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Without any equality signs the given terms can't be considered to be equations.

Q: Are the lines With the equations y2 x-1 and y-2x 1 perpendicular?

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if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)

The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.

In a graph, you first must find the slopes of both lines using the rise over run method or by using the x2-x1 over y2-y1 method known as slope formula. The lines will be perpendicular if the slope of one line is opposite reciprocal of the other. the opposite reciprocal of 1/2 is -2

Using algebra to determine if two lines are perpendicular to one another we first must determine each line's slope. Select two known points on each line to determine the slope for the line. The Point-slope form of a linear equation is (Y1-Y2) = m(X1-X2). Therefore The slope m = (Y1-Y2)/(X1-X2) We will use these points to generate the slope equation. Line A Line B Point 1 Point 2 Point 1 Point 2 X1,Y1 X2,Y2 A1,B1 A2,B2 If the product of the slopes of two lines = -1 then the two lines are perpendicular. Using the point slope form above the equation would look like this: [(Y1-Y2)/(X1-X2)] X [(A1-A2)/(B1-B2)] = m(line A) X m(line B) Example Line A Line B Point 1 Point 2 Point 1 Point 2 0,0 3,3 3,-3 0,0 Using the above formula [(0-3)/(0-3)] X [(3-0)/(-3-0)] = [-3/-3] X [3/-3] = 1 x -1 = -1 These two lines are perpendicular.

Point-slope form is written as: y-y1=m(x-x1), where (x1, y1) is a point on the line and m is the slope (hence the name, point-slope form).

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if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)

The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.

In a graph, you first must find the slopes of both lines using the rise over run method or by using the x2-x1 over y2-y1 method known as slope formula. The lines will be perpendicular if the slope of one line is opposite reciprocal of the other. the opposite reciprocal of 1/2 is -2

Using algebra to determine if two lines are perpendicular to one another we first must determine each line's slope. Select two known points on each line to determine the slope for the line. The Point-slope form of a linear equation is (Y1-Y2) = m(X1-X2). Therefore The slope m = (Y1-Y2)/(X1-X2) We will use these points to generate the slope equation. Line A Line B Point 1 Point 2 Point 1 Point 2 X1,Y1 X2,Y2 A1,B1 A2,B2 If the product of the slopes of two lines = -1 then the two lines are perpendicular. Using the point slope form above the equation would look like this: [(Y1-Y2)/(X1-X2)] X [(A1-A2)/(B1-B2)] = m(line A) X m(line B) Example Line A Line B Point 1 Point 2 Point 1 Point 2 0,0 3,3 3,-3 0,0 Using the above formula [(0-3)/(0-3)] X [(3-0)/(-3-0)] = [-3/-3] X [3/-3] = 1 x -1 = -1 These two lines are perpendicular.

You need to more precise in your question. There is no guarantee two lines interection in 3-d space. If they do intersect, they must lie in the same plane. And that plane would only need to be 2-D, aside from the trivial case where the lines are the same. If the two lines do intersect then x value for the two lines will be equal, as well as the y value & z value, in the equations of the two lines. Using parametric equations, If line1 is given by: x = x1 + a*t, y = y1 + b*t, and z = z1 + c*t; and line 2 is: x = x2 + d*s, y = y2 + e*s, and z = z2 + f*s, where x1,y1,z1 & x2,y2,z2 are starting points [a,b,c] & [d,e,f] are direction vectors, and s & t are the parameters. Since the x values must be equal, set x1+a*t = x2+d*s, and y1+b*t = y2+e*s, then you have two equations and two unknowns. Solve for t and s, then substitute into the z equations to find the z coordinate (they will both come up with the same z-value if indeed the lines do intersect). Even if both equations use 't' as the parameter, you need to treat them as two independent variables(so change one of them to s), since the lines could be changing differently as the parameter variable changes.

Point-slope form is written as: y-y1=m(x-x1), where (x1, y1) is a point on the line and m is the slope (hence the name, point-slope form).

Regula-Falsi Method evaluates using assumed variables like "a", "b", f(a), f(b) Secant Method Directly works with x1, x2, f(x1), f(x2) Difference is in the Assignment pattern only, otherwise both are used to find root of Non-Linear equations using the same procedure which is: x1= [a * f(b) - b * f(a)]/[f(b)-f(a)] x1= [x0 * f(x1) - x1 * f(x0)]/[f(x1)-f(x0)] Thank You :-)

You solve this type of problem using the following steps. 1) Write your original equation in slope-intercept form, that is, solved for "y". (The line is already in that form in this case). You can read off the slope directly: in an equation of the form: y = mx + b m is the slope. 2) Calculate the slope of the perpendicular line. Since the product of the slopes of perpendicular lines is -1, you can divide -1 by the slope you got in part (1). 3) Use the generic equation y - y1 = m(x - x1), for a line that has a given slope "m" and passes through point (x1, y1). Replace the given coordinates (variables x1 and y1). Simplify the resulting equation, if required.

Equation of a line may be written as y = mx + c. m is called the slope of the line. c is the point where the line crosses the y axis. If two points are given: (x1, y1) and (x2, y2), m is calculated as the y difference divided by the x difference: m = (y2 - y1) / (x2 - x1) Once you find m, you can find c by putting in the values into the equation y=mx+c. For example, if you use (x1, y1), you can do this: y1 = m*x1 + c take m*x1 to the other side: y1 - m*x1 = c Then you get the value of c. Now you have both m and c, so you can write the equation of the line: y = mx + c Put the values of m and c in. Leave y and x as it is. a.net/math_problems/equations-of-lines-problems-with-solutions.html

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Assuming you want the equation of a straight line through (-2, -4) and (4, -1): y - y0 = m(x - x0) where m = (y1 - yo) ÷ (x1 - x0) ⇒ y - (-4) = (-1 - (-4))/(4 - (-2))(x - (-2)) ⇒ y + 4 = 3/6 (x + 2) ⇒ y = x/2 - 3 or 2y = x - 6 Equations of higher order (eq quadratics) have infinitely many equations.

xn=x1+(n-1)v^t and Pn=P1+(n-1)iP1