The equation is \(\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }\)
Further explanation
This case asking the end result in the form of a slope-intercept.
Step-1: find out the gradient.
10x + 2y = -2
We isolate the y variable on the left side. Subtract both sides by 10x, we get:
2y = - 10x - 2
Divide both sides by two
y = -5x -1
The slope-intercept form is \(\boxed{ \ y = mx + c \ }\), with the coefficient m as a gradient. Therefore, the gradient is m = -5.
If you want a shortcut to find a gradient from the standard form, implement this:
\(\boxed{ \ ax + by = k \rightarrow m = - \frac{a}{b} \ }\)
10x + 2y = −2 ⇒ a = 10, b = 2
\(\boxed{m = - \frac{10}{2} \rightarrow m = -5}\)
Step-2: the conditions of the two parallel lines
The gradient of parallel lines is the same \(\boxed{ \ m_1 = m_2 \ }\). So \(\boxed{m_1 = m_2 = -5}.\)
Final step: figure out the equation, in slope-intercept form, of the parallel line to the given line and passes through the point (0, 12)
We use the point-slope form.
\(\boxed{ \ \boxed{ \ y - y_1 = m(x - x_1)} \ }\)
Given that
y - 12 = - 5(x - 0)
y - 12 = - 5x
After adding both sides by 12, the results is \(\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }\)
Alternative steps
Substitutes m = -5 and (0, 12) to slope-intercept form \(\boxed{ \ y = mx + c \ }\)
12 = -5(0) + c
Constant c is 12 then arrange the slope-intercept form.
Similar results as above, i.e. \(\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }\)
Note:
\(\boxed{Standard \ form: ax + by = c, with \ a > 0}\)
\(\boxed{Point-slope \ form: y - y_1 = m(x - x_1)}\)
\(\boxed{Slope-intercept \ form: y = mx + k}\)
Learn more
- A similar problem https://brainly.com/question/10704388
- Investigate the relationship between two lines https://brainly.com/question/3238013
- Write the line equation from the graph https://brainly.com/question/2564656
Keywords: given line, the equation, slope-intercept form, standard form, point-slope, gradien, parallel, perpendicular, passes, through the point, constant
