Respuesta :
Answer:
1. [tex]-5x+3y+44=0[/tex]
2. [tex]2x+y-2=0[/tex]
3. [tex]2x+y-4=0[/tex]
Step-by-step explanation:
Standard form of a line is [tex]Ax+By+C=0[/tex].
If a line passing through two points then the equation of line is
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope, i.e.,[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex].
1.
The line passes through the points (7,-3) and (4,-8). So, the equation of line is
[tex]y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)[/tex]
[tex]y+3=\dfrac{-5}{-3}(x-7)[/tex]
[tex]y+3=\dfrac{5}{3}(x-7)[/tex]
[tex]3(y+3)=5(x-7)[/tex]
[tex]3y+9=5x-35[/tex]
[tex]-5x+3y+9+35=0[/tex]
[tex]-5x+3y+44=0[/tex]
Therefore, the required equation is [tex]-5x+3y+44=0[/tex].
2.
We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to [tex]2 x + y =-5[/tex].
Slope of the line : [tex]m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2[/tex]
Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).
Equation of line is
[tex]y-2=-2(x-0)[/tex]
[tex]y-2=-2x[/tex]
[tex]2x+y-2=0[/tex]
Therefore, the required equation is [tex]2x+y-2=0[/tex].
3.
We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to [tex]2x + y =-5[/tex].
From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).
Equation of line is
[tex]y-0=-2(x-2)[/tex]
[tex]y=-2x+4[/tex]
[tex]2x+y-4=0[/tex]
Therefore, the required equation is [tex]2x+y-4=0[/tex].
