Presentation on theme: "Writing equations given slope and point"— Presentation transcript:
1 Writing equations given slope and point 5.212/6/13
2 Solving EquationsIn any equation, we can solve for any variable that is not knownRewrite in terms of that variableFind a value if all the other values are knownCan use this to find equations
3 Point-Slope Form of the Equation of a Line The point-slope equation of a non-vertical line of slope m that passes through the point (x1, y1) isy – y1 = m(x – x1).
4 Solving With Slope and a Point Substitute known values into the equationRewrite the equation y = mx + b with slope and y-intercept values
5 Example: Writing the Point-Slope Equation of a Line Write the point-slope form of the equation of the line passing through (-1,3) with a slope of 4. Then solve the equation for y.Solution We use the point-slope equation of a line with m = 4, x1= -1, and y1 = 3.This is the point-slope form of the equation.y – y1 = m(x – x1)Substitute the given values. Simply.y – 3 = 4[x – (-1)]We now have the point-slope form of the equation for the given line.y – 3 = 4(x + 1)We can solve the equation for y by applying the distributive property.y – 3 = 4x + 4y = 4x + 7Add 3 to both sides.
6 Solving with Slope and Point Find the equation of the line passing through the point (-3, 0 ) and has slope of 1/3
7 Solving with Slope and a Point Find the equation of the line passing through point (-2, -1) with slope of -3
8 Solving with Slope and Point Find the equation of the line passing through the point (-3, 0 ) and has slope of 1/3
9 Solving with Slope and a Point Find the equation of the line passing through (3 , -4) and is parallel to the line y = -3x – 2
10 Slope and Perpendicular Lines 90°Two lines that intersect at a right angle (90°) are said to be perpendicular. There is a relationship between the slopes of perpendicular lines.Slope and Perpendicular LinesIf two non-vertical lines are perpendicular, then the product of their slopes is –1.If the product of the slopes of two lines is –1, then the lines are perpendicular.A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.
11 Perpendicular LinesPerpendicular lines have negative reciprocal slopes so if you need the slope of a line perpendicular to a given line, simply find the slope of the given line, take its reciprocal (flip it over) and make it negative.
12 Example: Finding the Slope of a Line Perpendicular to a Given Line Find the slope of any line that is perpendicular to the line whose equation is x + 4y – 8 = 0.Solution We begin by writing the equation of the given line in slope-intercept form. Solve for y.x + 4y – 8 = 0This is the given equation.4y = -x + 8To isolate the y-term, subtract x and add 8 on both sides.Slope is –1/4.y = -1/4x + 2Divide both sides by 4.The given line has slope –1/4. Any line perpendicular to this line has a slope that is the negative reciprocal, 4.
13 Example: Writing the Equation of a Line Perpendicular to a Given Line Write the equation of the line perpendicular to x + 4y – 8 = 0 that passes thru the point (2,8) in standard form.Solution: The given line has slope –1/4. Any line perpendicular to this line has a slope that is the negative reciprocal, 4.So now we need know the perpendicular slope and are given a point (2,8). Plug this into the point-slope form and rearrange into the standard form.y – 8 = 4[x – (2)]y – y1 = m(x – x1)y1 = 8m = 4x1 = 2y - 8 = 4x - 8-4x + y = 04x – y = 0 Standard form
14 4 -1 2 y = - x 1 Let's look at a line and a point not on the line Let's find the equation of a line parallel to y = - x that passes through the point (2, 4)y = - xWhat is the slope of the first line, y = - x ?(2, 4)1This is in slope intercept form so y = mx + b which means the slope is –1.So we know the slope is –1 and it passes through (2, 4). Having the point and the slope, we can use the point-slope formula to find the equation of the line4-12Distribute and then solve for y to leave in slope-intercept form.
15 4 1 2 y = - x What if we wanted perpendicular instead of parallel? Let's find the equation of a line perpendicular to y = - x that passes through the point (2, 4)y = - xThe slope of the first line is still –1.(2, 4)The slope of a line perpendicular is the negative reciporical so take –1 and "flip" it over and make it negative.412Distribute and then solve for y to leave in slope-intercept form.So the slope of a perpendicular line is 1 and it passes through (2, 4).