Determine the lines represented by each of the following equation: {x}^(2) + 2xy + {y}^(2) - 2x - 2y - 15 = 0 Please help! ​

Question

Determine the lines represented by each of the following equation:


`{x}^(2) + 2xy + {y}^(2) - 2x - 2y - 15 = 0`
Please help! ​

Answer 1

The lines are:

• x+y-5=0
• x+y+3=0

Explanation:

The given equation is:

`\tt x^2+2xy+y^2-2x-2y-15=0`

We can write this equation in the standard form of a quadratic equation as follows:

`\tt x^2+2xy+y^2-2x-2y-15=0`

`\tt (x^2+2xy+y^2)-2x-2y-15=0`

`\tt (x+y)^2-2x-2y-15=0`

Using formula of
`\boxed{\tt (a+b)^2=a^2+2ab+c^2}`

`\tt (x+y)^2-3^2-2x-2y-6=0`

`\tt (x+y+3)(x+y-3)-2(x+y+3)=0`

using formula of
`\tt a^2-b^2=(a+b)(a-b)`

`\tt (x+y+3)(x+y-3-2)=0`

`\tt (x+y+3)(x+y-5)=0`

This equation tells us that the two lines represented by the equation are:

• x+y-5=0
• x+y+3=0

Solving the first equation for x, we get:

x=-y+5

So, the first line has an intercept of (0, 5) and a slope of -1.

Solving the second equation for x, we get:

x=-y-3

So, the second line has an intercept of (0, -3) and a slope of -1.

Therefore, the two lines represented by the equation
`\tt x^2+2xy+y^2-2x-2y-15=0` are:

• x+y-5=0
• x+y+3=0

Answer 2

`y=-x+5`

`y=-x-3`

Explanation:

Given equation:

`x^2+2xy+y^2-2x-2y-15=0`

Rearrange the given equation to standard form ax² + bx + c = 0:

`x^2+2xy-2x+y^2-2y-15=0`

`x^2+(2y-2)x+(y^2-2y-15)=0`

Therefore:

• a = 1
• b = (2y - 2)
• c = (y² - 2y - 15)

Solve with the quadratic formula.

`\boxed{\begin{minipage}{4 cm}\underline{Quadratic Formula}\\\\$x=(-b \pm √(b^2-4ac))/(2a)$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}`

Substitute the values/expressions for a, b and c into the quadratic formula:

`x=(-(2y - 2) \pm √((2y - 2)^2-4(1)(y^2- 2y - 15)))/(2(1))`

`x=(-(2y - 2) \pm √((2y - 2)^2-4(y^2- 2y - 15)))/(2)`

`x=(-(2y - 2) \pm √(4y^2-8y+4-4y^2+8y +60))/(2)`

`x=(-(2y - 2) \pm √(64))/(2)`

`x=(-2y +2 \pm 8)/(2)`

`x=-y +1 \pm 4`

So, the two solutions are:

`x = -y+5`

`x = -y-3`

Therefore, the solutions in slope-intercept linear form are:

`\boxed{\boxed{\begin{aligned}y&=-x+5\\y&=-x-3\end{aligned}}}`