The equation of the straight line is [tex]x+y=2[/tex].
Given:
- The line with intercepts (a,0) and (0,b) has the equation [tex]\frac{x}{a} +\frac{y}{b} =1, a\neq 0, b\neq 0[/tex]
- Point on the line: (-2, 4)
- x-intercept: (a, 0)
- y-intercept: (0, a)
- [tex]a\neq 0[/tex]
To find: The equation of this line
It is given that a line with intercepts (a,0) and (0,b) has the equation [tex]\frac{x}{a} +\frac{y}{b} =1, a\neq 0, b\neq 0[/tex]
Now, it is given that the referred line has intercepts (a, 0) and (0, a). Then, using the above statement, the equation of this line can be written as,
[tex]\frac{x}{a} +\frac{y}{a}=1[/tex]. It is already given that [tex]a\neq 0[/tex]. So, we need not mention it again.
It is also given that the point (-2, 4) lies on this line. Then, the coordinates of this point must satisfy the equation of the line.
This implies that,
[tex]\frac{-2}{a} +\frac{4}{a} =1[/tex]
[tex]\frac{2}{a} =1[/tex]
[tex]a=2[/tex]
Now, put [tex]a=2[/tex] in the equation of the line, [tex]\frac{x}{a} +\frac{y}{a}=1[/tex] to get,
[tex]\frac{x}{2} +\frac{y}{2} =1[/tex]
[tex]x+y=2[/tex]
So, the equation of the line is [tex]x+y=2[/tex].
Learn more about equations of straight lines here:
https://brainly.com/question/18879008
