Find the point on the line 2 x + 7 y + 4 =0 which is closest to the point ( 0, 1 )
let P be that point
slope of given line is -2/7
so slope of line from P to (0,1) is 7/2
and since (0,1) is the y-intercept,
its equation is y = (7,2)x + 1
solve the two equations to find P
use substitution, i.e.,
2x + 7((7/2)x + 1) = -4
etc
i found x value of P how do I find the y value of P?
sub it back in the original equation.
To find the point on the line closest to a given point, we can use the concept of perpendicular distance.
Let's call the point on the line we're trying to find as (x, y). The equation of the given line is 2x + 7y + 4 = 0.
To find the point on the line closest to (0, 1), we need to find the distance between these two points and minimize it. The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula:
distance = √((x₂ - x₁)² + (y₂ - y₁)²)
In this case, (x₁, y₁) = (0, 1), and (x₂, y₂) = (x, y).
So, the distance between these two points is:
distance = √((x - 0)² + (y - 1)²) = √(x² + (y - 1)²)
To find the point on the line closest to (0, 1), we want to minimize this distance.
To do this, we can substitute the value of y from the equation of the given line into the distance formula. Let's rewrite the given equation in terms of y:
2x + 7y + 4 = 0
7y = -2x - 4
y = (-2x - 4) / 7
Now, substitute this value of y into the distance formula:
distance = √(x² + ((-2x - 4) / 7 - 1)²)
Next, square both sides of the equation to eliminate the square root:
distance² = x² + ((-2x - 4) / 7 - 1)²
We can then expand and simplify this equation to:
distance² = x² + (1/49)(4x² + 8x + 9)
Simplify further to:
distance² = x² + (4x² + 8x + 9) / 49
Now, differentiate the equation with respect to x and set it equal to zero to find the critical point where the distance is minimized:
d(distance²)/dx = 2x + (8/49)(2x + 1) = 0
Multiply through by 49 to eliminate the fraction:
98x + 8(2x + 1) = 0
98x + 16x + 8 = 0
114x = -8
x = -8/114
x ≈ -0.07
Now, substitute this value of x back into the equation of the given line to find the corresponding y-value:
2(-0.07) + 7y + 4 = 0
-0.14 + 7y + 4 = 0
7y = -4.14
y = -4.14 / 7
y ≈ -0.59
So, the point on the given line closest to (0, 1) is approximately (-0.07, -0.59).