Question

The straight line l has equation x-y =3

the curve c has equation 3x^2 -y^2+xy=9
l and c intersect at the points p and q
find the coordinates of the midpoint of pq

Answer 1

Answer: The Midpoint of P and Q are (-0.5, -3.5)

Step-by-step explanation: To find the coordinates of the intersection points of the line and the curve, we can substitute the equation of the line into the equation of the curve and solve for x and y.

From the equation of the line: x - y = 3, we can see that y = x - 3

Substituting this into the equation of the curve:

3x^2 - (x-3)^2 + x(x-3) = 9

Now we can simplify and group

3x^2 - x^2 + 6x - 9 + x^2 - 3x = 9

3x^2 + 3x - 9 = 9

3x^2 + 3x = 18

3x(x+1) = 18

x(x+1) = 6

We can see that x = 2 and x = -3 are the solutions.

Substituting the x value back into the original equation of the line to find the y coordinate:

x = 2 => y = 2 - 3 => y = -1

x = -3 => y = -3 - 3 => y = -6

So the intersection points are (2,-1) and (-3,-6)

To find the midpoint of these two points, we take the average of their x and y coordinates:

x = (2 + -3) / 2 = -0.5

y = (-1 + -6) / 2 = -3.5

so the midpoint of the two intersection points is (-0.5, -3.5)

Answer 2

Point,
`(p,q)`, is
`((-1)/(2) ,(-7)/(2) )`.

Explanation:

Given the equations
`x-y=3` and
`3x^(2) -y^(2) +xy=9` . Find where they intersect and find the midpoint,
`(p,q)`, of the two intersection points. To find find points of intersection, you can either set up these equations as a system and solve algebraically or you can use a graphing calculator to graph the functions and see where they cross.

I will set up these equations as a system of equations,

`\left \{ {{x-y=3} \atop {3x^(2) -y^(2) +xy=9}} \right.`

I will solve the top equation for
`x`, and substitute it into the bottom.

`x-y=3 = > x=3+y`

Now substitute
`x=3+y` for x into the bottom equation. Solve for y,

=>
`3x^(2) -y^(2) +xy=9`

=>
`3(3+y)^(2) -y^(2) +(3+y)y=9`

=>
`3(y^(2)+6y+9) -y^(2) +y^(2)+3y =9`

=>
`3y^(2)+18y+27 -y^(2) +y^(2)+3y =9`

=>
`3y^(2)+18y+27+3y =9`

=>
`3y^(2)+21y+27 =9`

=>
`(3y^(2)+21y+27 =9)/(3)`

=>
`y^(2)+7y+9 =3`

=>
`y^(2)+7y+6 =0`

=>
`(y+1)(y+6)=0`

=>
`y=-1` or
`y=-6`

Now plug these values for
`y` into
`x=3+y`.

When y=-1:

`x=3+(-1)`

=>
`x=2`

When y=-6:

`x=3+(-6)`

=>
`x=-3`

The solution to the system (points of intersection) is
`(-3,-6)` and
`(2,-1)`. I also attached a graph of the intersection points.

Now to find the midpoint of the two intersection points,
`midpoint=((x_(1)+x_(2) )/(2) , (y_(1)+y_(2) )/(2) )`

`(x_(1),y_(1)) = > (-3,-6)`

`(x_(2),y_(2)) = > (2,-1)`

=>
`midpoint=((-3+2)/(2) ,(-6+(-1))/(2) )`

=>
`((-1)/(2) ,(-7)/(2) )`

Thus, the point,
`(p,q)`, is
`((-1)/(2) ,(-7)/(2) )`.