Question

In the xy-plane, line l has a y-intercept of -13 and is perpendicular to the line with the equation y=-2/3x. If the point (10, b) is on the line l, what is the value of b?

Answer 1

Answer:

The value of $b$ is 2.

Step-by-step explanation:

The standard form of the equation of the line is of the form:

$y = m\cdot x + k$

Where:

$x$, $y$ - Independent and dependent variable, dimensionless.

$m$ - Slope, dimensionless.

$k$ - y-intercept, dimensionless.

Given that line $l$ is perpendicular to $y = -\frac{2}{3}\cdot y$, the slope is equal to:

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

Where $m_{\perp}$ is the slope of the perpendicular line, dimensionless.

If $m_{\perp} = -\frac{2}{3}$, then:

$m = -\frac{1}{\left(-\frac{2}{3}\right) }$

$m = \frac{3}{2}$

If $x = 0$ and $y = -13$, the y-intercept of the line $l$ is:

$-13 = \frac{3}{2}\cdot (0) +k$

$k = -13$

The equation of the line $l$ is $y = \frac{3}{2}\cdot x -13$. Given that $y = b$ and $x = 10$, the value of $b$ is:

$b = \frac{3}{2}\cdot (10)-13$

$b = 15-13$

$b = 2$

The value of $b$ is 2.