Respuesta :
[tex]\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-4}{7-(-3)}\implies \cfrac{-5}{7+3}\implies \cfrac{-5}{10}\implies -\cfrac{1}{2}[/tex]
[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-\cfrac{1}{2}[x-(-3)]\implies y-4=-\cfrac{1}{2}(x+3) \\\\\\ y-4=-\cfrac{1}{2}x-\cfrac{3}{2}\implies y=-\cfrac{1}{2}x-\cfrac{3}{2}+4\implies y=-\cfrac{1}{2}x+\cfrac{5}{2}[/tex]
In order to find the equation of the line, you must first find the gradient of the line
Gradient of a line (m)= (y2-y1)/(x2-x1)
.....Now you substitute the points...
= ((-1)-4)/(7-(-3))
= (-5)/(10)
= -1/2
Therefore the gradient is negative a half (-1/2)
Now we can find the equation of the line. You can use one of the two points to find the equation of the line
Equation of a line: y-y1=m(x-x1)
... I chose to use (7,-1) as my points....
y-(-1)=-1/2(x-7)
y+1=-x/2+7/2
y=-x/2+(7/2)-1
1) y=-x/2+5/2
..... if you would like you can multiply everything by 2 to have while numbers instead of fractions ....
2) 2y=-x+5
... if you are uncomfortable of having x as negative, you can transpose it carrying both the -x and the 5 to the side of the 2y. NOTE that the sign changes and the equation will be equal to 0....
3) 2y+x-5=0
1,2 and 3 are all correct answers, just in different forms
Gradient of a line (m)= (y2-y1)/(x2-x1)
.....Now you substitute the points...
= ((-1)-4)/(7-(-3))
= (-5)/(10)
= -1/2
Therefore the gradient is negative a half (-1/2)
Now we can find the equation of the line. You can use one of the two points to find the equation of the line
Equation of a line: y-y1=m(x-x1)
... I chose to use (7,-1) as my points....
y-(-1)=-1/2(x-7)
y+1=-x/2+7/2
y=-x/2+(7/2)-1
1) y=-x/2+5/2
..... if you would like you can multiply everything by 2 to have while numbers instead of fractions ....
2) 2y=-x+5
... if you are uncomfortable of having x as negative, you can transpose it carrying both the -x and the 5 to the side of the 2y. NOTE that the sign changes and the equation will be equal to 0....
3) 2y+x-5=0
1,2 and 3 are all correct answers, just in different forms
