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andreyandreev [35.5K]
2 years ago
13

Halle la ecuación del eje x y la ecuación del eje y, usando dos puntos que se encuentren sobre los mismos.

Mathematics
1 answer:
lisov135 [29]2 years ago
8 0

Una linea recta ( cualquier eje coordenado es una línea recta) queda definida si se conocen dos puntos que están sobre ella.

Solución:

Ecuación del eje x      y = 0

Ecuación del eje y      x = 0

Para darle respuesta a la pregunta podemos seguir el siguiente procedimiento:

  • Escogemos dos puntos arbitrarios sobre el eje x, por ejemplo

P ( 2 ; 0 )  y Q ( 5 ; 0 )         ( todos los puntos sobre el eje x tienen coordenada y = 0.

Según la cual  m = (y₂ - y₁)/ ( x₂ - x₁ )    m = 0

  • Usamos la ecuación pendiente-Intercepto

y = m×x + b         donde m es la pendiente y b el intercepto con el eje y

y entonces tenemos:

  • m = 0    b ( 0 ; 0 )
  • Por sustitución en la ecuación pendiente-intercepto

y = 0

Procediendo de forma similar obtendremos la ecuación del eje y

P´( 0 ; 4 )     Q´( 0 : 8 ) entonces

y = m×x + b

En este caso, la pendiente no es definida ( tang 90° ) y b es de nuevo el punto b ( 0 ; 0).

A partir de  que todos y cada uno de los puntos sobre el eje y son  de valor 0 para x, concluímos que  ecuación del eje y es

x = 0

Enlaces de interés:brainly.com/question/21135669?

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Viefleur [7K]

i) We have to find the determinant of A.

We can do this as:

\det(A)=|\begin{bmatrix}{3} & {6} \\ {1} & {-2}\end{bmatrix}|=3*(-2)-6*1=-6-6=-12

ii) We have to find if the matrix A is singular.

Singular matrix have determinant equal to 0.

This is not the case for A, as its determinant is -12. Then, A is not a singular matrix.

iii) We have to find the values of x and y so that AB = C.

We have to write the matrix multiplication and we will obtain a system of linear equations:

We can now solve the system of equations by adding 3 times the second equation to the first equation:

\begin{gathered} 3(x-2y)+(3x+6y)=3(-3)+(-3) \\ 3x-6y+3x+6y=-9-3 \\ 6x+0y=-12 \\ x=\frac{-12}{6} \\ x=-2 \end{gathered}

We can now use the second equation to find the value of y:

\begin{gathered} x-2y=-3 \\ x+3=2y \\ y=\frac{x+3}{2} \\ y=\frac{-2+3}{2} \\ y=\frac{1}{2} \end{gathered}

The values are x = -2 and y = 1/2.

iv) When we want to multiply two matrices, the required condition is that the number of columns of the matrix on the left is equal the number of rows of the matrix on the right.

In the case of A(2x2) and B(2x1), when we do A*B this condition is satisfied.

But when we try to multiply BA, the number of columns of B is not equal to the number of rows of A, so the matrix mulitplication is not possible.

Answer:

i) det(A) = -12

ii) A is not singular because singular matrices have determinant equal to zero.

iii) x = -2 and y = 1/2.

iv) Is not possible because the number of columns of the first matrix has to be equal to the number of rows of the second matrix.

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