y – y 1 = m(x – x 1 )
description
Transcript of y – y 1 = m(x – x 1 )
y – y1 = m(x – x1)
Equation of a Straight Line
This allows use to find the equation of a straight line if we know its slope and one coordinate it passes through.
x coordinateslopey coordinate
Find the equation of the line
(a) passing through (-2 , 3) with slope = 3
(b) Passing through (4 , 7) with slope = 2/3
y – y1 = m(x – x1) y – 3 = 3(x - -2)
y – 3 = 3x + 6
y = 3x + 9
y – y1 = m(x – x1) y – 7 = 2/3(x – 4)
3y – 21 = 2x - 8
3y = 2x + 13
Find the equation of the line
1.passing through (2 , 1) with slope = 2
2. Passing through (3 , 2) with slope = 3
3. Passing through (-4 , 2) with slope = 5
4. Passing through (-2 , -3) with slope = -3
5. Passing through (5 , 3) with slope = 2/5
Find the equation of a straight line parallel to 2y = 4x + 5 which passes through (-1 , 3)
y = 2x + 5/3
2y = 4x + 5 y – y1 = m(x – x1)
y – 3 = 2(x - -1)
y – 3 = 2x + 2
y = 2x + 5
Find the equation of a straight line perpendicular to 3y = x - 2 which passes through (4 , 5)
y = x/3 - 2/3
3y = x - 2 y – y1 = m(x – x1)
y – 5 = -3(x - 4)
y – 5 = -3x + 12
y + 3x = 17
m = 1/3
Perp = -3
1. Find the equation of a straight line parallel to y = 2x + 5 which passes through (4 , 5)
2. Find the equation of a straight line parallel to 2y + 6x = 18 which passes through (2 , -4)
3. Find the equation of a straight line parallel to 3y = 4x - 2 which passes through (3 , 2)
4. Find the equation of a straight line perpendicular to y = 2x + 3 which passes through (1 , 3)
Find the equation of the line passing through the points (4 , 5) and (-2, 3).
m = 3 - 5
-2 - 4= -2/-6 = 1/3
y – 5 = 1/3(x – 4)
3y – 15 = x - 4
3y = x + 11
Find the equation of the line passing through the points
1) (2 , 3) and (8 , 6)
2) (-3, 1) and (7 , 6)