Question

find the joint equation of the line through (1,2) and parallel to the lines x-2y=5 and x=3y-4

Answer 1

The joint equation is  $x^2-5xy+8x+6y^2-19y+15=0$

Therefore $(x-2y+3)(x-3y+5)=x^2-5xy+8x+6y^2-19y+15=0$

Step-by-step explanation:

Given parallel lines are x-2y=5 and x=3y-4

To find the joint equation of the line passing through the point (1,2)

Given lines becomes

x-2y=5 and x=3y-4

x-2y-5=0 and x-3y+4=0

Let the line x-2y-5=0 be written as x-2y+p=0

Let the line x-3y+4=0 be written as x-3y+p=0

x-2y+p=0 passes through (1,2)

1-2(2)+p=0

1-4+p=0

-3+p=0

p=3

Substitute p=3 in x-2y+p=0

we have $x-2y+3=0\hfill (1)$

Similarly x-3y+p=0 passes through (1,2)

1-3(2)+p=0

1-6+p=0

-5+p=0

p=5

Substitute p=5 in x-3y+p=0

we have $x-3y+5=0\hfill (2)$

Now the joint equation of equations (1) and (2) is given below

• $(x-2y+3)(x-3y+5)=0$
• $x(x-3y+5)+(-2y)(x-3y+5)+3(x-3y+5)=0$ ( using the distributive property )
• $x(x)+x(-3y)+x(5)-2y(x)-2y(-3y)-2y(5)+3(x)+3(-3y)+3(5)=0$
• $x^2-3xy+5x-2xy+6y^2-10y+3x-9y+15=0$
• $x^2-5xy+8x+6y^2-19y+15=0$ ( here adding the like terms )
• Therefore $(x-2y+3)(x-3y+5)=x^2-5xy+8x+6y^2-19y+15=0$

The joint equation is  $x^2-5xy+8x+6y^2-19y+15=0$