1. The stright line y=x+3 cuts the curve x^2 + y^2 = 29 at two points.
a) Prove that x^2 + 3x - 10 = 0
b) Hence find the coordinates of the points.

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 $$\frac{1}{2}$$

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

Now we have to find the roots of this funtion

$$x^2+3x-10=0$$

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

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

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

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

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

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

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

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

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

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

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

then our function will be

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

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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

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

;)

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