Mathematics, 06.10.2019 08:00, jasperc2002owy5b5

Write the point-slope form of the line that passes through (5,5) and is perpendicular to a line with a slope of 1/4 include all of your work in your final answer.

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The point-slope form of the line that passes through (5,5) and is perpendicular to a line with a slope of $\frac{1}{4}$ is 4x + y -25 = 0

Solution:

The point slope form of the line that passes through the points $\left(x_{1} y_{1}\right)$ and perpendicular to the line with a slope of “m” is given as

$\bold{y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)}$ ---- eqn 1

Where “m” is the slope of the line. $x_{1} \text { and } y_{1}$ are the points that passes through the line.

From question, given that slope “m” = $\frac{1}{4}$

Given that the line passes through the points (5,5).Hence we get

$x_{1}=5 ; y_{1}=5$

By substituting the values in eqn 1 , we get the point slope form of the line which is perpendicular to the line having slope $\frac{1}{4}$ can be found out.

y - 5 = -4(x - 5)

y - 5 = -4x + 20

on simplifying the above equation, we get

y - 5 + 4x -20 = 0

4x + y - 25 = 0

hence the point slope form of given line is 4x + y - 25 = 0

answered: Guest

step-by-step explanation:

a difference is the result of subtraction. while each of these expressions has subtraction in it, a difference will be the result when subtraction is the last operation applied in the order of operations for the expression. this means subtraction must be outside of the parenthesis. only b has this.

answered: Guest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

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step-by-step explanation:

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Write the point-slope form of the line that passes through (5,5) and is perpendicular to a line with...