Perpendicular lines intersect at a point, thereby creating right angles at that point of intersection. Multiplying the slopes of those two lines result in a product of -1, which constitutes a negative reciprocal relationship between them.
Line 2: y = ⅕x - 3 is in slope-intercept form.
Line 4: y + 1 = -5 (x + 2) is in point-slope form. We need to transform this into slope-intercept form, y = mx + b:
y + 1 = -5 (x + 2)
y + 1 = -5x - 10
y + 1 - 1 = -5x - 10 - 1
y = -5x - 11 (This is the slope-intercept form).
Multiplying the slope of Line 2 with the slope of Line 4 results in a product of -1:
Line 2 (m1) slope: ⅕
Line 4 (m2) slope: - 5
m1 × m2 = ⅕ × - 5 = -1.
Therefore, Lines 2 and 4 are perpendicular lines.
Interestingly, Lines 1 and 3 also form perpendicular lines, but not by the definition of negative reciprocal relationship between two slopes.
Line 1: y = 2 is a horizontal line with a zero slope, and its y-intercept is given by (0, b). In this case, the y-intercept is (0, 2).
Line 3: x = -4 is a vertical line with an undefined slope, represented by the equation, x = a. Its x-intercept is (-4, 0).
We cannot multiply the slopes of Lines 1 and 3 because Line 1 has a zero slope, and Line 3 has an undefined slope. Thus, Lines 1 and 3 do not have negative reciprocal slopes. However, if their lines are graphed, it will show that they are perpendicular from each other. Because the vertical line will always have the same x-coordinate, -4, which means that at some point it will cross the y-coordinate, 2. In other words, Lines 1 and 3 intersect at point (-4, 2).
Therefore, Lines 2 and 4 are perpendicular from each other.
The same goes with Lines 1 and 3-- they are also perpendicular from each other at point (-4, 2).
