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

       → $\frac{y-b}{x-a} =\frac{q-b}{p-a}$

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

Equation of line in intercept form is →$\frac{x}{a} + \frac{y}{b} =1$, 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

→ $\frac{x}{\frac{5}{3}} + \frac{y}{\frac{2}{3}}=1$

 → $\frac{3x}{5} + \frac{3y}{2} =1$  

→ 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

Answer-

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

$\Rightarrow 0 =3x-5$

$\Rightarrow 3x=5$

$\Rightarrow x=\dfrac{5}{3}$

So, coordinate of M is $(\dfrac{5}{3},\ 0)$

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

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

$\Rightarrow 3y=2$

$\Rightarrow y=\dfrac{2}{3}$

So, coordinate of N is $(0,\ \dfrac{2}{3})$

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

$\Rightarrow \dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$

$\Rightarrow \dfrac{y-0}{\frac{2}{3}-0}=\dfrac{x-\frac{5}{3}}{0-\frac{5}{3}}$

$\Rightarrow \dfrac{y}{\frac{2}{3}}=\dfrac{x-\frac{5}{3}}{-\frac{5}{3}}$

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

$\Rightarrow -5y=2(x-\dfrac{5}{3})$

$\Rightarrow -5y=2x-\dfrac{10}{3}$

$\Rightarrow 2x+5y-\dfrac{10}{3}=0$

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

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