# Writing equations given slope and point

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Writing equations given slope and point
5.2 12/6/13

Solving Equations In any equation, we can solve for any variable that is not known Rewrite in terms of that variable Find a value if all the other values are known Can use this to find equations

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) is y – y1 = m(x – x1).

Solving With Slope and a Point
Substitute known values into the equation Rewrite the equation y = mx + b with slope and y-intercept values

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 + 4 y = 4x + 7 Add 3 to both sides.

Solving with Slope and Point
Find the equation of the line passing through the point (-3, 0 ) and has slope of 1/3

Solving with Slope and a Point
Find the equation of the line passing through point (-2, -1) with slope of -3

Solving with Slope and Point
Find the equation of the line passing through the point (-3, 0 ) and has slope of 1/3

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

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 Lines If 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.

Perpendicular Lines Perpendicular 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.

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 = 0 This is the given equation. 4y = -x + 8 To isolate the y-term, subtract x and add 8 on both sides. Slope is –1/4. y = -1/4x + 2 Divide 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.

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 = 8 m = 4 x1 = 2 y - 8 = 4x - 8 -4x + y = 0 4x – y = 0 Standard form

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 = - x What is the slope of the first line, y = - x ? (2, 4) 1 This 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 line 4 -1 2 Distribute and then solve for y to leave in slope-intercept form.

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 = - x The 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. 4 1 2 Distribute 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).

Homework: 5.2 worksheet #’s 1-15 Due Monday, 12/9