Complete This Schedule to Determine Line 3 and Line 6 of Form M1

They are planted such that six bushes form an equilateral triangle with three bushes on each side,equidistantly,and 7th one in the centre. The sides of the triangle are three straight lines with three rose bushes on each line.The lines joining each of the verticies to the mid bush of the respective opposite sides form the other three straight lines with seventh bush in the centre.

Through the point of [math](3,-4)[/math] and with the same [math]x[/math]-intercept as the line [math]2x+3y=4[/math]. How do you write this out in standard form?The simplest way, I think, is to find the coordinates of the [math]x[/math]-intercept and then treat it as an ordinary problem of finding the standard form of a line through two given points.The [math]x[/math]-intercept is the value of [math]x[/math] when [math]y=0[/math]. So we just plug in [math]y=0[/math] and solve for [math]x[/math]:[math]2x + 3\times 0 = 4\quad\Rightarrow\quad x=2\tag*{}[/math]We’ve now reduced the problem to finding the standard form of a line through the points [math](3,-4)[/math] and [math](2,0)[/math]. My preferred approach here is to find the slope-intercept form of the line equation ([math]y=mx+b[/math]) and convert it to standard form. (This doesn’t always work. See the note below for why and what to do about it.) The slope [math]m[/math] is the ratio of the change in [math]y[/math] coordinate between the two points to the change in [math]x[/math] coordinate. It doesn’t matter which order you use for the two points, so long as you are consistent between the numerator and denominator:[math]m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{0 - (-4)}{2 - 3} = -4\tag*{}[/math]With the slope in hand, we can use either point to then solve for the [math]y[/math]-intercept [math]b[/math]. We just plug in the values for [math]x[/math], [math]y[/math], and [math]m[/math] into the slope-intercept form and solve for the [math]y[/math]-intercept. Let’s use the point [math](3,-4)[/math]:[math]y = mx +b\quad\Rightarrow\quad -4 = (-4) \times 3 + b\quad\Rightarrow\quad b = 8\tag*{}[/math]So the equation in slope-intercept form is:[math]y = -4 x + 8\tag*{}[/math]and in standard form this becomes:[math]\boxed{4x + y = 8}\tag*{}[/math]The approach of first finding the slope-intercept form doesn’t always work because the line may be vertical. That is, the [math]x[/math] coordinates of the two points may be equal and you end up dividing by zero when trying to find the slope. But this is actually a simpler case, because standard form equation for a vertical line is:[math]x = C\tag*{}[/math]All we have to do then is pick one of the points, use the [math]x[/math] coordinate for [math]C[/math] and we’re done.

You need to find the slope (m) of the line. Remember that lines that are perpendicular have a slopes that are opposite reciprocals of each other.To find the slope of 5x - 3y = 6 you can use the formula: m = -A/B for the standard form Ax + By = CA = 5 and B = -3 so slope = -5/-3 which simplifies to 5/3.If you don't have that formula memorized, just solve the equation for y to use slope-intercept form.5x - 3y =6-3y = -5x + 6y = (-5/-3)x + (6/-3)y = (5/3)x - 2Slope = 5/3Opposite of 5/3 is -(5/3) and the reciprocal of 5/3 is 3/5, so the perpendicular slope is -(3/5).Find the equation of a line that passes through the point (0,0) which is the origin and has the slope of -(3/5).The next step is a personal preference but I prefer starting with point-slope form and going from there.y - y1 = m (x - x1)y1 = 0, x1 = 0, m = -(3/5)y - 0 = -(3/5) ( x - 0)y = -(3/5)x(3/5)x + y = 03x + 5y = 0The power of mathematics is noting patterns. See the similarities in the question and solution (5x - 3y = 6 and 3x + 5y = 0) ? Is this always the case for this type of problem? What if the lines were parallel? What if the new line was going through a point that was not the origin? I won't spoil your discovery with explanations so I encourage you to explore the possibilities!

Slope intercept requires two pieces of information a slope, m, and a y-intercept, b, it is written y = mx + b.You are fortunate because you are told the line must pass through the origin, (0,0). The y-intercept is always the y value when x is 0, but we already know this value is 0.Next we must find the slope. To do this isolate y in your standard equation.3x + 2y = 6, subtract 3x from both sides2y = -3x + 6 divide both sides by 2y = (-3/2)x + 3, so the slope, or m is (-3/2).y = (-3/2)xCheck: 0 = (-3/2)(0) => 0 = 0

-y=-x+3 → y=x-3. So the slope is the coef. of x (that is, 1) and the y-intercept is -3.That’s the usual was they present the line: y=f(x). What the *? The only line parallel to (y=x-3) and has a y-intercept of -3 is that line! (x-y coordinate system)

A vector equation for [math]y=3x+5\,\,[/math] is[math]\hspace{7ex}\langle x,y\rangle=\langle 0,5\rangle+t\langle1,3\rangle\,.[/math]There is no Cartesian equation for a line in 3-space. The symmetric equations for the given line are[math]\hspace{8ex}\dfrac{x-2}{-1}=\dfrac{y+1}{5}=\dfrac{z-3}{6}\,.[/math]

This is my 2000th Quora answer!We can do calculus on algebraic curves using only middle school math. I’ll go slow so all you college students can keep up.Let’s write[math]f(x,y) = x^3 - y^2 [/math]so our curve is[math]f(x,y)=0[/math]Now, the middle school Taylor expansion around [math](r,s)[/math] is just a bit of algebra:[math]f(x,y)=f(r+(x-r),s+(y-s)) = (r+(x-r))^3 - (s+(y-s))^2 [/math][math]f(x,y) = r^3 + 3r^2(x-r) + 3r(x-r)^2 + (x-r)^3 - s^2 - 2s(y-s) - (y-s)^2[/math]The essence of differential calculus is that if [math]x-r[/math] is small [math](x-r)^2[/math] is smaller still. Let’s sort by degree:[math]f(x,y) = r^3 - s^2 + 3r^2(x-r) - 2s(y-s) + 3r(x-r)^2 - (y-s)^2 + (x-r)^3[/math]We get the tangent line by ignoring all the square and higher terms in [math]x-r[/math] and [math]y-s.[/math] (We get the tangent conic by ignoring just the cubic term at the end, but it’s embarrassing when a middle schooler can do things the college calculus student can’t, so let’s not go there.) Note [math]r^3-s^2=f(r,s)=0[/math] since by assumption [math](r,s)[/math] is on the curve. So our tangent line is[math]0 = 3r^2(x-r) - 2s(y-s)[/math]The normal line is perpendicular to the tangent. It flips the coefficients on [math]x-r[/math] and [math]y-s,[/math] negating one, but still has to go through [math](r,s)[/math]. So it must be[math]0 = 2s(x-r) + 3r^2(y-s)[/math]Let’s try these at [math](r,s)=(1,1).[/math]Tangent: [math]0=3(x-1)-2(y-1)=3x-2y-1[/math]Normal: [math]0=2(x-1)+3(y-1)=2x+3y-5[/math]It’s my 2000th answer, so let’s go there. The tangent conic at [math](1,1)[/math] is a hyperbola: [math] 0= 3(x-1) - 2(y-1) + 3(x-1)^2 - (y-1)^2.[/math] Pretty good for middle school!Check: Alpha