangle = (1/2)ab
= (1/2)(a)(5a/(a-3) )
= 5a^2/(2a-6)
d(Area)/da = ( (2a-6)(10a) - 5a^2 (2))/(2a-6)^2
= (20a^2 - 60a - 10a^2)/(2a-6)^2
= 0 for a min of area
10a^2 - 60a = 0
a(10a - 60) = 0
a = 0 or a = 6
if a = 6, the point A is (6,0) and the slope of the line is
(0-5)/(6-3) = -5/3
equation:
y-5 = (-5/3)(x-3)
y = (-5/3)x + 10
check: according to my equation the y-intercept should be B(0,10)
from my b = 5a/(a-3)
b = 30/(3) = 10
looks good
Yes i did exactly the same but website is not accepting my answer, not sure why
You did not do exactly the same, your slope was -3/5, different from mine.
(another example of why I prefer traditional class room instructions, where the learning of math is not just "answer based" )
To find the values of A and C in the equation of the line, we need to consider the point (3, 5). Let's break down the steps:
Step 1: Determine the slope
You correctly stated that the slope of the line is -3/5. This can be denoted as B.
Step 2: Determine the y-intercept
Since the line passes through the point (3, 5), we can substitute the coordinates into the equation of a line to solve for the y-intercept (A).
The equation becomes: 5 - A = (-3/5)(3 - C)
Step 3: Determine the x-intercept
Next, we need to determine the x-intercept or the value of C. To do this, we consider that the line cuts off the least area from the first quadrant. This means the line must intersect the x-axis at a point with the smallest x-coordinate in the first quadrant.
Since the x-axis has a y-coordinate of 0, we substitute y = 0 into the equation of the line:
0 - A = (-3/5)(x - C)
Simplifying further, we have: -A = (-3/5)(x - C)
Step 4: Find the point of intersection
We can substitute the x-coordinate of the point of intersection on the x-axis as (C, 0) into the equation from Step 3:
-0 = (-3/5)(C - C)
Simplifying, we get: 0 = 0
Step 5: Solve for A
From Step 4, we obtain a trivial equation, indicating that A can be any value. This means that A is a free parameter.
Step 6: Write the equation of the line
Now, we can write the equation of the line in the form y - A = B(x - C). Substituting the known values:
y - A = (-3/5)(x - C)
Since A can be any value, we can rewrite the equation as:
y = (-3/5)(x - C) + A
And that's the equation of the line passing through the point (3, 5) that cuts off the least area from the first quadrant.