Question

11. Through (-3,-5), perpendicular to -2x - 5y = -19

Answer 1

Answer:

11) D. y=5/2x+5/2 , 12) B. y=8/5x+69/5, 14) A. y=-9/5x-67/5

Step-by-step explanation:

11) The function of the perpendicular line can be found in terms of its slope and a given point by this formula:

$y-y_{o} = m_{\perp}\cdot (x-x_{o})$

Where:

$x_{o}$, $y_{o}$ - Components of the given point, dimensionless.

$m_{\perp}$ - Slope, dimensionless.

Besides, a slope that is perpendicular to original line can be calculated by this expression:

$m_{\perp} = -\frac{1}{m}$

Where $m$ is the slope of the original line, dimensionless.

The original slope is determined from the explicitive form of the given line:

$-2\cdot x - 5\cdot y = -19$

$2\cdot x +5\cdot y = 19$

$5\cdot y = 19 - 2\cdot x$

$y = \frac{19}{5} -\frac{2}{5}\cdot x$

The original slope is $-\frac{2}{5}$, and the slope of the perpendicular line is:

$m_{\perp} = -\frac{1}{\left(-\frac{2}{5}\right) }$

$m_{\perp} = \frac{5}{2}$

If $x_{o} = -3$, $y_{o} = -5$ and $m_{\perp} = \frac{5}{2}$, then:

$y-(-5) = \frac{5}{2}\cdot [x-(-3)]$

$y + 5 = \frac{5}{2}\cdot x +\frac{15}{2}$

$y = \frac{5}{2}\cdot x +\frac{5}{2}$

The right answer is D.

12) The function of the parallel line can be found in terms of its slope and a given point by this formula:

$y-y_{o} = m_{\parallel}\cdot (x-x_{o})$

Where:

$x_{o}$, $y_{o}$ - Components of the given point, dimensionless.

$m_{\parallel}$ - Slope, dimensionless.

Its slope is the slope of the given, which must be transformed into its explicitive form:

$-8\cdot x + 5\cdot y = 89$

$5\cdot y = 89 +8\cdot x$

$y = \frac{89}{5}+\frac{8}{5} \cdot x$

The slope of the parallel line is $\frac{8}{5}$.

If $x_{o} = -8$, $y_{o} = 1$ and $m_{\parallel} = \frac{8}{5}$, then:

$y-1 = \frac{8}{5}\cdot [x-(-8)]$

$y-1 = \frac{8}{5}\cdot x +\frac{64}{5}$

$y = \frac{8}{5}\cdot x +\frac{69}{5}$

The correct answer is B.

14) The function of the perpendicular line can be found in terms of its slope and a given point by this formula:

$y-y_{o} = m_{\perp}\cdot (x-x_{o})$

Where:

$x_{o}$, $y_{o}$ - Components of the given point, dimensionless.

$m_{\perp}$ - Slope, dimensionless.

Besides, a slope that is perpendicular to original line can be calculated by this expression:

$m_{\perp} = -\frac{1}{m}$

Where $m$ is the slope of the original line, dimensionless.

The original slope is determined from the explicitive form of the given line:

$-5\cdot x +9\cdot y = 49$

$9\cdot y = 49+5\cdot x$

$y = \frac{49}{9} +\frac{5}{9}\cdot x$

The original slope is $\frac{5}{9}$, and the slope of the perpendicular line is:

$m_{\perp} = -\frac{1}{m}$

$m_{\perp} = -\frac{1}{\frac{5}{9} }$

$m_{\perp} = -\frac{9}{5}$

If $x_{o} = -8$, $y_{o} = 1$ and $m_{\perp} = -\frac{9}{5}$, then:

$y-1 = -\frac{9}{5}\cdot [x-(-8)]$

$y-1 = -\frac{9}{5}\cdot x-\frac{72}{5}$

$y = -\frac{9}{5}\cdot x -\frac{67}{5}$

The correct answer is A.