Question

find the equation of the line passing through the intersection of 5x-3y=1 and 2x+3y-23=0 and perpendicular to the line 5x-3y-1

Answer 1

hope this will solve your problem

Answer 2

Answer:

Equation of the perpendicular line is $y=(-\frac{3}{5})x+\frac{781}{105}$

Step-by-step explanation:

The given system of equations is

5x - 3y = 1 ---------(1)

2x + 3y = 23 -----------(2)

We have to find the equation of a line passing through point of intersection of both the equations and perpendicular to line 5x - 3y = 1

3y = 5x - 1

y = $\frac{5}{3}x-\frac{1}{3}$

Therefore, slope of this line $m_{1}=\frac{5}{3}$

Let the equation of the perpendicular line is y = mx + b

where m = slope or gradient of the line

b = y-intercept of the line

By the property of perpendicular lines,

$m\times m_{1}=(-1)$

$m\times (\frac{5}{3})=(-1)$

$m=-\frac{3}{5}$

Now we will find the solution of the system of equations

By adding equations (1) and (2)

(5x - 3y) + (2x + 3y) = 1 + 23

7x = 24

x = $\frac{24}{7}$

From equation (2)

$2(\frac{24}{7})+3y=23$

$3y=23-\frac{48}{7}$

$3y=\frac{113}{7}$

$y=\frac{113}{21}$

From equation (1)

$5x-3(\frac{113}{21})=1$

$5x-\frac{113}{7}=1$

$5x=\frac{120}{7}$

$x=\frac{24}{7}$

Now the perpendicular line having slope $(-\frac{3}{5})$ will be

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

This line passes through $(\frac{24}{7},\frac{113}{21})$

$\frac{113}{21}=(-\frac{3}{5})(\frac{24}{7})+b$

$\frac{113}{21}=-\frac{72}{35}+b$

b = $\frac{113}{21}+\frac{72}{35}$

$b=\frac{781}{105}$

Now equation will be

$y=(-\frac{3}{5})x+\frac{781}{105}$

Learn more about the slope-intercept form of the equation of a line from https://brainly.in/question/8648499