To find the equation of a line passing through two points, we first calculate the slope using the formula (y2 - y1) / (x2 - x1). Given the points (1, 11) and (-2, 2), the slope is (2 - 11) / (-2 - 1) = -9 / -3 = 3. Next, we use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substituting (1, 11) as the point and 3 as the slope, we get the equation y - 11 = 3(x - 1). Simplifying, we get y = 3x + 8 as the equation of the line.
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Oh, dude, it's like super easy. So, you can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is your point and m is the slope. First, find the slope using (y2 - y1) / (x2 - x1) = (2 - 11) / (-2 - 1) = -9 / -3 = 3. Then plug in the point (1, 11) and the slope into the equation: y - 11 = 3(x - 1). And there you have it, the equation of the line!
Points: (1, 11) and (-2, 2)
Slope: (11-2)/(1--2) = 3
Equation: y-11 = 3(x-1) => y = 3x+8
Equation: y-2 = 3(x--2) => y = 3x+8
Plug both points into the equation of a line, y =m*x + b and then solve the system of equations for m and b to get equation of the line through the points.
The equation of a line through (-1,7) and (-4,9) is y=-(2/3)*x+19/3
y -x + 1
The equation in slope-intercept form would be y=-2x
The general equation of a line isy = mx + cIf the gradient is 3 then m = 3 so thaty = 3x + cSince the point (4,1) lies on this line, these coordinates must satisfy the equation.So 1 = 3*4 + cie c = -11Therefore, the equation is y = 3x - 11