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Felix Turner · Answer 1 · 2015-08-01T08:41:07+0000

First question

line:~~~~y=x+3

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

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

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

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

$\boxed{y_1=5}$

y_2=x_2+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

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

point 1

13=2a+b

point 2

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

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

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The third one, we have

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