Example Question - angle with axis
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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 the Equation of a Line with a Given Point and Angle
<p>The equation of a line in point-slope form is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line and \((x_1, y_1)\) is a point on the line.</p> <p>We are given the point \((1, 2)\) and that the line makes a \(30^\circ\) angle with the x-axis. The slope \(m\) of the line is the tangent of the angle it makes with the x-axis, thus \(m = \tan(30^\circ) = \frac{\sqrt{3}}{3}\).</p> <p>Substituting the point and the slope into the point-slope form equation, we get:</p> <p>\(y - 2 = \frac{\sqrt{3}}{3}(x - 1)\)</p> <p>Multiplying both sides by 3 to clear the fraction, we get:</p> <p>\(3(y - 2) = \sqrt{3}(x - 1)\)</p> <p>Expanding, we have:</p> <p>\(3y - 6 = \sqrt{3}x - \sqrt{3}\)</p> <p>Adding 6 and \(\sqrt{3}\) to both sides gives us the final equation:</p> <p>\(3y = \sqrt{3}x + 6 - \sqrt{3}\)</p> <p>Or we can express it in standard form \(Ax + By = C\) by rearranging terms:</p> <p>\(\sqrt{3}x - 3y = -6 + \sqrt{3}\)</p>
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