The path of a train A is given by the equation $3x + 4y - 12 = 0$ and the path of another train B is given by the equation $6x + 8y - 48 = 0$. Represent this situation graphically.

Given:

The path of the train A is given by the equation $3x\ +\ 4y\ -\ 12\ =\ 0$ and the path of another train B is given by the equation $6x\ +\ 8y\ -\ 48\ =\ 0$.

To do:

We have to represent the above situation graphically.

Solution:

The given pair of equations are:

$3x\ +\ 4y\ -\ 12\ =\ 0$....(i)

$4y=12-3x$

$y=\frac{12-3x}{4}$

$6x\ +\ 8y\ -\ 48\ =\ 0$....(ii)

$8y=48-6x$

$y=\frac{48-6x}{8}$

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation (i),

If $x=0$ then $y=\frac{12-3(0)}{4}=\frac{12}{4}=3$

If $y=0$ then $0=\frac{12-3x}{4}$

$12=3x$

$x=\frac{12}{3}=4$

For equation (ii),

If $x=0$ then $y=\frac{48-6(0)}{8}=\frac{48}{8}=6$

If $y=0$ then $0=\frac{48-6x}{8}$

$\Rightarrow 6x=48$

$\Rightarrow x=\frac{48}{6}=8$

The above situation can be plotted graphically as below:

The line AB represents the equation $6x+8y-48=0$ and the line PQ represents the equation $3x+4y-12=0$.

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Updated on: 10-Oct-2022

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