TwoPoint Form is used to generate the Equation of a straight line passing through the two given points. (y-y1/y2-y1 = x-x1/x2-x1) Examples: Find the equation of the line joining the points (3, 4) and (2, -5) 2021-11-14. Oskar lundgrens väg 3 i mölnlycke; The slope of any line is the inclination of the line with x axis; and;

In this very article, we are going to discuss various forms of the equation of a line. A coordinate plane consists of an infinite number of points. If we consider a point Px,y in a 2d plane and a line named it as N. Then what we will determine is that the point we consider lies on the line L or it lies above or below of the line. That’s when straight-line comes into this scenario. Here we will include the important topic related to the equation of a line in different forms. Forms of the Equation of the LineBased on the parameters known for the straight line, there are 5 forms of the equation of a line that is used to determine and represent a line's equationPoint Slope Form –This form requires a point on the line and the slope of the line. The referred point on the line is x1,y1 and the slope of the line is m. The point is a numeric value and represents the x coordinate and the y coordinate of the point and the slope of the line m is the inclination of a line with the positive m can have a positive, negative, or zero slope. Hence, the equation of a line is as follows y - y11 = m x - x11Two Point Form –This form is a further explanation of the point-sloon of a line passing through the two points - x11, y11, and x22, y22 is in this wayy−y1=y2−y1x2−x1x−x1y−y1=y2−y1x2−x1x−x1Slope Intercept Form –The slope-intercept form of the line is y = mx + c. And here, 'm' is the slope of the line and 'c' is the y-intercept of a line. This line cuts the y-axis at the point 0, c, where c is the distance of this point on the y-axis from the slope-intercept form is an important form and has great applications in the different topics of = mx + cIntercept Form –The equation of a line in this form is formed with the x-intercept a and the y-intercept b. The line cuts the x-axis at a point a, 0, and the y-axis at a point0, b, and a, b are the respective distances of these points from the origin. While these two points can be substituted in a two-point form and simplified to get this intercept form of the equation of a intercept form of the equation of the line explains the distance at which the line cuts the x-axis and the y-axis from the Form –The normal form is based on the line perpendicular to the given line, which passes through the origin, is known as the the parameters of length of the normal is 'p' and the angle made by this normal is 'θ' with the positive x-axis is useful to form the equation of a line. The normal form of the equation of the line is in this wayxcosθ + ysinθ = PDifferent Forms of the Equation of a Straight LineA. Equation of Line Parallel to the y-axisEquation of a straight line which is parallel to the y-axis at a distance of a’ then the equation of y-axis will be x=a here a’ is a coordinate in the plane.Consider this example Equation of line parallel to y-axis for coordinate 7,8 is x=8 B. Equation of Line Parallel to the x-axisEquation of a straight line if the straight line is parallel to the x-axis the equation will be y=a where a’ is an arbitrary understand one can consider this example, consider this a point 9,10 Equation of line parallel to the x-axis is x=9 C. Point- slope Form of an EquationLet a line passing through a particular point QX1, Y1 and PX, Y be any point present in the mentioned slope of a line= Y - Y1/X – X2And by the definition m is the slope,Hence, m = Y - Y1/X – X2On comparing Y – Y1 = mX – X1 is the required point-slope form equation of a line D. Equation of the Line in Two-point FormConsider an arbitrary constant Px,y present in the line L and the Line L passes through two points Ax1,y1 and Bx2,y2. We consider m’ as the slope of the line y2-y1 / x2- x1Then the equation of the line isy2-y1 = mx2-x1Substituting the value of m we gety-y1={ y2- y1/ x2-x1}x-x1Equation of the required line in two point form is y - y1= y2- y1/ x2 - x1x -x1.E. Equation of a Line in Intercept FormLet AB line cuts intercept on the x-axis at a, 0 and on the y-axis at 0, bFrom two-point form y = -b/a x – a y = b/a a – x x/ a + y/b = 1 is the required equation of line in intercept formExampleConsider finding the equation of a line which has made an intercept of 4 in x axis and has made a cut of y-axis in the graphSolutionSo, b = -3 and a = 4 x/4 + y/-3 = 1 3x – 4y = 12 hence the required equation of a line in intercept formSlope-intercepts Form of a LineConsider a line L whose slope be m which cuts an intercept on the y-axis at the distance of a’. hence the point is 0, aHence, the required equation is y – a = mx – 0 y = mx + a which is the required equation of a the equation of a line which has a slope of -1 and has an intercept of 4 units in the positive section of the m = -1 and a = -4Substituting this value in y = mx + a we get y = -x – 4 x + y + 4 = 0Solved ExamplesExampleDetermine the equation of a line which passes through the point -4, -3 and it is parallel to the m = 0, X1 = -4, Y1 = the above equation Y + 3 = 0X + 4 Y = -3 is the required equationExampleFind the equation of the line joining by the points 4,-2 and -1,3.Solution here the two given points are X1,Y1 = -1,3 and X2,Y2= 4,-2Equation of line in two point form is y – 3 = { 3 – - 2/ -1 – 4 } x+1 - x – 1 = y – 3 x + y – 2 = 0.

Findstep-by-step Linear algebra solutions and your answer to the following textbook question: Prove that (x1, y1), (x2, y2) , and (x3, y3) are collinear points if and only if [x1 y1 1, x2 y2 1, x3 y3 1] = 0.

Math Physics Chemistry Graphics Others Area Fun Love Sports Engineering Unit Weather Health Financial Currency Two Point Form is used to generate the Equation of a straight line passing through the two given points. Formula Two point Form y-y1/y2-y1 = x-x1/x2-x1 Examples Find the equation of the line joining the points 3, 4 and 2, -5. x1 = 3, y1 = 4, x2 = 2, y2 = -5 Apply Formula y-y1/y2-y1 = x-x1/x2-x1 y-4/-5-4 = x-3/2-3 y-4/-9 =x-3/-1 -1y-4 = -9x-3 1y-4 = 9x-3 y-4 = 9x – 27 y-9x = -27 + 4 y-9x = -23 9x-y=23 Therefore equation of the line is 9x-y=23 AdBlocker Detected!To calculate result you have to disable your ad blocker first. y(x) = Y1 + S* (x-X1) The line passing through two points (X1,Y1) and (X2,Y2) is. y (x) = Y1 + (Y2-Y1)* (x-X1)/ (X2-X1) The line crosses the y-axis at. Y0 = (X2*Y1-X1*Y2)/ (X2-X1) Alternate form of the line on the xy plane is. (X2-X1)*y - (Y2-Y1)*x = X2*Y1-X1*Y2 = constant. Share. Improve this answer.

Prévia do material em textoCurso de Álgebra Linear Abrangência Graduação em Engenharia e Matemática - Professor Responsável Anastassios H. Kambourakis Exercícios de Álgebra Linear - Lista 02 – Espaços vetoriais 1. No conjunto V={x , y / x , y ∈IR}. Definimos as operações de * Adição x1 , y1 + x2 , y2 = x1 + x2 , 0; *Multiplicação kx , y = kx , ky, ∀ k ∈IR. Verificar se, nessas condições, V é um espaço Vetorial. Dizemos que um conjunto V é um espaço vetorial quando neste conjunto vale as oito propriedades, a de adição e a de multiplicação. Adição A1 u+v=v+u x1,y1+x2,y2 = x2,y2+ x1,y1 x1+x2 , 0 = x2+x1 , 0, Vale A1 A2 u+v+w = u+v+w x1,y1+[x2,y2+x3,y3] = [x1,y1+x2,y2]+x3,y3 x1,y1+[x2+x3 , 0] = x1+x2 , 0+x3 , y3 x1+x2+x3 , 0 = x1+x2+x3 , 0, Vale A2 A3 u+0 = u x1,y1+0,0 = x1,y1 x1+0,y1+0 = x1,y1 x1,0 = x1,y1 , não vale A3 A4 u+-u = 0 x1,y1+-x1,-y1 = 0,0 0,0=0,0, Vale A4 Multiplicação M1 λku = λku λ[kx1,y1] = λkx1,y1 λkx1,ky1 = λkx1, λky1 λkx1, λky1 = λkx1, λky1, Vale M1 M2 ku+v = ku+kv K[x1,y1+x2,y2] = kx1,y1+kx2,y2 K[x1+x2 , 0] = kx1,ky1+kx2,ky2 kx1+kx2 , 0 = kx1+kx2 , 0, Vale M2 M3 λ+ku = λu+ku λ+kx1,y1 = λx1,y1+kx1,y1 [λ+kx1,λ+ky1] = λx1, λy1+kx1,ky1 λx1+kx1,λy1+ky1 ≠ λx1+kx1,0, Não vale M3 M4 1u = u 1x1,y1 = x1,y1 x1,y1 = x1,y1 Vale M4 , Para ser um espaço vetorial, é necessário satisfazer as oito propriedades, e como não valem a A3 e M3, não é um espaço vetorial. 2. No conjunto dos pares ordenados de números reais , se definirmos a operação de adição como x1 , y1 + x2 , y2 = x1 + x2 , y1+ y2, e a operação de Multiplicação como kx , y = x , ky, o conjunto V assim definido não é um espaço quais das 8 propriedades não são válidas; Adição A1 u+v=v+u x1,y1+x2,y2 = x2,y2+ x1,y1 x1+x2 , y1+y2 = x2+x1 , y2+y1 , Vale A1 A2 u+v+w = u+v+w x1,y1+[x2,y2+x3,y3] = [x1,y1+x2,y2]+x3,y3 x1,y1+[x2+x3 , y2+y3] = x1+x2 , y1+y2+x3 , y3 x1+x2+x3 , y1+y2+y3 = x1+x2+x3 , y1+y2+y3, Vale A2 A3 u+0 = u x1,y1+0,0 = x1,y1 x1+0,y1+0 = x1,y1 x1,y1 = x1,y1, Vale A3 A4 u+-u = 0 x1,y1+-x1,-y1 = 0,0 0,0=0,0, Vale A4 Multiplicação M1 λku = λku λ[kx1,y1] = λkx1,y1 λx,ky1 = x, λky1 x, λky1 = x, λky1, Vale M1 M2 ku+v = ku+kv K[x1,y1+x2,y2] = kx1,y1+kx2,y2 K[x1+x2 , y1+y2] = x1,ky1+x2,ky2 x1+x2 , ky1+ky2 = x1+x2 , ky1+ky2, Vale M2 M3 λ+ku = λu+ku λ+kx1,y1 = λx1,y1+kx1,y1 [x1,λ+ky1] = x1, y1+x1,ky1 x1,λy1+ky1 ≠ 2x1, λy1+ ky1, Não vale M3 M4 1u = u 1x1,y1 = x1,y1 x1,y1 = x1,y1 , vale M4 Não é um Espaço Vetorial e a propriedade que não vale é a M3. 3. Considerando os Espaços Vetoriais U e V sobre IR, provar que o conjunto W=UxV={u,v / u ∈ U e v ∈ V} é um espaço vetorial em relação às operações; Adição u1 , v1 + u2 , v2 = u1 + u2 , v1 + v2 e Multiplicação ku ,v =ku ,kv. Adição A1 u+v=v+u u1,v1+u2,v2 = u2,v2+ u1,v1 u1+u2 , v1+v2 = u2+u1 , v2+v1 , Vale A1 A2 u+v+w = u+v+w u1,v1+[u2,v2+u3,v3] = [u1,v1+u2,v2]+u3,v3 u1,v1+[u2+u3 , v2+v3] = u1+u2 , v1+v2+u3 , v3 u1+u2+u3 , v1+v2+v3 = u1+u2+u3 , v1+v2+v3, Vale A2 A3 u+0 = u u1,v1+0,0 = u1,v1 u1+0,v1+0 = u1,v1 u1,v1 = u1,v1, Vale A3 A4 u+-u = 0 u1,v1+-u1,-v1 = 0,0 0,0=0,0, Vale A4 Multiplicação M1 λku = λku λ[ku1,v1] = λku1,v1 λku1,kv1 = λku1, λkv1 λku1, λkv1 = λku1, λkv1, Vale M1 M2 ku+v = ku+kv K[u1,v1+u2,v2] = ku1,v1+ku2,v2 K[u1+u2 , v1+v2] = ku1,kv1+ku2,kv2 ku1+ku2 , kv1+kv2 = ku1+ku2 , kv1+kv2, Vale M2 M3 λ+ku = λu+ku λ+ku1,v1 = λu1,v1+ku1,v1 [λ+ku1,λ+kv1] = λu1, λv1+ku1,kv1 λu1+ku1, λv1+kv1 = λu1+ku1, λv1+kv1, Não vale M3 M4 1u = u 1u1,v1 = u1,v1 u1,v1 = u1,v1 , vale M4, portanto é um espaço vetorial. 4. No conjunto dos pares ordenados de números reais, definirmos a operação de adição como x1 , y1 + x2 , y2 = 2x1 –2y1 , -x1+ y1, e a operação de Multiplicação como kx, y = 3ky, -kx. Com estas operações, verificar se V é espaço vetorial sobre IR. Adição A1 u+v=v+u x1,y1+x2,y2 = x2,y2+ x1,y1 2x1+-2y1 , -x1+y1 ≠ 2x2-y2 , -x2+y2 , não vale A1 A2 u+v+w = u+v+w x1,y1+[x2,y2+x3,y3] = [x1,y1+x2,y2]+x3,y3 x1,y1+[2x2+2y2 , -x2+y2] = [2x1-2y1 , -x1+y1]+x3 , y3 2x1-2y1 , -x1+y1 ≠ [2[2x1-2y1-2-x1+y1], não vale A2 A3 u+0 = u x1,y1+0,0 = x1,y1 2x1-2y1 , -x1+y1 ≠ x1,y1, não vale A3 A4 u+-u = 0 x1,y1+-x1,-y1 = 0,0 2x1-2y1 , -x1+y1 ≠ 0,0, não vale A4 Multiplicação M1 λku = λku λ[kx1,y1] = λkx1,y1 3λky1,- λkx1 = 3λky1, -λkx1, Vale M1 M2 ku+v = ku+kv K[x1,y1+x2,y2] = kx1,y1+kx2,y2 K[2x1-2y1 , -x1+y1] = 3ky1,-kx1+3ky2,-kx2 3K-x1+y1 , -k2x1-2y1 ≠ [23ky1-2-kx1 , -3ky1-kx1], não vale M2 M3 λ+ku = λu+ku λ+kx1,y1 = λx1,y1+kx1,y1 [3λ+ky1 , -λ+kx1] = 3λy1, λx1+3ky1 , -kx2 [3λ+ky1 , -λ+kx1] ≠ [23λy1-2-λx1 , -3λky1 –λx1, não vale M3 M4 1u = u 1x1,y1 = x1,y1 3y1,-x1 ≠ x1,y1 , não vale M4 Assim sendo não é um Espaço Vetorial 5. Verificar se são Sub-espaços Vetoriais os seguintes subconjuntos do Espaço Vetorial do IR3 e , em caso negativo, identificar para cada caso, qual item da definição de sub-espaço vetorial não é atendido. Para ser um sub-espaço do R3, devemos ter satisfeitas as seguintes condições i o vetor nulo ∈ IR3, ii o vetor soma u1+u2 de dois vetores de W, ∈ W, iii o vetor obtido pelo produto de um real por um vetor u, ∈ a Uku ,também ∈ W. a W={x, y, z ∈ IR3 / x = 0} i 0=0,0,0 ∈ W ii w1=x1,y1,z1∈ W w1=0,y1,z1 w2=x2,y2,z2∈ W w2=0,y2,z2 w1+w2=0,y1,z1+ 0,y2,z2 = 0, y1+y2 , z1+z2 ∈ W iii kw=k0,y,z=0,ky,kz ∈ W Portanto w é um sub-espaço de R 3 . b W={x, y, z ∈ IR3 / x ∈ Z} i 0=0,0,0 ∈ W ii w1=x1,y1,z1∈ W w1=x1,y1,z1,com x1 ∈ Z w2=x2,y2,z2∈ W w2=x2,y2,z2, com x2 ∈ Z w1+w2= x1,y1,z1+ x2,y2,z2 = x1+x2, y1+y2 , z1+z2, com x1+x2∈ Z, ∈ W iii kw=kx,y,z= kx,ky,kz,não vale pois k∈R e x∈ Z, kx pode w, então w não é um sub-espaço. c W={x, y, z ∈IR3 / y é Irracional} i 0=0,0,0 w, pois y é irracional, então w não é subespaço. d W={x, y, z ∈IR3 / x −3z = 0} i 0=0,0,0 ∈ W, pois 0-30=0, 0=0 ii w1=x1,y1,z1∈ W w1= x1-3z1=0 w2=x2,y2,z2∈ W w2= x2-3z2=0 w1+w2= x1,y1,z1+ x2,y2,z2 = x1+x2, y1+y2 , z1+z2 / x1+x2+-3z1+z2=0 x1+x2+-3z1-3z2=0 x1-3z1+ x2-3z2=0 0+0=0, ∈ W iii kw1=kx,y,z=kx,ky,kz / kx-3kz=0 kx-3z=0 k0=0 0=0, portanto w é um sub-espaço. e W={x, y, z ∈IR3 / a x + b y + c z = 0, com a, b, c ∈ IR} i 0=0,0,0 ∈ W, a0+b0+c0=0 0=0 ii w1=x1,y1,z1∈ W w1= ax1+by1+cz1=0 w2=x2,y2,z2∈ W w2= ax2+by2+cz2=0 w1+w2= x1,y1,z1+ x2,y2,z2 = x1+x2, y1+y2 , z1+z2 / ax1+x2+by1+y2+cz1+z2 =0 ax1+ax2+by1+by2+ cz1+cz2 =0 ax1+by1+cz1+ ax2+by2+cz2=0 0=0, ∈ W iii kw1=kx,y,z=kx,ky,kz /kax+kby+kcz=0 kax+kby+kcz=0 kax+by+cz=0 k0=0 0=0, portanto w é um sub-espaço. f W={x, y, z ∈IR3 / x = 1} i 0=0,0,0 w, pois 1+0+0≠0, então w não é subespaço g W={x, y, z∈ IR3 / x2 + y + z =0} i 0=0,0,0 ∈ W, pois 02+0+0=0 ii w1=x1,y1,z1∈ W w1= x1 2 +y1+z1=0 w2=x2,y2,z2∈ W w2= x2 2 +y2+z2=0 w1+w2= x1,y1,z1+ x2,y2,z2 = x1+x2, y1+y2 , z1+z2 / x1+x2 2 + y1+y2 +z1+z2=0 x1 2 +2 x2 2 + y1+y2 +z1+z2=0 x1 2 +y1+z1+ x2 2 +y2+z2+ w, portanto não é sub-espaço. h W={x, y, z ∈IR3 / x ≤ y ≤ z } i 0=0,0,0 ∈ W, pois 0 0 0 ii w1=x1,y1,z1∈ W x1 y1 z1 w2=x2,y2,z2∈ W x2 y2 z2 w1+w2= x1,y1,z1+ x2,y2,z2 = x1+x2, y1+y2 , z1+z2 / x1+x2 y1+y2 z1+z2 x1+y1+ z1+y2 x2+y2+z2, ∈ W iii kw=kx,y,z=kx,ky,kz/ kx ky kz, w pois nada garante que kx ky kz, pois k é um número real qualquer, portanto w não é um sub-espaço. i W={x, y, z ∈IR3 / x + y ∈ Q} i 0=0,0,0 ∈ W, pois 0+0=0 ∈ Q ii w1=x1,y1,z1∈ W x1+y1 ∈ Q w2=x2,y2,z2∈ W x2+y2 ∈ Q w1+w2= x1,y1,z1+ x2,y2,z2 = x1+x2, y1+y2 , z1+z2/ x1+x2 y1+y2 ∈ Q x1+y1 x2+y2 ∈ Q, ∈ W iii kW=kx,ky,kz/ kx+ky ∈ W kx+y ∈ W, W, pois kx não será necessariamente um número racional. 6. Verificar se é um Espaço Vetorial o conjunto dos vetores W do IR 5 tais que W= { 0, x2 , x3 , x4 , x5 , com xi ∈ IR}. O conjunto w de vetores do R 5 , é um espaço vetorial sobre IR, se estiverem definidas nesse conjunto as seguintes operações fechadas de adição de vetores e multiplicação por um número real. A1 u+v = v+u 0, x2 , x3 , x4 , x5 +0, y2 , y3 , y4 , y5 =0, y2 , y3 , y4 , y5 +0, x2 , x3 , x4 , x5 0, x2 +y2, x3+y3 , x4 + y4 , x5 +y5 = 0, y2 + x2, y3 + x3, y4 + x4, y5 + x5 vale A1 A2 u+v+w=u+v+w 0,x2,x3,x4, x5+[0,y2,y3,y4,y5+0,z2,z3,z4,z5]= [0,x2,x3,x4, x5+0,y2,y3,y4,y5]+0,z2,z3,z4,z5 0,x2,x3,x4 x5+ 0, y2 + z2, y3 + z3, y4 + z4, y5 + z5= 0, x2 +y2, x3+y3 , x4 + y4 , x5 +y5+ 0,z2,z3,z4,z5 0, x2+y2+z2, x3+y3+z3, x4+y4+z4, x5+y5+z5=0, x2+y2+z2, x3+y3+z3, x4+y4+z4, x5+y5+z5 Vale A2 A3u+0=u 0,x2,x3,x4 x5+0,0,0,0,0= 0,x2,x3,x4 x5 0, x2 +0, x3+0 , x4 + 0 , x5 +0 A4u+-u=0 0,x2,x3,x4 x5+ 0,-x2,-x3, ,-x5 =0,x2,x3,x4 x5 0,x2-x2,x3-x3,x4-x4, x5-x5=0,0,0,0,0 valeA4 M1 λku = λku λ [k0,x2,x3,x4 x5] = λk. 0,x2,x3,x4 x5 λ 0,kx2,kx3,k x4,k x5] =0, λkx2, λkx3, λkx4 ,λkx5 0, λkx2, λkx3, λkx4 ,λkx5= 0, λkx2, λkx3, λkx4 ,λkx5 M2 ku+v = ku+kv K[0, x2 , x3 , x4 , x5 +0, y2 , y3 , y4 , y5 ]=K0, x2 , x3 , x4 , x5 +k0, y2 , y3 , y4 , y5 k0, x2 +y2, x3+y3 , x4 + y4 , x5 +y5= 0, kx2, kx3,kx4 ,kx5 +0, ky2 , ky3 ,ky4 , ky5 0, kx2 +ky2, kx3+ky3 , kx4 + ky4 , kx5 +ky5= 0, kx2 +ky2, kx3+ky3 , kx4 + ky4 , kx5 +ky5 Vale M2 M3 λ+ku = λu+ku λ+k. 0,x2,x3,x4 x5 = λ0,x2,x3,x4 x5+k0,x2,x3,x4 x5 0, λ+k. x2, λ+k. x3, λ+k. x4 , λ+k. x5= λ0, λ x2, λ x3, λ x4 ,λ x5+k0, λ x2, λ x3, λ x4 ,λ x5 0, λx2+k x2, λx3+k x3, λx4+k x4, λx5+k x5= 0, λx2+k x2, λx3+k x3, λx4+k x4, λx5+k x5 vale M3 M4 1u = u 10,x2,x3,x4 x5 =0,x2,x3,x4 x5 0,1x2,1x3, 1x4, 1x5 =0,x2,x3,x4 x5 0,x2,x3,x4 x5 =0,x2,x3,x4 x5

X1 X2, Y1, & Y2 Classifications Class X and Y capacitors are also given a number to represent their impulse test rating. The most common are X1 (tested to 4,000 volts), X2 (2,500 volts), Y1 (8,000 volts) and Y2 (5,000 volts). HKeZr.

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