Point-Slope Form 11-4 Warm Up Problem of the Day Lesson Presentation Pre-Algebra

Problem of the Day Without using equations for horizontal or vertical lines, write the equations of four lines that form a square. Possible answer: y = x + 2, y = x – 2, y = –x + 2, y = –x – 2

Warm Up Write the equation of the line that passes through each pair of points in slope-intercept form. 1. (0, –3) and (2, –3) 2. (5, –3) and (5, 1) 3. (–6, 0) and (0, –2) 4. (4, 6) and (–2, 0) y = –3 x = 5 y = – x – 2 1 3 y = x + 2

WARM-UP Find the SLOPE for a line with this equation: y = 4x - 2 Then you graph it, making certain the line travels through (-1,-6) M = 4 Remember M is the Slope

Learn to find the equation of a line given one point and the slope.

We have been talking about finding equations that will lead to a desired line! If you wanted to travel from one coordinate on a plane to another… let’s say in auto pilot, you would need to tell the computer…. Telling it by using an equation because computers only understand math!

Here is another way to find an equation: point-slope form

The point-slope of an equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1). Point on the line Point-slope form y – y1 = m (x – x1) (x1, y1) slope

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula: y – y1 = m(x – x1) Don't let the subscripts scare you, Tyson! They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted (Drew, are you listening?) in the formula. Here's how you use the point-slope formula:

Find the equation of the straight line that has slope m = 4 and passes through the point (–1, –6). They've given me m = 4, x1 = –1, and y1 = –6.  I'll plug these values into the point-slope form, and solve for "y=": y – y1 = m(x – x1) y – (–6) = (4)(x – (–1)) y + 6 = 4(x + 1) (Distributive Property again!) y + 6 = 4x + 4 Subtract 6 from each side y = 4x – 2

Find the equation of the line that passes through the points (–2, 4) and (1, 2). Well, if I have two points on a straight line, I can always find the slope; that's what the slope formula is for. Now I have the slope and two points. I know I can find the equation (by solving first for "b") if I have a point and the slope. So I need to pick one of the points (it doesn't matter which one), and use it to solve for b. Using the point (–2, 4), I get: y = mx + b 4 = (– 2/3)(–2) + b 4 = 4/3 + b 4 – 4/3 = b 12/3 – 4/3 = b b = 8/3 ...so  y = ( – 2/3 ) x + 8/3. WE DID THIS A COUPLE DAYS AGO

Find the equation of the line that passes through the points (–2, 4) and (1, 2). Given two points, I can always find the slope: Then I can use either point as my (x1, y1), along with this slope Ive just calculated, and plug in to the point-slope form. Using (–2, 4) as the (x1, y1), I get: y – y1 = m(x – x1) y – (4) = ( – 2/3 )(x – (–2)) y – 4 = ( – 2/3 )(x + 2) y – 4 = ( – 2/3 ) x – 4/3 y = ( – 2/3 ) x – 4/3 + 4 y = ( – 2/3 ) x – 4/3 + 12/3 y = ( – 2/3 ) x + 8/3

Additional Example 1: Using Point-Slope Form to Identify Information About a Line Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. A. y – 7 = 3(x – 4) y – y1 = m(x – x1) The equation is in point-slope form. y – 7 = 3(x – 4) Read the value of m from the equation. m = 3 (x1, y1) = (4, 7) Read the point from the equation. The line defined by y – 7 = 3(x – 4) has slope 3, and passes through the point (4, 7).

Rewrite using subtraction instead of addition. Additional Example 1B: Using Point-Slope Form to Identify Information About a Line 1 3 B. y – 1 = (x + 6) y – y1 = m(x – x1) 1 3 y – 1 = (x + 6) y – 1 = [x – (–6)] 1 3 Rewrite using subtraction instead of addition. m = 1 3 Why is the “1” positive and the “-6” negative? (x1, y1) = (–6, 1) The line defined by y – 1 = (x + 6) has slope , and passes through the point (–6, 1). 1 3

Try This: Example 1 Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. A. y – 5 = 2 (x – 2) y – y1 = m(x – x1) The equation is in point-slope form. y – 5 = 2(x – 2) Read the value of m from the equation. m = 2 (x1, y1) = (2, 5) Read the point from the equation. The line defined by y – 5 = 2(x – 2) has slope 2, and passes through the point (2, 5).

Try This: Example 1B 2 3 B. y – 2 = (x + 3) y – y1 = m(x – x1) 2 3 y – 2 = (x + 3) y – 2 = [x – (–3)] 2 3 Rewrite using subtraction instead of addition. m = 2 3 (x1, y1) = (–3, 2) The line defined by y – 2 = (x + 3) has slope , and passes through the point (–3, 2). 2 3

Additional Example 2: Writing the Point-Slope Form of an Equation Write the point-slope form of the equation with the given slope that passes through the indicated point. A. the line with slope 4 passing through (5, -2) y – y1 = m(x – x1) Substitute 5 for x1, –2 for y1, and 4 for m. [y – (–2)] = 4(x – 5) y + 2 = 4(x – 5) The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4(x – 5).

Additional Example 2: Writing the Point-Slope Form of an Equation B. the line with slope –5 passing through (–3, 7) y – y1 = m(x – x1) Substitute –3 for x1, 7 for y1, and –5 for m. y – 7 = -5[x – (–3)] y – 7 = –5(x + 3) The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5(x + 3).

Try This: Example 2A Write the point-slope form of the equation with the given slope that passes through the indicated point. A. the line with slope 2 passing through (2, –2) y – y1 = m(x – x1) Substitute 2 for x1, –2 for y1, and 2 for m. [y – (–2)] = 2(x – 2) y + 2 = 2(x – 2) The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2(x – 2).

Try This: Example 2B B. the line with slope -4 passing through (-2, 5) y – y1 = m(x – x1) Substitute –2 for x1, 5 for y1, and –4 for m. y – 5 = –4[x – (–2)] y – 5 = –4(x + 2) The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4(x + 2).

Additional Example 3: Entertainment Application A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet. As x increases by 30, y increases by 20, so the slope of the line is or . The line passes through the point (0, 18). 20 30 2 3

Additional Example 3 Continued y – y1 = m(x – x1) Substitute 0 for x1, 18 for y1, and for m. 2 3 y – 18 = (x – 0) 2 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = x. Substitute 150 for x to find the value of y. 2 3 y – 18 = (150) 2 3 y – 18 = 100 y = 118 The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward.

Try This: Example 3 A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet. As x increases by 45, y increases by 15, so the slope of the line is or . The line passes through the point (0, 15). 15 45 1 3

Try This: Example 3 Continued y – y1 = m(x – x1) Substitute 0 for x1, 15 for y1, and for m. 1 3 y – 15 = (x – 0) 1 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 15 = x. Substitute 300 for x to find the value of y. 1 3 y – 15 = (300) 1 3 y – 15 = 100 y = 115 The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward.

3. the line with slope 4 passing through (3, 5) Lesson Quiz Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. 1. y + 6 = 2(x + 5) 2. y – 4 = – (x – 6) Write the point-slope form of the equation with the given slope that passes through the indicated point. 3. the line with slope 4 passing through (3, 5) 4. the line with slope –2 passing through (–2, 4) (–5, –6), 2 2 5 (6, 4), – 2 5 y – 5 = 4(x – 3) y – 4 = –2(x + 2)