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Sagot :
The equation in slope-intercept form of the line that passes through the point (-1, 2) and is perpendicular to the line equation, [tex]2 y = 2 x - 1[/tex] is: [tex]y = -x + 1[/tex]
Recall:
Equation of a line in slope-intercept form is: [tex]y = mx + c[/tex], where, slope = m, and y-intercept = c.
Point-slope form is: [tex]y - b = m(x - a)[/tex]
The slope value of a line will be the negative reciprocal of the slope of the line it is perpendicular to. i.e. if the slope of a line is a, the slope of the line that it is perpendicular to will be -a.
Given:
- The line passes through point (-1, 2)
- Equation of line it is perpendicular to is: [tex]2 y = 2 x - 1[/tex]
To write the equation of the line that passes through (-1, 2), we need to find the slope value.
- First, rewrite [tex]2 y = 2 x - 1[/tex] and find its slope.
[tex]\frac{2y}{2} = \frac{2x}{2} - \frac{1}{2} \\\\y = x - \frac{1}{2}[/tex]
- The slope (m) of [tex]2 y = 2 x - 1[/tex] is therefore 1.
- Thus, the slope (m) of the line that passes through (-1, 2) will be the negative reciprocal of 1 = -1.
- Write the equation of the line by substituting (a, b) = (-1, 2) and m = -1 into [tex]y - b = m(x - a)[/tex]:
- Thus:
[tex]y - 2= -1(x - (-1))\\\\y - 2 = -1(x + 1)[/tex]
- Rewrite this in the form of [tex]y = m x + c[/tex]
- Thus:
[tex]y - 2 = -x - 1\\\\y = -x - 1 + 2\\\\y = -x + 1[/tex]
Therefore, the equation in slope-intercept form of the line that passes through the point (-1, 2) and is perpendicular to the line equation, [tex]2 y = 2 x - 1[/tex] is [tex]y = -x + 1[/tex]
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