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Chapter 1:Qua
END-OF-CHAPTER REVIEW EXERCISE 1
1
A curve has equation y = 2xy + 5 and a line has equation 2x + 5y = 1.
The curve and the line intersect at the points A and B. Find the coordinates of the midpoint
of the line AB​

Sagot :

Answer:

            (¹/₂, 0)

Step-by-step explanation:

2x + 5y = 1     ⇒    2x = 1 - 5y

y = 2xy + 5

y = (1 - 5y)y + 5

y = y - 5y² + 5

5y² = 5

y² = 1

 y₁ = 1               ∨           y₂ = -1

2x₁ = 1 -5(1)                   2x₂ = 1 - 5(-1)

2x₁ = 1 - 5                      2x₂ = 1 + 5

2x₁ = -4                         2x₂ = 6

 x₁ = -2                           x₂ = 3

Midpoint:

[tex]M=\left(\frac{x_1+x_2}2\ ,\ \frac{y_1+y_2}2\right)=\left(\frac{-2+3}2\ ,\ \frac{1+(-1)}2\right)=\left(\frac12\ ,\ 0\right)[/tex]

The coordinates of the midpoint of the line AB is (0, 1/2).

The equation of the curve is given by y = 2xy + 5.

The equation of the line is given by 2x + 5y = 1.

We have to first find the point of intersection between the curve and the line and then find the coordinates of the midpoint of line AB where A and B is the point of intersection between the curve and the line.

What is the formula for finding the coordinates of a point that divides a given line in the ratio m:n?

If C(x, y) divides a line AB in m:n then we have,

[tex]x = \frac{mx_2 +nx_1}{m+n},~~~y =\frac{my_2+ny_1}{m+n}[/tex]

And if C(x,y) is the midpoint then m = n.

we have,

[tex]x = \frac{x_2 +x_1}{2},~~~y =\frac{y_2+y_1}{2}[/tex]

Where x and y are the coordinates of the midpoint o the line AB.

Find the point of intersection between the line and the curve.

y = 2xy + 5............(1)

2x + 5y = 1..............(2)

From (2) we have,

2x = 1 - 5yx = (1-5y) / 2..........(3)

Substituting (3) in (1).

[tex]y =2\frac{(1-5y)}{2}y + 5\\\\y =\frac{2y-10y^2}{2} + 5\\\\2y = 2y - 10y^2 + 10\\\\10y^2 = 10\\\\y^2 = 1[/tex]

So we have,

y = 1 and y = -1

Puttin y = 1 in (3) we get,

x = (1 - 5) / 2 = - 4 / 2 = -2

Puttin y = -1 in (3) we get,

x = {1-5(-1)} / 2 = (1+5) / 2 = 6 / 2 = 3

Now we have two points of intersection A( 1, -2 ) and B( -1, 3 ).

Finding the coordinates of the midpoint of the line AB.

we have,

[tex]A(1, -2) = A (x_1, y_1)~~and~~B(-1, 3) = B(x_2, y_2)[/tex]

Substituting in the given equation.

[tex]x = \frac{x_2 +x_1}{2},~~~y =\frac{y_2+y_1}{2}\\\\x = \frac{-1 +1}{2},~~~y =\frac{3+(-2)}{2}\\\\x = 0,~~~y=\frac{1}{2}[/tex]

So the coordinates of the midpoint is (x,y) = ( 0, 1/2 ).

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