Exam (elaborations) MATH 114N

1. (3 points) Find the equation of the line through the points (−1, 5) and (2, −4) in slope- intercept form. Solution. To find the equation, we first use the points (x1, y1) = (−1, 5) and (x2, y2) = (2, −4) to find the slope, and then use point-slope form. First: m = y2 − y1 = −4 − 5 = −9 = −3 x2 − x1 Now, we use point-slope form: 2 − (−1) 3 y − y1 = m(x − x1) y − 5 = −3(x − (−1)) y − 5 = −3x − 3 + 5 = + 5 y = −3x + 2 So the equation in slope-intercept form is 2. (3 points) Line L1 has the equation 2x − 5y + 3 = 0. Line L2 is perpendicular to L1. Line L3 is parallel to L1. Find the slopes of L1, L2, and L3. Solution. To find the slope of the line L1, we put the equation in slope-intercept form: 2 3 y = x + 5 5 — 3) So the slope of L1 is . Since L2 is perpindicular to L1, its slope is the negative recip- rocal of m1, m2 = − 5 . Since L3 is parallel to L3, its slope is the same as the slope of L1, m3 = m1 = 2 . In summary: ..............................................CONTINUED...........................