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Question: The equation of the line \[x + y = 2\] in double intercept form, the \[x - \]intercept, and the \[y ...

The equation of the line x+y=2x + y = 2 in double intercept form, the xx - intercept, and the yy - intercept, respectively are
(a) x2y2=1\dfrac{x}{2} - \dfrac{y}{2} = 1, a=2a = 2, and b=2b = - 2
(b) x2+y2=1 - \dfrac{x}{2} + \dfrac{y}{2} = 1, a=2a = - 2, and b=2b = 2
(c) x2+y2=1\dfrac{x}{2} + \dfrac{y}{2} = 1, a=2a = 2, and b=2b = 2
(d) x2y2=1 - \dfrac{x}{2} - \dfrac{y}{2} = 1, a=2a = - 2, and b=2b = - 2

Explanation

Solution

Here, we will rewrite the given equation to get the equation in double intercept form. Then, we will use the formula for the equation of the line in double intercept form to find the values of the xx - intercept and the yy - intercept of the line.
Formula Used:
The double intercept form of a line is given by the equation xa+yb=1\dfrac{x}{a} + \dfrac{y}{b} = 1, where aa is the xx - intercept of the line and bb is the yy - intercept of the line.

Complete step by step solution:
We will rewrite the given equation of the line in its double intercept form.
Dividing both sides of the equation x+y=2x + y = 2 by 2, we get
x+y2=22\Rightarrow \dfrac{{x + y}}{2} = \dfrac{2}{2}
Simplifying the equation, we get
x+y2=1\Rightarrow \dfrac{{x + y}}{2} = 1
Rewriting the equation by splitting the L.C.M., we get
x2+y2=1\Rightarrow \dfrac{x}{2} + \dfrac{y}{2} = 1
We can observe that this is the double intercept form of the given line.
Now, we will find the values of the xx - intercept and the yy - intercept of the line.
Comparing the double intercept form of the line x2+y2=1\dfrac{x}{2} + \dfrac{y}{2} = 1 to xa+yb=1\dfrac{x}{a} + \dfrac{y}{b} = 1, we get
a=2a = 2 and b=2b = 2.
\therefore We get the xx - intercept and the yy - intercept of the line as a=2a = 2 and b=2b = 2 respectively.

Thus, the correct option is option (c).

Note:
The xx - intercept of a line is the point at which it touches the xx - axis. It is given by the point (a,0)\left( {a,0} \right). Similarly, the yy - intercept of a line is the point at which it touches the yy - axis. It is given by the point (0,b)\left( {0,b} \right). We can easily find the yy - intercept by substituting x=0x = 0 in the given equation.
x+y=2 0+y=2 y=2\begin{array}{l}x + y = 2\\\ \Rightarrow 0 + y = 2\\\ \Rightarrow y = 2\end{array}
So the yy - intercept of the line is (0,2)\left( {0,2} \right).
Now we can find xx - intercept by substituting y=0y = 0 in the given equation.
x+y=2 x+0=2 x=2\begin{array}{l}x + y = 2\\\ \Rightarrow x + 0 = 2\\\ \Rightarrow x = 2\end{array}
So the xx - intercept of the line is (2,0)\left( {2,0} \right).