Seea solution process below: Explanation: To find the x-intercept: Substitute \displaystyle{0} for \displaystyle{y} and solve for \displaystyle{x} : \displaystyle\frac{{1}}{{2}}{x}+{2}{y}=-{2} w=1/3(x+y-z) 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 Computeanswers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history x-x1)/(x2-x1) = (y-y1)/(y2-y1) we get: (x-(-2))/(4-(-2)) = (y-3)/(-1-3) that is: x+2/6 = y-3/-4 -4(x+2) = 6(y-3)-4x - 8 = 6y - 18 Thus the equation of the line is: 4x + 6y - 10 = 0 The way to verify this it to substitute the two values above in place of x and y and check that the equation is true. 过点Ax1,y1)和B(x2,y2)两点的直线方程是() - ——[选项] A. y−y1 y2−y1= x−x1 x2−x1 B. y−y1 x−x1= y2−y1 x2−x1 C. (y2-y1)(x-x1)-(x2-x1)(y-y1)=0 D. (x2-x1)(x-x1)-(y2-y1)(y-y1)=0 下列叙述中正确的是() - ——[选项] A. 点斜式y-y1=k(x-x1)适用于过点(x1,y1)且不垂直x轴的任何直线 B. y−y1 x−x1=k表示过点P1(x1,y1)且斜率为k的直线方程 Justifyyour answers. Transcribed Image Text: (X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6) (p). c (x, y, z) = (cx + 6c - 6, cy + 6c - 6, cz + 6c - 6) The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. . 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. There are three major forms of linear equations point-slope form, standard form, and slope-intercept form. We review all three in this are three main forms of linear equals, start color ed5fa6, m, end color ed5fa6, x, plus, start color 1fab54, b, end color 1fab54y, minus, start color 7854ab, y, start subscript, 1, end subscript, end color 7854ab, equals, start color ed5fa6, m, end color ed5fa6, left parenthesis, x, minus, start color 7854ab, x, start subscript, 1, end subscript, end color 7854ab, right parenthesisA, x, plus, B, y, equals, Cwhere start color ed5fa6, m, end color ed5fa6 is slope and start color 1fab54, b, end color 1fab54 is the y-interceptwhere start color ed5fa6, m, end color ed5fa6 is slope and start color 7854ab, left parenthesis, x, start subscript, 1, end subscript, comma, y, start subscript, 1, end subscript, right parenthesis, end color 7854ab is a point on the linewhere A, B, and C are constantsExampleA line passes through the points left parenthesis, minus, 2, comma, minus, 4, right parenthesis and left parenthesis, minus, 5, comma, 5, right parenthesis. Find the equation of the line in all three forms listed of the forms require slope, so let's find that \text{slope}=\maroonC m &= \dfrac{\Delta y}{\Delta x}\\\\ &=\dfrac{5-4}{-5-2}\\\\ &=\dfrac{9}{-3} \\\\ &=\maroonC{-3} \end{aligned}Now we can plug in start color ed5fa6, m, end color ed5fa6 and one of the points, say start color 7854ab, left parenthesis, minus, 5, comma, 5, right parenthesis, end color 7854ab, to get point-slope form, y, minus, start color 7854ab, y, start subscript, 1, end subscript, end color 7854ab, equals, start color ed5fa6, m, end color ed5fa6, left parenthesis, x, minus, start color 7854ab, x, start subscript, 1, end subscript, end color 7854ab, right parenthesisy−y1=mx−x1y−5=−3x−−5y−5=−3x+5\begin{aligned} y-\purpleD{y_1}&=\maroonC mx-\purpleD{x_1} \\\\ y-\purpleD{5}&=\maroonC{-3}x-\purpleD{-5} \\\\ y-\purpleD{5}&=\maroonC{-3}x+\purpleD{5} \end{aligned}Solving for y, we get slope-intercept form, y, equals, start color ed5fa6, m, end color ed5fa6, x, plus, start color 1fab54, b, end color 1fab54y−5=−3x+5y−5=−3x−15y=−3x−10\begin{aligned} y-{5}&=\maroonC{-3}x+{5} \\\\ y-5&=\maroonC{-3}x-15 \\\\ y&=\maroonC{-3}x\greenD{-10} \end{aligned}And adding 3, x to both sides, we get standard form, A, x, plus, B, y, equals, Cy, plus, 3, x, equals, minus, 10Want to practice the different forms yourself? Check out this a more in-depth review of each form? Check out these review articlesSlope-intercept form reviewPoint-slope form reviewStandard form review 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. Chirag - that line of code plot[x0i,x1i],[y0i,y1i]is using the square brackets to concatenate two elements together to create two 1x2 arrays. These arrays, or coordinates, are then used to plot a line with an origin of x0i,y0i and an end point of x1i,y1i. Put a break point at this line and run the above code. When the debugger pauses at this line, look at the inputs coordinates and see how they are used to draw the line on the figure for each iteration of the loop. Página Inicial > Cálculo > Listas de Cálculo > EDOs LinearesExercícios Resolvidos de EDOs LinearesVer TeoriaEnunciadoPasso 1Oiee! Essa questão parece muito sinistra, mas não precisa se preocupar! Com o nosso passo a passo vamos perceber que ela não é um monstro de 7 cabeças. Temos aqui uma EDO linear de primeira ordem que tem esse formato aqui y ' = A x y = B x Show! Nossa equação é y ' - x y = 1 - x 2 e x 2 2 Comparando essas equações temos que A x = - x B x = 1 - x 2 e x 2 2 Vamos partir para o método. Passo 2Vamos começar calculando a ∫ A x d x ∫ A x d x = ∫ - x d x ∫ A x d x = - x 2 2 Passo 3E, como I x = e ∫ A x d x I x = e - x 2 2 Não tem muito o que mexer, vamos deixar assim mesmo! Passo 4Agora vamos passar para o próximo passo que é calcular ∫ I x B x d x ∫ I x B x d x = ∫ e - x 2 2 1 - x 2 e x 2 2 d x Como a gente tem dois e elevados a alguma coisa vamos juntar eles e somar os expoentes ∫ e - x 2 2 + x 2 2 1 - x 2 d x Opa, eles vão zerar, que beleza! Então vamos ficar com ∫ e 0 1 - x 2 d x Como e 0 = 1 , que nos dá ∫ 1 - x 2 d x Podemos separar em duas integrais ∫ 1 d x - ∫ x 2 d x E resolvendo teremos ∫ I x B x d x = x - x 3 3 Passo 5Agora que já achamos todas os nossos coeficientes, vamos lembrar a fórmula que vai dar a nossa solução geral. y x = 1 I x ∫ I x ⋅ B x d x + C Substituindo o que encontramos nos outros passo e lembrando que C é uma constante real. y x = 1 e - x 2 2 x - x 3 3 + C Lembrando que se temos algo assim 2 x - 2 Podemos escrever como 2 x 2 Então, podemos passar esse e - x 2 2 para cima mudando o sinal do expoente, ficando com y x = e x 2 2 x - x 3 3 + C Passo 6Show achamos a equação geral, mas a nossa jornada ainda não acabou, porque temos um Problema de Valor Inicial, que diz que y 0 = 0 , ou seja, quando x = 0 , temos que y = 0 . Então, vamos substituir esses valores na nossa equação para encontrar o valor da constante C . 0 = 1 . 0 - 0 3 3 + C C = 0 Agora a gente pega a solução geral que tínhamos e substitui o valor de C que acabamos de encontrar. Logo a solução do PVI será y x = e x 2 2 x - x 3 3 Só uma observação antes de terminar não é sempre que a nossa constante vai dar zero beleza? Nesse caso deu por coincidência, mas ele pode ser qualquer outro valor, por isso não podemos esquecer dele 😊 RespostaVer TambémVer tudo sobre CálculoLista de exercícios de EDOs Lineares

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