Algebra

1. Algebra 5.5 Point-Slope Form

2. Point-Slope Form • A formula used to find the linear equation when given a point on the line and the slope of the line. • Point-Slope Form y – y1 = m(x – x1) The green variables will always remain variables. All others will be replaced with numbers.

3. Let’s try. ( -1, 8) m = 7 y – y1 = m(x – x1) y – 8 = 7(x + 1) You try. (3, -5) m = ½ y – y1 = m(x – x1) y + 5 = ½ (x – 3) Write an equation in point-slope form of the line that passes through the given point and has the given slope.

4. Let’s try. (3, 7) m = -3 y – y1 = m(x – x1) y – 7 = -3(x – 3) y – 7 = -3x + 9 +7 +7 y = -3x + 16 Does the graph of this line go thru (3, 7) and have a slope of -3? You try. (-2, 4) m = -½ y – y1 = m(x – x1) y – 4 = -½ (x + 2) y – 4 = -½ x – 1 +4 +4 y = -½ x + 3 Does the graph of this line go thru (-2, 4) and have a slope of -½ ? Write the point-slope form of the equation of the line that passes through the point and has the given slope. Then rewrite the equation in slope-intercept form.

5. Write the equation in point-slope form and then rewrite into slope-intercept form of the line that is parallel to the line y = ¾ x – 5 and passes through the point (-2, 4). The line is parallel to a line with a slope of ¾ , thus its slope must be ¾ also. Now, we know m = ¾ and it passes through (-2, 4). y – y1 = m(x – x1) y - 4 = ¾ (x + 2) y - 4 = ¾ x + 1 ½ +4 +4 y = ¾ x + 5 ½

6. You try! Write in point-slope form and then rewrite into slope-intercept form the equation of the line that is parallel to the line y = 4x – 7 and passes through the point (-1, -7). The line is parallel to a line with a slope of 4 , thus its slope must be 4 also. Now, we know m = 4 and it passes through (-1, -7). y – y1 = m(x – x1) y + 7 = 4 (x + 1) y + 7 = 4x + 4 -7 -7 y = 4x - 3

7. HW P.303-304 (30-47) P. 282-283 (24-29, 32-37) Write in slope-int form