Question

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

Answer 1

firstly , we should find out intersecting point of given lines 5x -3y = 1 and 2x + 3y = 23.
5x - 3y = 1 ------(1)
2x + 3y = 23 ------(2)

add equations (1) and (2) ,
7x = 24 ⇒ x = 24/7 , put it in equation (1)
3y = 120/7 -1 ⇒y = 113/21

Hence, unknown line is passing through the point (24/7 , 113/21)
now, unknown line is perpendicular upon 5x - 3y = 1
Means, slope of unknown line ×{ slope of 5x - 3y = 1} = -1
Let slope of unknown line is m
∴ m × 5/3 = -1 ⇒ m = -3/5

Now, equation of unknown line is ,
(y - 113/7) = -3/5(x - 24/7)
⇒5y - 565/7 + 3x - 72/7 = 0
⇒ 3x + 5y - 91= 0

Hence, answer is 3x + 5y = 91

Answer 2

At the point of intersection of two lines there exist a common and equal point for the two lines. The two Equations are thus solve simultaneously to get the coordinates of the common point.

5x-3y=1.......i
2x+3y=23.......ii

Lets eliminate x by multiplying equation i by 2 and ii by 5 then subtracting i from ii.

10x-6y=2......iii
10x-15y=115.......iv

Subtracting iii from iv we get :

21y =113

y=113/21

Substituting the value of y in equation i we get:

5x - 3× (113/21)=1

5x=120/7

x= 24/7

The common point : (24/7, 113/21)

The line is perpendicular to 5x-3y-1=0

Writing the equation in the form: y=mx + c where m is the gradient, and c is the y intercept we have :

y=5/3x - 1/3

The grad =5/3

For perpendicular lines the product of their gradient is equal to - 1

5/3 × m=-1

m =-3/5

The equation of the line :

(y-113/21)/(x-24/7)=-3/5

5(y-113/21)=-3(x-24/7)

5y - 565/21=-3x + 72/7

Writing this in the form y=mx + c we get the equation of the line as:

y= - 3/5x + 781/105

Converting to decimals we have :

y= - 0.6x + 7.44