Question

Calculate the coordinates of the points where the line y = 1 - 2x cuts the curve x^2+ y^2 = 2

Answer 1

`(1, -1)` and
`(-(1)/(5), (7)/(5))`

In decimal form this would be (1, -1) and (-0.2, 1.4)

Explanation:

Basically here we are trying to solve a system of 2 equations. The first one is a straight line and the second is a circle. This can be done graphically and graph is attached. There will be two points where the line cuts the circle so two solution sets

The equations are

Line Equation
`y = 1 - 2x` (1) and

Circle Equation
`x^2+ y^2 = 2` (2)

`y=1-2x`

Plug this value of y into the second equation for the circle

`x^2 + (1-2x)^2= 2`

Apply the perfect squares formula
`\left(a-b\right)^2=a^2-2ab+b^2`

`(1-2x)^2 = 1^2-2\cdot \:1\cdot \:2x+\left(2x\right)^2\\= 1-4x+4x^2\\\\`

So equation (2) becomes

`1-4x+4x^2 +x^2 = 2 \\\\\\5x^2 -4x - 1 = 0`

Solve using the quadratic formula

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

Here a is the coefficient of x ², b is the coefficient of x and c is the constant

So

`x_(1,\:2)=(-\left(-4\right)\pm √(\left(-4\right)^2-4\cdot \:5\left(-1\right)))/(2\cdot \:5)`

Now,

`√(\left(-4\right)^2-4\cdot \:5\left(-1\right)) = √(\left(-4\right)^2+4\cdot \:5\cdot \:1)`

=
`√(16+20) = √(36) = 6`

Therefore

`x_(1,\:2)=(-\left(-4\right)\pm \:6)/(2\cdot \:5)`

`x_1=(-\left(-4\right)+6)/(2\cdot \:5)` =
`(-\left(-4\right)+6)/(2\cdot \:5) =`
`(10)/(2\cdot \:5)`
`= 1`

`x_2 = (-\left(-4\right)-6)/(2\cdot \:5)`
`= (4-6)/(2\cdot \:5)` =
`-(1)/(5)`

To find corresponding values for
`y_1 y_2` plug these values of
`x_1, x_2` into the line equation

`y_1 = 1- 2(1) = -1`

`y_2 = 1 - 2(-(1)/(5)) = 1 + (2)/(5) = (7)/(5)`

So the two points of intersection are

`(1, -1)` and (
`(-(1)/(5), (7)/(5))`.

In decimal form this would be (1, -1) and (-0.2, 1.4)

See attached graph