Every line has an equation that can be written in the general form Ax + By + C = 0, where A, B, and C are real numbers, and A and B are not both 0.
First, we'll find the slope of the line:
m = (y2 - y1)/(x2 - x1) = (9 - 4)/(9 - 3) = 5/6. Now,
(y - y1) = m(x - x1) Point-slope form;
(y - 4) = (5/6)(x - 3)
y - 4 = (5/6)x - 15/6 Subtract (5/6)x and add 15/6 to both sides;
y - 3/2 - (5/6)x = 0 arrange terms;
-(5/6)x + y - 3/2 = 0 Multiply by -1 both sides;
(5/6)x - y + 3/2 = 0
Suppose the equation of the line is y = mx + c where m and c need to be determined. The slope of the line = (difference in y-coordinates of the two given points)/(difference in x-coordinates of the two given points) = (-6 - 2)/(0 - 4) = -8/-4 = 2 So m = 2 ie the equation of the line becomes: y = 2x + c where c still needs to be determined. The point (0, -6) is on the line. That is, when x = 0, y = -6. Substituting in the equation, -6 = 2*0 + c so that c = -6 and the equation of the line is y = 2x - 6
A lateral is the correct answer (if your doing the crossword puzzle I find to be very popular)
by synthetic division and quadratic equation
Equation: y-6 = 2/5(x-10)Solution: 5y = 2x+10
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what is the slope of the line containing points (5-,-2) and (-5,3)? 2
the slope is the 'm' in y=mx+b so even if the points aren't given, if there is an equation, then you can find the slope. for example, if you have an equation like this: y=2x+5 the slope is 2 and the y-intercept is 5.
Assuming the point is (3, -6) and the slope 1, the equation is x - y - 9 = 0
Use the equation; y=mx+b where m is the slope Use your 2 points as y and b (intercept)
Suppose the equation of the line is y = mx + c where m and c need to be determined. The slope of the line = (difference in y-coordinates of the two given points)/(difference in x-coordinates of the two given points) = (-6 - 2)/(0 - 4) = -8/-4 = 2 So m = 2 ie the equation of the line becomes: y = 2x + c where c still needs to be determined. The point (0, -6) is on the line. That is, when x = 0, y = -6. Substituting in the equation, -6 = 2*0 + c so that c = -6 and the equation of the line is y = 2x - 6
To find the equation of a line given two points, we first need the coordinates of both points. Assuming the points are (-4, 3) and (7, 5), we calculate the slope (m) using the formula ( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{7 - (-4)} = \frac{2}{11} ). Next, using the point-slope form ( y - y_1 = m(x - x_1) ), we can take one of the points, say (-4, 3), to find the equation: ( y - 3 = \frac{2}{11}(x + 4) ). Simplifying this gives the equation of the line.
If the slope is 2/3 and the coordinate is (2, -1) then the straight line equation is 3y=2x-7
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
If you have two equations give AND one parametric equation why do you need to find yet another equation?
I suggest that the simplest way is as follows:Assume the equation is of the form y = ax2 + bx + c.Substitute the coordinates of the three points to obtain three equations in a, b and c.Solve these three equations to find the values of a, b and c.
You solve an equation containing y.
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.