Question

What are the intercepts of $y = 2x - 4$ ?

Answer 1

In this question, we have to find out the intercepts of a given equation.
The equation of a line which cuts off intercepts a and b respectively from the $x$ and $y$ axes is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ ,
$a = {\text{ }}x$ -intercept value of the linear equation.
$b = {\text{ }}y$ -intercept value of the linear equation.
To find the intercepts we will compare the given equation with $\dfrac{x}{a} + \dfrac{y}{b} = 1$ and finding the value of $a$ and $b$ we will get the required intercepts.

Complete step by step solution:
We have to find out the intercepts of $y = 2x - 4$ .
We know that the intercept form of a straight line equation is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ ……i),
Where
$a = {\text{ }}x$ -intercept value of the linear equation.
$b = {\text{ }}y$ -intercept value of the linear equation.
Now we can write the given equation as $2x - y = 4$
Or, $\dfrac{{2x}}{4} - \dfrac{y}{4} = 1$
Or, $\dfrac{x}{{\dfrac{4}{2}}} + \dfrac{y}{{ - 4}} = 1$
$\dfrac{x}{2} + \dfrac{y}{{ - 4}} = 1$ .……...ii)
Comparing equation i) and ii) we get,
$a = {\text{ }}x$ -intercept = $2$ .
$b = {\text{ }}y$ -intercept = $- 4$ .
Hence, the intercepts of $y = 2x - 4$ are $2$ and $- 4$ .

Note: Linear equation:
Any equation which can be put in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are real numbers, $a$ and $b$ are not both zero, is called a linear equation in two variables.
Another method:
We have to find out the intercepts of $y = 2x - 4$ .
This is the equation of a line. When the line crosses the $x$ -axis, the $y$ -coordinate at this point will be zero. By substituting $y = 0$ into the equation will give us the $x$ -intercept.
i.e. $0 = 2x - 4$
$2x = 4$
Therefore, $x = \dfrac{4}{2} = 2$
Similarly, when the line crosses the $y$ -axis, the $x$ -coordinate at this point will be zero and substituting $x = 0$ into the equation will give us the y-intercept.
i.e. $y = 2 \times 0 - 4$
Hence, $y = - 4$
Hence, the intercepts of $y = 2x - 4$ are $2$ and $- 4$ .