Question

Find the equation of the line that passes through (-1,2) and is perpendicular to 2 y = 2 x − 1 . Leave your answer in the form y = m x + c

Answer 1

The equation in slope-intercept form of the line that passes through the point (-1, 2) and is perpendicular to the line equation, $2 y = 2 x - 1$ is: $y = -x + 1$

Recall:

Equation of a line in slope-intercept form is: $y = mx + c$, where, slope = m, and y-intercept = c.

Point-slope form is: $y - b = m(x - a)$

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: $2 y = 2 x - 1$

To write the equation of the line that passes through (-1, 2), we need to find the slope value.

• First, rewrite  $2 y = 2 x - 1$ and find its slope.

$\frac{2y}{2} = \frac{2x}{2} - \frac{1}{2} \\\\y = x - \frac{1}{2}$

• The slope (m) of $2 y = 2 x - 1$ 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  $y - b = m(x - a)$:

• Thus:

$y - 2= -1(x - (-1))\\\\y - 2 = -1(x + 1)$

• Rewrite this in the form of $y = m x + c$

• Thus:

$y - 2 = -x - 1\\\\y = -x - 1 + 2\\\\y = -x + 1$

Therefore, the equation in slope-intercept form of the line that passes through the point (-1, 2) and is perpendicular to the line equation, $2 y = 2 x - 1$ is $y = -x + 1$

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