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Help with line equations?

Help with line equations?-example-1
Help with line equations?-example-1
Help with line equations?-example-2
Help with line equations?-example-3

1 Answer

5 votes
First question


line:~~~~y=x+3


curve:~~~~x^2+y^2=29

now we have to replace the y of curve by the y of line, therefore


x^2+(x+3)^2=29


x^2+x^2+6x+9=29


2x^2+6x+9-29=0


2x^2+6x-20=0

we can multiply each member by
(1)/(2)


\boxed{\boxed{x^2+3x-10=0}}

Now we have to find the roots of this funtion


x^2+3x-10=0

Sum and produc or Bhaskara

Then we find two axis


\boxed{x_1=2~~and~~x_2=-5}

now we replace this values to find the Y, we can replace in curve or line equation, I'll prefer to replace it in line equation.


y=x+3


y_1=x_1+3


y_1=2+3


\boxed{y_1=5}


y_2=x_2+3


y_2=-5+3


\boxed{y_2=-2}

Therefore


\boxed{\boxed{P_1(2,5)~~and~~P_2(-5,-2)}}
_______________________________________________________________

The second question give to us


y=ax+b


P_1(2,13)


P_2(-1,-11)

We just have to replace the value then we'll get a linear system.

point 1


13=2a+b

point 2


-11=-a+b

then our linear system will be


\begin{Bmatrix}2a+b&=&13\\-a+b&=&-11\end{matrix}

I'll multiply the second line by -1 and I'll add to first one


\begin{Bmatrix}2a+(a)+b+(-b)&=&13+(-11)\\-a+b&=&-11\end{matrix}


\begin{Bmatrix}3a&=&24\\-a+b&=&-11\end{matrix}


\begin{Bmatrix}a&=&8\\-a+b&=&-11\end{matrix}

therefore we can replace the value of a, at second line


\begin{Bmatrix}a&=&8\\-8+b&=&-11\end{matrix}


\begin{Bmatrix}a&=&8\\b&=&-11+8\end{matrix}


\boxed{\boxed{\begin{Bmatrix}a&=&8\\b&=&-3\end{matrix}}}

then our function will be


\boxed{\boxed{y=8x-3}}

_________________________________________________________________

The third one, we have


line:~~~~y=3x-4


curve:~~~~y=x^2-2x-4

This resolution will be the same of our first question.

Let's replace the y of curve by the y of line


3x-4=x^2-2x-4


0=x^2-2x-4-3x+4

therefore


\boxed{\boxed{x^2-5x=0}}

now we have to find the roots of this function.


x^2-5x=0

put x in evidence


x*(x-5)=0


\boxed{x_1=0~~and~~x_2=5}

then


y=3x-4


y_1=3x_1-4


y_1=3*0-4


\boxed{y_1=-4}


y_2=3x_2-4


y_2=3*5-4


y_2=15-4


\boxed{y_2=11}

our points will be


\boxed{\boxed{P_1(0,-4)~~and~~P_2(5,11)}}

I hope you enjoy it ;)
User Felix Turner
by
8.4k points

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