Question

The line y =3x-5 meet x-axis at the point M. The line 3y+2x=2 meets y-axis at point N. Find the equation of the line joining M and N in the form ax + by + c = 0 where: a,b,c are integers.

Answer 1

Solution: As given line y =3x-5 meet x-axis at the point M.

On x axis y coordinate is zero.

Put y =0 in above equation, we get →x = 5/3

∴ Coordinate of M is (5/3,0).

As, also given , line 3y+2x=2 meets y-axis at point N.

On y axis , x coordinate is zero.

Substituting , x=0 in above equation, gives y =2/3.

Coordinate of point N is (0,2/3).

Equation of line passing through two points (a,b) and (p,q) is given by

→
`(y-b)/(x-a) =(q-b)/(p-a)`

Or as X intercept = 5/3, and Y intercept = 2/3

Equation of line in intercept form is →
`(x)/(a) + (y)/(b) =1`, where a and b is X intercept and y intercept respectively.

So, line passing through (5/3,0) and (0,2/3) is given by

→
`(x)/((5)/(3))  +  (y)/((2)/(3))=1`

→
`(3x)/(5) + (3y)/(2) =1`

→ 6 x + 15 y =10 [Taking LCM of 5 and 2 which is 10]

→ 6 x + 15 y -10=0, which is equation of the line joining M and N in the form ax + by + c = 0 where: a,b,c are integers.

Answer 2

The line equation is,

`\boxed{\boxed{6x+15y-10=0}}`

Solution-

The line
`y =3x-5` meets x-axis at the point M, i.e M is the x-intercept of this line. At the x-intercept y=0, so

`\Rightarrow 0 =3x-5`

`\Rightarrow 3x=5`

`\Rightarrow x=(5)/(3)`

So, coordinate of M is
`((5)/(3),\ 0)`

The line
`3y+2x=2` meets y-axis at point N, i.e N is the y-intercept of this line. At the y-intercept x=0, so

`\Rightarrow 3y+2(0)=2`

`\Rightarrow 3y=2`

`\Rightarrow y=(2)/(3)`

So, coordinate of N is
`(0,\ (2)/(3))`

The line joining M and N can be found out by applying two point formula of straight line,

`\Rightarrow (y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)`

`\Rightarrow (y-0)/((2)/(3)-0)=(x-(5)/(3))/(0-(5)/(3))`

`\Rightarrow (y)/((2)/(3))=(x-(5)/(3))/(-(5)/(3))`

`\Rightarrow -(5)/(3)y=(2)/(3)(x-(5)/(3))`

`\Rightarrow -5y=2(x-(5)/(3))`

`\Rightarrow -5y=2x-(10)/(3)`

`\Rightarrow 2x+5y-(10)/(3)=0`

As it is given that all the coefficients are integers, so multiplying with 3

`\Rightarrow 6x+15y-10=0`