1. Use point-slope form to write the equation of a line that has a slope of 2/3 and passed through (-3,-1). Write your final equation in slope-intercept form. Is the answer y=2/3+-1
2. Write an equation of the line that passed through (2,-1) and is parallel to the graph of y= 5x-2. Write your final equation in slope-intercept form. I can't figure out this question.
3. Write an equation of the line that passes through (3,5) and is perpendicularto the graph of y= -3x+7. Write your final equation in slope-intercept form,
This one is hard for me too, please I just need a little help.
Thank you for your time
1) (Y+1)= 2/3 *(x+3)
y=2/3 x +1
2) slope 5
y=5x+b put in the point (2,-1) and solve for b.
3. perpendicular: slope is neg reciprocal, or 1/3
y=1/3 * x + b Put in (3,5) and solve for b.
1. To write the equation of a line using point-slope form, we use the formula: y - y1 = m(x - x1), where (x1, y1) represents a point on the line and m represents the slope. In this case, the slope is 2/3 and the point (-3, -1) is given. Substituting these values into the point-slope formula, we get: y - (-1) = (2/3)(x - (-3)). Simplifying, we have y + 1 = (2/3)(x + 3). To write the equation in slope-intercept form (y = mx + b), we need to isolate y. Distributing the 2/3, we get y + 1 = (2/3)x + 2. Subtracting 1 from both sides gives y = (2/3)x + 1. So, the correct answer is y = (2/3)x + 1.
2. To find the equation of a line parallel to another line, we need to know that parallel lines have the same slope. In this case, the given line has a slope of 5. We are also given the point (2, -1) that the new line must pass through. Using the point-slope form equation: y - y1 = m(x - x1), we substitute the values into the equation: y - (-1) = 5(x - 2). Simplifying, we have y + 1 = 5(x - 2). Expanding and rearranging, we get y + 1 = 5x - 10. Subtracting 1 from both sides gives y = 5x - 11. Therefore, the equation of the line parallel to y = 5x - 2 and passing through the point (2, -1) is y = 5x - 11.
3. To find the equation of a line perpendicular to another line, we need to know that perpendicular lines have slopes that are negative reciprocals of each other. In this case, the given line has a slope of -3. To find the negative reciprocal of -3, we take the reciprocal (flipping the fraction) and change the sign. So, the negative reciprocal of -3 is 1/3. We are also given the point (3, 5) that the new line must pass through. Using the point-slope form equation: y - y1 = m(x - x1), we substitute the values into the equation: y - 5 = (1/3)(x - 3). Simplifying, we have y - 5 = (1/3)x - 1. Multiplying through by 3 to eliminate the fraction, we get 3(y - 5) = x - 3. Expanding and rearranging, we have 3y - 15 = x - 3. Adding 15 to both sides gives 3y = x + 12. Dividing by 3, we have y = (1/3)x + 4. Therefore, the equation of the line perpendicular to y = -3x + 7 and passing through the point (3, 5) is y = (1/3)x + 4.