Question

Y=kx+b A(7;3),B(-2;4). Решите

Answer 1

$k = \frac{y2 - y1}{x2 - x1} = \frac{4 - 3}{ - 2 - 7} = - \frac{1}{9} \\k = \frac{y - y1}{x - x1} = \frac{y - 3}{x - 7} = - \frac{1}{9} \\ 9y - 27 = - x + 7 \\ 9y = - x + 7 + 27 \\ 9y = - x + 34 \\ y = - \frac{1}{9} x + \frac{34}{9} \\ y = - \frac{1}{9} x + 3 \frac{7}{9}$

Answer 2

To find the equation of a straight line that passes through two points, A(7, 3) and B(-2, 4), we can use the slope-intercept form of the line equation: y = mx + b, where m is the slope and b is the y-intercept.

Step 1: Calculate the slope (m) using the formula: m = (y2 - y1) / (x2 - x1)

Where (x1, y1) = A(7, 3) and (x2, y2) = B(-2, 4):

m = (4 - 3) / (-2 - 7) m = 1 / -9 m = -1/9

Step 2: Use the slope (m) and one of the points (A or B) to find the y-intercept (b). Let's use point A(7, 3):

y = mx + b 3 = (-1/9) * 7 + b 3 = -7/9 + b

Step 3: Solve for b:

b = 3 + 7/9 b = (27 + 7) / 9 b = 34 / 9

Now that we have the slope (m = -1/9) and the y-intercept (b = 34/9), we can write the equation of the line:

y = (-1/9)x + 34/9

So, the equation of the line passing through points A(7, 3) and B(-2, 4) is y = (-1/9)x + 34/9.