Example Question - line equation
Here are examples of questions we've helped users solve.
Question on Sequence Terms from Binomial Expansion and Line Equations
<p>For the binomial expansion part, to find the 4th term from the end in the expansion of $\left( \frac{3}{2} - \frac{x^3}{6} \right)^7$, note that the 4th term from the end is equivalent to the 4th term from the beginning (or the $T_4$ term) because the binomial is symmetric.</p> <p>To find $T_4$, we use the binomial theorem which states $T_{k+1} = ^nC_k \cdot a^{n-k} \cdot b^{k}$.</p> <p>$T_4 = ^7C_3 \left( \frac{3}{2} \right)^{7-3} \left( -\frac{x^3}{6} \right)^3$</p> <p>$T_4 = 35 \cdot \left( \frac{3}{2} \right)^4 \cdot \left( -\frac{x^3}{6} \right)^3$</p> <p>$T_4 = 35 \cdot \frac{81}{16} \cdot -\frac{x^9}{216}$</p> <p>$T_4 = -\frac{35 \cdot 81 \cdot x^9}{16 \cdot 216}$</p> <p>$T_4 = -\frac{35 \cdot 81 \cdot x^9}{6^3 \cdot 6}$</p> <p>$T_4 = -\frac{35 \cdot 81 x^9}{6^4}$</p> <p>$T_4 = -\frac{945 \cdot x^9}{1296}$</p> <p>For the equation of lines passing through (1,2) and making an angle $30^\circ$ with the y-axis, the slope of the line is the tangent of the angle with the x-axis. Since the line makes a $30^\circ$ angle with the y-axis, it makes a $60^\circ$ angle with the x-axis. Hence, the slope $m$ is $\tan(60^\circ) = \sqrt{3}$.</p> <p>The equation of a line in slope-intercept form is $y = mx + b$.</p> <p>To find $b$, substitute $(1,2)$ (x, y) into the equation:</p> <p>$2 = \sqrt{3}(1) + b$</p> <p>$b = 2 - \sqrt{3}$</p> <p>Therefore, the equation of the line is:</p> <p>$y = \sqrt{3}x + (2 - \sqrt{3})$</p>
Finding Equation of Parallel Line Passing Through a Point
To find the equation of line r that is parallel to line q and passes through the point (-6, 1), we start by determining the slope of line q. The equation of line q is given in slope-intercept form as: y = -5 - 1/2(x + 2) In slope-intercept form, which is y = mx + b, m represents the slope and b represents the y-intercept. Based on line q's equation, the slope (m) is -1/2. Since line r is parallel to line q, line r will have the same slope as line q. Therefore, the slope of line r will also be -1/2. Using the point-slope form of a line equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope, we can substitute the slope and the point through which line r passes. The point (-6, 1) will be our (x1, y1), and our slope (m) will be -1/2. y - y1 = m(x - x1) y - 1 = -1/2(x - (-6)) y - 1 = -1/2(x + 6) Now, we solve for y to get the equation in slope-intercept form. y = -1/2 * x - 1/2 * 6 + 1 y = -1/2 * x - 3 + 1 y = -1/2 * x - 2 So, the equation of line r in slope-intercept form is: y = -1/2x - 2 This equation uses numbers written as simplified proper fractions, improper fractions, or integers, as requested.
Method to Find Line Equation Given Intersecting Points
很抱歉，图片中并没有提供直接的图像来帮助求解。不过，根据题目描述中的信息，我们可以找到解决这个问题的方法。 题目描述了一条直线 L 通过点 P(-10, -9) 并且分别与 x 轴和 y 轴相交于点 H 和点 K。我们知道，直线与 x 轴的交点会有形如 (a, 0) 的坐标，与 y 轴的交点会有形如 (0, b) 的坐标，其中 a 和 b 分别是 x 轴和 y 轴上的截距。 题目还给出了一个条件，即 3OH = 2OK。这里的 OH 和 OK 分别是原点到交点 H 和 K 的距离。 我们可以用点到原点的距离的公式来表示 OH 和 OK： OH = √(a^2 + 0^2) = |a| OK = √(0^2 + b^2) = |b| 因此，3|a| = 2|b| ，由于 a 和 b 是截距，我们可以假设二者都是正值，所以可以直接写作： 3a = 2b 接下来要做的是用一个已知点 (-10, -9) 和截距点 (a, 0) 和 (0, b) 来找到直线 L 的方程。我们可以用两点式来写出这个直线方程： (y - y1)/(y2 - y1) = (x - x1)/(x2 - x1) 把点 P(-10, -9) 和 (a, 0) 代入此方程（这里假设直线 L 与 x 轴相交于点（a, 0））： (y - 0)/(-9 - 0) = (x - a)/(-10 - a) 简化得到： y/-9 = (x - a)/(-10 - a) 进一步简化得到： 9(x - a) = -y(10 + a) 我们也可以用点 P(-10, -9) 和 (0, b) 来写出方程： (-9 - 0)/(y2 - y1) = (-10 - 0)/(x2 - x1)，即 (-9)/b = (-10)/a 现在我们有了两个方程： 1. 3a = 2b 2. 9a = 10b 由第二个方程我们可以得到 b = 9a/10。将这个代入第一个方程得到： 3a = 2(9a/10) 3a = 18a/10 10 = 6a a = 10/6 = 5/3 将 a 的值代入 b = 9a/10 得到： b = 9(5/3)/10 b = 45/30 b = 3/2 所以，点 H 的坐标是 (5/3, 0)，点 K 的坐标是 (0, 3/2)。 这个解答假设了截距都是正值，实际上因为给出的点 P(-10, -9) 都有负坐标，截距应该也是负值。这里的解答是为了示范解决问题的方法，如果有具体图像信息，答案可能会有所不同。
Understanding a Line Equation in Slope-Intercept Form
It seems you've provided an image with an equation of a line: \( y = \frac{1}{2}x - 4 \). This equation is already in slope-intercept form, which is given by \( y = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept. From the given equation: - The slope (\( m \)) of the line is \( \frac{1}{2} \). - The y-intercept (\( b \)) of the line is \( -4 \). If you need to solve for a specific value, you would need to substitute that value for \( x \) and solve for \( y \), or vice versa. However, since there is no specific request on what to solve for, this is as much as can be said about the given equation. If you have a specific question about this equation, such as finding a point on the line or graphing the line, please provide further information.
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