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The slope of the line is -10/9 and the value of y-intercept of the line is 50/3.
Slope of the line:
A slope of a line is the change in y coordinate with respect to the change in x coordinate.
Given,
The point (9,10) that cuts off the least area from the first quadrant
Here we need to find the slope and y intercept of the line.
A line is passing through a point (9,10) cuts off least area in first quadrant.
The equation of the line passing through point (9,10) and has slope m can be expressed as,
=> y - 10 = m (x - 9)
=> y = m(x - 9) + 10
=> y - mx = - 9m + 10
Divide the terms by - 9m + 10, then we get,
=> (y/- 9m+10) - (mx/ - 9m+10) = 1
=> (mx/9m+10) - (y/9m+10) = 1
It can be rewritten as,
[tex]\implies\frac{x}{\frac{9m-10}{m}} -\frac{y}{10-9m}[/tex]
We know that,
The equation of straight line in intercept form is
=> x/a + y/b = 1
So, the intercept made by the line on x-axis i.e. a and on y-axis i.e. b are:
a = 9m-10/m
b = 10 - 9m
The area of the triangle made by the line in first quadrant can be expressed as,
=> A = 1/2 x (9m-10/m) x (10 - 9m)
=> A = -1/2 (9m-10)²/m-----------------------(1)
Differentiate the equation (1) with respect to m.
[tex]\frac{dA}{dm}=-\frac{1}{2}[\frac{2m(9m-10).9-(9m-10)^2}{m^2}][/tex]
[tex]\frac{dA}{dm}=-\frac{9m-10}{2m^2}[2m.9-(9m-10)][/tex]
[tex]\frac{dA}{dm}=-\frac{9m-10}{2m^2}[18m-9m-10)][/tex]
when we simplify it,
=> dA/dm = -(9m-10)/2m² [9m-10]
Let dA/dm = 0, then
=> 0 = -1/2m² (9m-10)(9m-10)
=> 0 = 1/m² (10-9m)(9m-10)
=> 0 = (10 - 9m) (9m -10)
Therefore, the value of m is,
=> 10 - 9m = 0
=> -9m = -10
=> m = 10/9
And
=> 9m - 10 =0
=> 9m = 10
=> m = 10/9
The line will be in first quadrant and forms a triangle with axes if the slope is negative.
Substitute m = -10/9
Then, b = 10-9m
Then,
=> y = 10 - 6(-10/9)
=> y = 10 + 20/3
=> y = 50/3
So, the slope of the line is -10/9 and the value of y-intercept of the line is 50/3.
To know more about Slope of the Line here.
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