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Learn how to find the equation of a line passing through a point with a given slope using point-slope form. Practice writing equations for parallel and perpendicular lines.
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EXAMPLE 2 Write an equation given the slope and a point Write an equation of the line that passes through (5, 4) and has a slope of – 3. SOLUTION Because you know the slope and a point on the line, use point-slope form to write an equation of the line. Let (x1, y1) = (5, 4) and m = – 3. y –y1=m(x –x1) Use point-slope form. y –4=–3(x –5) Substitute for m,x1, and y1. y – 4 = – 3x + 15 Distributive property y = – 3x + 19 Write in slope-intercept form.
EXAMPLE 3 Write equations of parallel or perpendicular lines Write an equation of the line that passes through (–2,3) and is (a) parallel to, and (b) perpendicular to, the line y= –4x + 1. SOLUTION a. The given line has a slope of m1 = –4. So, a line parallel to it has a slope of m2 = m1 = –4. You know the slope and a point on the line, so use the point-slope form with (x1,y1) = (– 2, 3) to write an equation of the line.
EXAMPLE 3 Write equations of parallel or perpendicular lines y –y1=m2(x –x1) Use point-slope form. y –3=–4(x –(– 2)) Substitute for m2, x1, andy1. y – 3 =– 4(x + 2) Simplify. y – 3 =– 4x – 8 Distributive property y =– 4x – 5 Write in slope-intercept form.
1 1 1 1 1 b. A line perpendicular to a line with slope m1 = – 4 has a slope of m2 = – = . Use point-slope form with (x1, y1) = (– 2, 3) 2 4 4 4 2 1 m1 y –3= (x – (–2)) 1 y – 3 = (x +2) 4 1 4 y – 3 = x + y = x + EXAMPLE 3 Write equations of parallel or perpendicular lines y –y1=m2(x –x1) Use point-slope form. Substitute for m2, x1, andy1. Simplify. Distributive property Write in slope-intercept form.
for Examples 2 and 3 GUIDED PRACTICE GUIDED PRACTICE 4.Write an equation of the line that passes through (– 1, 6) and has a slope of 4. SOLUTION Because you know the slope and a point on the line, use the point-slope form to write an equation of the line. Let (x1, y1) = (–1, 6) and m=4 y –y1 = m(x –x1) Use point-slope form. Substitute for m, x1, andy1. y – 6 = 4(x – (– 1)) Distributive property y – 6 = 4x + 4 Write in slope-intercept form. y = 4x + 10
for Examples 2 and 3 GUIDED PRACTICE GUIDED PRACTICE 5.Write an equation of the line that passes through (4, –2) and is (a) parallel to, and (b) perpendicular to, the line y = 3x – 1. SOLUTION The given line has a slope of m1 = 3. So, a line parallel to it has a slope of m2 = m1 = 3. You know the slope and a point on the line, so use the point - slope form with (x1,y1) = (4, – 2) to write an equation of the line. y –y1 = m2(x –x1) Use point-slope form. y – (– 2) = 3(x – 4) Substitute for m2, x1, andy1. y + 2 = (x – 4) Simplify. y + 2 = 3x – 12 Distributive property y = 3x – 14 Write in slope-intercept form.
4 1 2 1 1 1 1 3 3 3 3 3 3 3 y – (– 2) =– (x – 4) y + 2 =– (x – 4) y + 2 =–x– y = – x – for Examples 2 and 3 GUIDED PRACTICE GUIDED PRACTICE b. A line perpendicular to a line with slope m1 = 3 has a slope of m2 = – = – 1 m1 Use point - slope form with (x1, y1) = (4, – 2) y –y1 = m2(x –x1) Use point-slope form. Substitute for m2, x1, andy1. Simplify. Distributive property Write in slope-intercept form.