Question

which is the equation for the line perpendicular to y=-5/3x+11 1/3 and containing P(-2,3)?

A. y-2=-3/5(x-3)
B. y=-5/3x+4 1/3
C. y=-3/5x+4 1/5
D. y=3/5x+4 1/5

Answer 1

Answer:Option D

Step-by-step explanation:

The given equation is $y=\frac{-5}{3}x+11 \frac{1}{3}$

The slope m of the line is $-\frac{5}{3}$

Slope of line perpendicular is reciprocal with opposite sign.

Slope of  perpendicular m=$\frac{3}{5}$

Equation of line passing through point P(-2,3) and having a slope m=$\frac{3}{5}$ is

y-3=$\frac{3}{5}(x+2)$

y-3=$\frac{3}{5}x+ \frac{6}{5}$

Adding 3 both sides

y=$\frac{3}{5}x+4 \frac{1}{5}$

Option D

Answer 2

The equation of the line perpendicular to $y=-\frac{5}{3}x+11\frac{1}{3}$ and passes through a point $P(-2,3)$ is $\boxed{\dfrac{3}{5}x+4\dfrac{1}{5}}$.

Further explanation:  

The linear equation of the line is $y=mx+b$ where $m$ is the slope of the line and $b$ is the $y$-intercept of the line.

If the slope of the line $1$ is $m_{1}$ and the slope of the line $2$ is $m_{2}$ which is perpendicular to line $1$. Then the relation between two slopes is expressed as follows:

$\boxed{m_{1}\times m_{2}=-1}$    

Therefore, we can say that the perpendicular slopes of the line are negative reciprocals of each other.

The point slope form equation of the line passing through a point $(x_{1},y_{1})$ is as follows:

$\boxed{y-y_{1}=m(x-x_{1})}$

Given:

The given equation is $y=-\frac{5}{3}x+11\frac{1}{3}$ and the point is $P(-2,3)$.

Step 1:

First we find the slope of the given equation.

The given equation is $y=-\frac{5}{3}x+11\frac{1}{3}$.

Compare the given equation with standard linear equation. It is obnserved that the slope and the $y$-intercept of the equation is as follows:

$m=-\dfrac{5}{3}\\ b=11\dfrac{1}{3}$

Step 2:

Now find the slope of the line that is perpendicular to the given equation of the line.

Consider $m_{1}$ as the slope of the given equation of the line and $m_{2}$ as the slope of the line which is perpendicular to the given line.

Substitute the value of $m_{1}$ in the relation between two slopes.

$\begin{aligned}m_{1}\times m_{2}&=-1\\-\dfrac{5}{3}\times m_{2}&=-1\\m_{2}&=\dfrac{3}{5}\end{aligned}$

Step 3:

Now, we find the equation of the line containing the given point $P(-2,3)$.

Substitute $x_{1}=-2$, $y_{1}=3$ and the slope $m_{2}=\frac{3}{5}$ in the point slope form  equation of the line.

$\begin{aligned}y-y_{1}&=m(x-x_{1})\\y-3&=\dfrac{3}{5}(x-(-2))\\y&=3+\dfrac{3}{5}(x-(-2))\\y&=3+\dfrac{3}{5}x+\dfrac{6}{5}\\y&=\dfrac{3}{5}x+\dfrac{6}{5}+3\end{aligned}$

Further solution of the above equation is calculated as,

$\begin{aligned}y&=\dfrac{3}{5}x+\dfrac{6+15}{5}\\y&=\dfrac{3}{5}x+\dfrac{21}{5}\\y&=\dfrac{3}{5}x+4\dfrac{1}{5}\end{aligned}$  

Therefore, the equation of the line perpendicular to $y=-\frac{5}{3}x+11\frac{1}{3}$ and passes through a point $P(-2,3)$ is $\boxed{\dfrac{3}{5}x+4\dfrac{1}{5}}$.

Learn more:  

1. Learn more about the equation in the slope intercept form https://brainly.com/question/1473992

2. Learn more about the equation of the circle https://brainly.com/question/1506955

3. Learn more about the line segment https://brainly.com/question/909890

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Straight lines

Keywords: Slope, intercept, equation, perpendicular normal, angle between two line, point, coordinate, passes, parallel, cartesian plane, quadrilateral, distance of a point from a line, two point form, slope-intercept form.