Challenge Question: Write the equation of the line that passes through the point (-5, 1) and is PERPENDICULAR to the line y=(-5/3)x-7 Once you have the slope, count it from the first point and get your second point.
To find the equation of the line that is perpendicular to y = (-5/3)x - 7, we need to find the negative reciprocal of the slope of the given line. The slope of the given line is -5/3, so the negative reciprocal would be 3/5.
Let's use the point-slope form of a linear equation to find the equation of the line passing through the point (-5, 1) with a slope of 3/5.
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope.
Plugging in the values, we have:
y - 1 = (3/5)(x - (-5))
y - 1 = (3/5)(x + 5)
Now, let's simplify the equation:
y - 1 = (3/5)x + (3/5)(5)
y - 1 = (3/5)x + 3
To get the equation in slope-intercept form (y = mx + b), we can add 1 to both sides of the equation:
y = (3/5)x + 3 + 1
y = (3/5)x + 4
Therefore, the equation of the line that passes through the point (-5, 1) and is perpendicular to y = (-5/3)x - 7 is y = (3/5)x + 4.