Question

The given line passes through the points (−4, −3) and (4, 1).




What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the point (−4, 3)?

Answer 1

The slope of the line that passes through points $(x_1,y_1) \text{ and } (x_2,y_2)$ is

$\dfrac{y_2-y_1}{x_2-x_1}.$

For the points (-4,-3) and (4,1) the slope is

$\dfrac{1-(-3)}{4-(-4)}=\dfrac{4}{8}=\dfrac{1}{2}.$

Two perpendicular lines have slopes that form product of -1, then the slope of perpendicular line is $-\dfrac{1}{\frac{1}{2}}=-2.$

The  line that passes through the point (−4, 3) and has slope -2 has equation:

$y-3=-2(x+4),\\y=-2x-8+3,\\y=-2x-5.$

Answer: y=-2x-5.

Answer 2

The equation of a line passing through the point $\left( { - 4,3} \right)$, and perpendicular to line joining the points $\left( {4,1} \right)$ and $\left( { - 4, - 3} \right)$ is ${\bf{y - 3 =-2\left( {x + 4} \right)}$.

Further explanation:

The equation of the line passing through point $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is as follows:

$y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$                    ……(1)

The given line passing through the points $\left( {4,1} \right)$ and $\left( { - 4, - 3} \right)$.

Consider the point $\left( {4,1} \right)$ as $\left( {{x_1},{y_1}} \right)$ and $\left( { - 4, - 3} \right)$ as $\left( {{x_2},{y_2}} \right)$.

Substitute 4 for ${x_1}$, 1 for ${y_1}$, -4 for ${x_2}$ and -3 for ${y_2}$ in equation (1).

$\begin{aligned}y - 1 &= \frac{{ - 3 - 1}}{{ - 4 - 4}}\left( {x - 4} \right) \hfill \\y - 1& = \frac{{ - 4}}{{ - 8}}\left( {x - 4} \right) \hfill \\y - 1 &= \frac{1}{2}\left( {x - 4} \right) \hfill \\ \end{aligned}$

The point slope form of the equation passing through point $\left( {{x_1},{y_1}} \right)$ is as follows:

$y - {y_1} = m\left( {x - {x_1}} \right)$                            ……(2)

Here,  $m$ is the slope of the line.

Compare the equation (2) with $y - 1 = \dfrac{1}{2}\left( {x - 4} \right)$ to obtain the slope ${m_1}$ of line.

Thus, the slope ${m_1}$ of the first line is $\dfrac{1}{2}$.

The slope ${m_2}$ of the other line perpendicular to first line is obtained as follows:

${m_1} \times {m_2} =-1$                                                ……(3)

Substitute $\dfrac{1}{2}$ for ${m_1}$ in equation (3).

$\begin{aligned}\frac{1}{2} \cdot {m_2} &=-1 \hfill \\{m_2} &=-2 \hfill \\ \end{aligned}$

Since, the second line passing through point $\left( { - 4,3} \right)$, therefore, the equation of the second line is obtained as follows:

Substitute -4 for ${x_1}$, 3 for ${y_1}$ and -2 for $m_2$  in equation (2).

$\begin{aligned}y - 3 &=-2\left( {x - \left( { - 4} \right)} \right) \hfill \\y - 3 &=-2\left( {x + 4} \right) \hfill \\ \end{aligned}$

Thus, the equation of the line passing through point $\left( { - 4,3} \right)$ and perpendicular to line $y - 1 = \dfrac{1}{2}\left( {x - 4} \right)$ is   ${\bf{y - 3 =-2\left( {x + 4} \right)}$.

The graph is attached below in figure 1.

Learn more:  

1. Which function has an inverse that is also a function? {(–1, –2), (0, 4), (1, 3), (5, 14), (7, 4)} {(–1, 2), (0, 4), (1, 5), (5, 4), (7, 2)} {(–1, 3), (0, 4), (1, 14), (5, 6), (7, 2)} {(–1, 4), (0, 4), (1, 2), (5, 3), (7, 1)}  

https://brainly.com/question/1632445  

2. A given line has the equation 10x + 2y = −2. what is the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12)? y = ( )x + 12  

https://brainly.com/question/1473992  

3. what are the domain and range of the function f(x) = 3x + 5?  

https://brainly.com/question/3412497  

Answer Details :

Grade: Junior high School.

Subject: Mathematics.

Chapter: Coordinate geometry.

Keywords:

lines, straight lines, (4,1),function, point slope, equation of line passing, perpendicular line, graph, domain, intervals, intercepts, function value, intercepts of lines, slope, slope intercept form, continuous, range, point, line segment.