Theparametric equations x = x1 + (x2 - X1), y = y1 + (Y2 - Y1)t where Osts i describe the line segment that joins the points P1(X1,Yı) and P2(X2, Y2). Use a graphing device to draw the triangle with vertices A(1, 1), B(4,4), C(1, 6). Find the parametrization, including endpoints, and sketch to check. (Enter your answers as a comma-separated $\frac{243 x + 243 y + \left(x_{1} - y_{1}\right) \left(- 9 x_{1} y_{1} + 27 x_{2} + 27 y_{2}\right)}{\left(3 x + 3 y\right) \left(- 9 x_{1} y_{1} + 27 x_{2} + 27 y Question The parametric equations x = X1 + (x2 - X1)t, y = Y1 + (y2 - Y1)t where Osts i describe the line segment that joins the points P1(X1, Y1) and P2(x2, Y2). Draw the triangle with vertices A(1, 1), B(5, 4), C(1, 6). Find the parametrization, including endpoints, and sketch to check. (Enter your answers as a comma-separated list of YesNo Maybe. 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: 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). SolutionThe correct option is A Lie on a straight line Explanation for the correct option : As the common ratio of x 1, x 2, x 3 is same as y 1, y 2, y 3, so they can be written as x 1 = a, x 2 = a r, x 3 = a r 2 a n d y 1 = b, y 2 = b r, y 3 = b r 2 So, the points will be . 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. 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 I have a dataframe df with XY combinations as follows > df df X1 Y1 X2 Y2 1 1 16 4 -1 2 2 15 5 -2 3 3 14 6 -3 4 4 13 7 -4 and want to reshape dfto df2by merging X1 and X2to a new variable X adding NA where Y1 or Y2 is left without value. The result would look like this > df2 X Y1 Y2 1 1 16 NA 2 2 15 NA 3 3 14 NA 4 4 13 -1 5 5 NA -2 6 6 NA -3 7 7 NA -4 What is the most efficient way to do this? asked Jan 24, 2020 at 1753 You can use dplyrfull_join df2 <- dplyrfull_joindf[, c"X1", "Y1"], df[, c"X2", "Y2"], by = c"X1" = "X2" namesdf2[1] <- "X" df2 X Y1 Y2 1 1 16 NA 2 2 15 NA 3 3 14 NA 4 4 13 -1 5 5 NA -2 6 6 NA -3 7 7 NA -4 answered Jan 24, 2020 at 1808 dave-edisondave-edison3,6467 silver badges19 bronze badges Using merge from base R mergedf[c'X1', 'Y1'], df[c'X2', 'Y2'], = 'X1', = 'X2', all = TRUE answered Jan 24, 2020 at 1825 akrunakrun871k37 gold badges535 silver badges655 bronze badges Jon P. asked • 01/07/15 problem continued..."and P2x2,y2. Draw the triangle with vertices A1, 1, B4, 3, C1, 7. Find the parametrization, including endpoints, and sketch to check. Enter your answers as a comma-separated list of equations. Let x and y be in terms of t."1A to B2B to C3A to CI don't know where to start on this problem, I do not know what is asking me to find either. I get parametric equations and how they work but this question confuses me. 1 Expert Answer Jon, The statement above makes sense but like you I don't see how that relates to Sorry seems like something is missing. Jim Still looking for help? Get the right answer, fast. OR Find an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. Among all the subjects, mathematics is the most complex subject for most people. The reason behind that is every formula seems complicated initially, but when it is understood properly, mathematics becomes the easiest subject. Every person has their own way of explaining a certain thing and every person has their own pace of learning things. Mathematics gets easier and more complicated depending on the person explaining it. Every formula in mathematics has it own importance and upon changing it even in the slightest manner, it can change everything about it; therefore we have to pay our full attention while learning mathematics. Mathematics has many topics and for every one of them, there is a formula. One of the topics is called Slope. A slope is a numerical measure of a line’s horizontal inclination. The slope of a ray, line or any line segment is basically the ratio of the vertical to the horizontal distance between two points, this geometry is called analytic geometry. A slope can also be called a Tangent or a Gradient. To find the slope of the straight line the formula is written like m=y2-y1/x2-x1 and it is the right way of putting the values. You can’t change the formula m=x2-x1/y2-y1 because it might result in complete failure as it isn’t the right way. Check out this video to learn how to use the formula in a problem. The difference between y2,y1,x2,x1 and x2,x1,y2,y1 is that both of these are used for different situations. To find the slope y2,y1,x2,x1 is used which is written like m=y2-y1/x2-x1 and to find the distance between two points x2,x1,y2,y1 is used which is written like d=√x2-x1²+y2-y1². You can merely switch the values of x1 and y2 with x2 and y2 respectively. Have a quick look at this video for a better understanding How to find the equation of a line If we don’t want to get technical, you can say that y2,y1,x2,x1, and x2,x1,y2,y1 have merely switched their positions. If you know the formulas to find the slope and to find the distance between two points, it doesn’t matter if y2,y1,x2,x1 is written like x2,x1,y2,y1 or vice versa. What does y2 y1 x2 x1 mean?Do x1 y1 and x2 y2 numbers matter?What is y1 x1 y2 x2 called?What happens when you change the formula?To ConcludeOther Articles What does y2 y1 x2 x1 mean? You will find the y2 y1 x2 x1 formula in almost every mathematics book and every one of them describes this the same way. As you must know, a rectangular or Cartesian plane has two lines that intersect at right angles at the point O which is called the origin. The horizontal axes are called the x-axis and the vertical axes are called the y-axis. As every problem has its own formula, to find the slope you have to use a formula which is written as m=y2-y1/x2-x1, you can only change the values of x1 and y1 with x2 and y2 respectively, anymore changes can result in complete failure. Moreover, the slope of a straight line can be positive, negative, zero, or undefined. If y2 – y1 and x2 – x1 have the same signs then the slope of the straight line will be positive. Do x1 y1 and x2 y2 numbers matter? Wrong coordinates will result in wrong answers. Yes, they do matter, to know what are the coordinates. This way it is easier to put the values in the formula. For example, 3,9 and 7,8 are the coordinates, so we can see that the value of x1 is 3, y1 is 9, x2 is 7, and y2 is 8. This way it gets easier to put the values in a formula in their right places as each coordinate has its own place. Without x1 y1 and x2 y2, you might make mistakes by putting in the wrong coordinates which will, of course, result in wrong answers. Here is the table for different formulas that contains y2,y1,x2,x1 and x2,x1,y2,y1. Name of the FormulaFormulaTo find the distance/length between two pointsd=√x2-x1²+y2-y1²To find the slopem=y2-y1/x2-x1Formulas and their uses What is y1 x1 y2 x2 called? Slopes have many formulas. y1 x1 y2 x2 is called a Slope, although some may refer to them as Gradient. Mathematics can sometimes be challenging as the topic of slope can have many similar formulas. We can mistakenly change the formula which can result in wrong answers. x1 y1 and x2 y2 are the right way which makes y1 x1 and y2 x2 wrong. When you are given a problem that can be 3,9 and 7,8 you have to put the values in a formula, for example, the formula of slope which is m=y2-y1/x2-x1, now how do you know which is the value of x1 x2 and y1 y2. Well, x1 y1 and x2 y2 is the way to know that, basically, the value of x1 is 3, y1 is 9, x2 is 7, and last but not least y2 is 8. What happens when you change the formula? In mathematics, we can’t just change formulas because that can create different outcomes. We can in some cases make changes to the formula, but we aren’t supposed to add anything that doesn’t belong there. For example, in the formula of finding the distance/length between two points d=√x2-x1²+y2-y1² you can merely change the position of x1 and y1 with x2 and y2 respectively. Changing the formula will often result in the wrong answers. If you change the formula by adding in different things, there are a number of outcomes that you can get Wrong but right but wrong answer. These are the reasons why we can’t change formulas as we want. Although you can change them if you are using them for a different problem, we have to seek help from a mathematician as mathematics is quite complex. To Conclude Mathematics tends to get easier or more complicated depending on the person explaining it. As we know, there are many topics in mathematics, and one of them is called Slope. A slope is a numerical measure of a line’s horizontal inclination. The slope/Gradient/Tangent of a ray, line, or any line segment is the ratio of the vertical to the horizontal distance between two points. The difference between y2,y1,x2,x1 and x2,x1,y2,y1 is both of these are used in different situations. To find the slope y2,y1,x2,x1 is used which is written as m=y2-y1/x2-x1 and to find the distance/length between two points x2,x1,y2,y1 is used which is written as d=√x2-x1²+y2-y1². You can’t change the formula because it can give wrong answers, you can only switch the values of x1 and y2 with x2 and y2 respectively. There are many formulas in mathematics and every one of them has its own importance. A rectangular or Cartesian plane has two lines that intersect at right angles at the point O which is known as the origin. The horizontal axes are called the x-axis and the vertical axes are called the y-axis. To know which value is put in a formula x1 y1 and x2 y2 helps immensely. For example, 3,9 and 7,8 are the coordinates, so the value of x1 is 3, y1 is 9, x2 is 7, and y2 is 8. The topic of the slope has many similar formulas. We can mistakenly change the formula which can result in wrong answers. x1 y1 and x2 y2 are the right way and y1 x1 and y2 x2 are wrong. We aren’t supposed to change formulas because it can result in different outcomes which can be both right and wrong. But, yes you can make a few changes within the formula, for instance, in d=√x2-x1²+y2-y1² you can switch x1 and y1 with x2 and y2 respectively, other than that you aren’t supposed to change anything else. Mathematics is difficult, but when you have a firm grasp on the formulas and their uses it can get much easier. Other Articles BASE VS NUCLEOPHILE UNDERSTANDING IMPORTANT FACTSCOORDINATION BONDING VS IONIC BONDING COMPARISON60 WATTS AND 240 OHM LIGHT BULB PHYSICS EXPLAINEDTHE DIFFERENCE BETWEEN A TRAPEZOID & A RHOMBUS Click here to learn more differences when you change the variables in the formula. Álgebra Exemplos Etapa 1Toque para ver mais passagens...Etapa dos dois lados da cada termo em por e para ver mais passagens...Etapa cada termo em por .Etapa o lado para ver mais passagens...Etapa dois valores negativos resulta em um valor o lado para ver mais passagens...Etapa para ver mais passagens...Etapa dois valores negativos resulta em um valor 2Reescreva na forma para ver mais passagens...Etapa forma reduzida é , em que é a inclinação e é a intersecção com o eixo 3Use a forma reduzida para encontrar a inclinação e a intersecção com o eixo para ver mais passagens...Etapa os valores de e usando a forma .Etapa inclinação da linha é o valor de , e a intersecção com o eixo y é o valor de .Inclinação intersecção com o eixo y Inclinação intersecção com o eixo y Etapa 4Qualquer reta pode ser representada graficamente usando-se dois pontos. Selecione dois valores e substitua-os na equação para encontrar os valores para ver mais passagens...Etapa a tabela dos valores e .Etapa 5Desenhe a reta no gráfico usando a inclinação e a intersecção com o eixo y, ou os intersecção com o eixo y

y y1 y2 y1 x x1 x2 x1