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3 years ago

Find the two intersection points

Mathematics

1 answer:

bogdanovich [222]3 years ago

6 0

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two x-coordinates.

To find the y-coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

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All vectors are in Rn. Check the true statements below:

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A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .