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2 years ago

What is the equation that represents the line that passes through the point (2,6) and has a slope of -5?

Mathematics

1 answer:

tiny-mole [99]2 years ago

3 0

Answer:

y = - 5x + 16

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - 5, thus

y = - 5x + c ← is the partial equation of the line

To find c substitute (2, 6) into the partial equation

6 = - 10 + c ⇒ c = 6 + 10 = 16

y = - 5x + 16 ← equation of line

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Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$

∴

$\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}$

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2 years ago