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Sagot :
Slope intercept form of line -
[tex]y - y_1 = m(x - x_1)[/tex]
y1 = y - coordinate of point
x1 = x - coordinate of point
m = slope of line
- [tex]y -(- 3) = - 6(x - (- 9))[/tex]
- [tex]y + 3 = - 6(x + 9)[/tex]
- [tex]y + 3 = - 6x - 54[/tex]
- [tex]y + 6x = - 54 - 3[/tex]
- [tex]y + 6x = - 57[/tex]
- [tex]6x + y + 57 = 0[/tex]
Correct option is : -
D. y + 3 = -6(x + 9)
Answer:
[tex]\textsf{d)} \quad y+3=-6(x+9)[/tex]
Step-by-step explanation:
Point-slope form of a linear equation:
[tex]\boxed{y-y_1=m(x-x_1)}[/tex]
Where:
- m is the slope.
- (x₁, y₁) is a point on the line.
Given information:
- Slope = -6
- Point = (-9, -3)
Substitute the given values into the formula:
[tex]\begin{aligned}y-y_1 & =m(x-x_1)\\\implies y-(-3) & =-6(x-(-9))\\y+3 & =-6(x+9)\end{aligned}[/tex]
Therefore, the equation that represents a line that passes through (-9, -3) and has a slope of -6 is:
[tex]\boxed{y+3=-6(x+9)}[/tex]
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