Which Of The Lines Graphed In The Diagram Represents The Equation 6x 4y = 8

Given the equation of a line 5x + 3y=15, at what point does the graph of the line intersect the y-axis?

First of all, we know that any point (x,y) on the y-axis has zero (0) as its x-coordinate; therefore, the x-coordinate of the point of intersection of the line whose equation is 5x + 3y = 15 and the y-axis is 0.We also know that the coordinates (x, y) of any point on the given line has to satisfy its equation:  5x + 3y = 15.Therefore, we can find the y-coordinate of the point of intersection of the line whose equation is 5x + 3y = 15 and the y-axis as follows:5x + 3y = 155(0) + 3y = 150 + 3y = 153y/3 = 15/3y = 5Therefore, the line whose equation is 5x + 3y = 15 intersects the y-axis at the point (0, 5).

How do I find an equation of 2 lines from a pair of straight lines?

So, you have a general 2nd degree equation as-[math]ax^2+by^2+2hxy+2gx+2fy+c=0[/math]As far as we're concerned, it could represent a circle, an ellipse, a parabola, a hyperbola or a pair of straight lines.We need to find a relation between [math]\:x\:[/math] and [math]\:y\:[/math] to find out what it represents. So, treat the equation as a quadratic in one variable.[math]ax^2+(2hy+2g)x+(by^2+2fy+c)=0[/math]Correlate this with a general quadratic equation,[math]AX^2+BX+C=0[/math]And use the Quadratic Formula-[math]x=\dfrac{-2hy-2g}{2a}\pm\:\dfrac{\sqrt{(2hy+2g)^2–4a(by^2+2fy+c)}}{2a}[/math][math]\Rightarrow x=\dfrac{-hy-g}{a}\pm\:\dfrac{\sqrt{h^2y^2+g^2+2hgy-aby^2–2afy-ac}}{a}[/math]Now, the term inside the square root needs to be a perfect square, only then can we have a linear relation between [math]\:x\:[/math] and [math]\:y.[/math] As you can see, the term there also happens to be a quadratic, but in [math]\:y[/math], and so we set it's determinant to zero.Examining this term a bit closely, we get-[math](h^2-ab)y^2+2(hg-af)y+(g^2-ac)=0[/math]Set the determinant to zero as discussed above:[math](hg-af)^2=(h^2-ab)(g^2-ac)[/math]Expand it, and you'll see that:[math]abc+2gfh-af^2-bg^2-ch^2=0,\:\: a\neq 0[/math]So, we conclude, as long as this condition is satisfied, the equation represents a pair of straight lines.Now that the term inside the square root is a perfect square, it's root can be found as follows-[math](h^2-ab)y^2+2(hg-af)y+(g^2-ac)=0[/math][math]\Rightarrow y=\dfrac{-2(hg-af)}{2(h^2-ab)}[/math]So the equation reduces to-[math]{\big(y+\dfrac{hg-af}{h^2-ab}\big)}^2=0[/math]Plug that into the equation in [math]\:x[/math]-[math]x=\dfrac{-hy-g\:\pm\big(y+\dfrac{hg-af}{h^2-ab}\big)}{a}[/math]Well, these are the equations you needed. You just need to expand them a bit further, which I'm not doing here.

How do I find the equation of the diameter of the circle given by x²+y²-12x+4y+6=0?

First, recognize that the given equation, x² + y² ‒ 12x + 4y + 6= 0, is the general form for the equation of a circle. To find the diameter d of the given circle, we’ll have to first convert the given equation, x² + y² ‒ 12x + 4y + 6 = 0, from its current  less informative and less useful general form into the more desired standard formfor the equation of a circle by using the method of “completing the square” on its x-terms and on its y-terms as follows: x² + y² ‒ 12x + 4y + 6 = 0 (given) Collecting and grouping the x-terms and y-terms together, we get:(x² ‒ 12x    ) + (y² + 4y    ) + 6= 0 “Completing the square” in each quadratic grouping, we have: (x² ‒ 12x + 36) + (y² + 4y + 4) + 6 = 0 + 36 + 4(x² ‒ 12x + 36) + (y² + 4y + 4) + 6 ‒ 6 = 0 + 36 + 4 ‒ 6(x² ‒ 12x + 36) + (y² + 4y + 4) = 34Factoring on the left side,  we have: (x ‒ 6)² + (y + 2)² = 34  is the standard form for the equation of                                                a circle i.e., (x ‒ h)² + (y ‒ k)² = r², with                                                center (h, k) and radius r, where, for this                                               problem, h = 6, k = ‒2, and radius r = √(34).The radius of a circle is a line segment joining  the center of the circle to a point on the circle, and the diameter of a circle is a line segment that passes through the center of the circle and joins two points on the circle; therefore,  the length of the diameter is twice the length of the radius for any given circle, and, consequently, the formula for finding the diameter of a circle is: d = 2r; therefore, the diameter of the given circle is found as follows:d = 2rd = 2√(34)  units is the exact diameter of the given circle.ord = 2(5.8309519) (the second factor is rounded to 7 decimal places)d = 11.662 units is the diameter of the given circle (rounded to 3 decimal places).Also, the equation for the diameter d of the given circle is: d = 2r,   but r = √(34) = √[(x‒ 6)² + (y + 2)²]; Substituting, we get: d= 2√[(x ‒ 6)² + (y + 2)²], where (x, y) is any point on the given circle.

How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?

There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.

How can one find the values of k for which the line 2x -k is tangent to the circle with the equation x^2 + y^2 = 5?

The simple way to do this is to clearly define what  it means for tangent so that finding the k values is the easiest. Problem Specific AnswerWe have [math]y = 2x - k[/math]  and [math]x^2 + y^2 = 5[/math] and the line is tangent to the circle. What it means for a 2 things to be tangent is that one point and only one point satisfies both equations at the same time. That means that in the above two equations,  only the point represented by  [math](x,y)[/math] satisfies both equations. We get:  [math]y = 2x - k[/math] and [math]x^2 + y^2 = 5 [/math]Substituting, we get [math]x^2 + (2x-k)^2 = 5.[/math] This equation must only have one solution. [math]5x^2 -4kx +(k^2 - 5) = 0.[/math]This quadratic equation must have its determinant equal to 0 in order for the two to be tangent. [math]b^2 - 4ac = 0[/math][math]16k^2 - 20k^2+100 = 0[/math][math]4k^2 = 100[/math][math]\boxed{k = \pm 5}[/math]Generic AnswerWe can generalize this solution for any line and circle.Line: [math]y = mx+b[/math]Circle: [math]x^2 + y^2 = r^2[/math]Substituting we get: [math]x^2 + (mx+b)^2 = r^2[/math][math]\Rightarrow (m^2 + 1)x^2 + 2mbx + (b^2 - r^2) = 0[/math]Taking the discriminant and making it equal to 0, we get[math]4m^2b^2 - 4(m^2 + 1)(b^2 - r^2) = 0[/math][math]4m^2b^2 - 4m^2b^2 + 4m^2r^2 - 4b^2 + 4r^2 = 0[/math][math]m^2r^2 + r^2 = b^2[/math][math]m, r,[/math] and [math]b[/math] must satisfy these equations in order to be tangent. We can check out previous answer by plugging in here: [math]m = 2; r = \sqrt{5}; b = -k.[/math][math]20 + 5 = k^2.[/math]Therefore, [math]\boxed{k = \pm 5}[/math]