(a) Find the gradient of the curve \(y=2x^{3}-7x+4\) when \(x=-2\)
(b) A is the point (7, 2) and B is the point (−5, 8).
(i) Calculate the length of AB.
(ii) Find the equation of the line that is perpendicular to AB and that passes through the point (−1, 3). Give your answer in the form \(y=mx+c\).
(iii) AB is one side of the parallelogram ABCD and
- \(\vec{BC}=\begin{pmatrix}-a\\ -b\end{pmatrix}\) where \(a>0\) and \(b>0\)
- The gradient of BC is 1
- \(\left | \vec{BC} \right |= 8\)
Find the coordinates of D.
▶️ Answer/Explanation
(a) Ans: 17
Differentiate to get \(\frac{dy}{dx}=6x^2-7\). At \(x=-2\), gradient is \(6(-2)^2-7=24-7=17\).
(b)(i) Ans: 13.4 or 13.41 to 13.42
Using distance formula: \(\sqrt{(-5-7)^2+(8-2)^2}=\sqrt{144+36}=\sqrt{180}\approx13.416\).
(b)(ii) Ans: \(y=2x+5\)
Gradient of AB is \(\frac{8-2}{-5-7}=-0.5\). Perpendicular gradient is 2. Using point-slope form: \(y-3=2(x+1)\).
(b)(iii) Ans: (5, 0)
Given gradient BC=1 and \(|\vec{BC}|=8\), solve \(-b/-a=1\) and \(a^2+b^2=64\) to get \(a=b=4\sqrt{2}\). Then \(\vec{AD}=\vec{BC}\), so D is \((7-4\sqrt{2}-4\sqrt{2}, 2-4\sqrt{2}-4\sqrt{2})\).
A line joins the points A(-3,8) and B(2, -2)
(a) Find the coordinates of the midpoint of AB.
(b) Find the equation of the line through A and B.
Give your answer in the form y = mx + c.
(c) Another line is parallel to AB and passes through the point (0, 7).
Write down the equation of this line.
(d) Find the equation of the line perpendicular to AB which passes through the point (1, 5).
Give your answer in the form ax + by + c = 0 where a, b and c are integers.
▶️ Answer/Explanation
(a) Ans: (–0.5, 3)
Midpoint formula: \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
Substitute A(-3,8) and B(2,-2): \(\left(\frac{-3+2}{2}, \frac{8+(-2)}{2}\right) = (-0.5, 3)\).
(b) Ans: y = –2x + 2
Slope \(m = \frac{-2-8}{2-(-3)} = \frac{-10}{5} = -2\).
Using point A(-3,8): \(y – 8 = -2(x + 3)\) simplifies to \(y = -2x + 2\).
(c) Ans: y = –2x + 7
Parallel lines have the same slope \(m = -2\).
Using point (0,7): \(y = -2x + 7\).
(d) Ans: x – 2y + 9 = 0
Perpendicular slope: \(m_{\perp} = \frac{1}{2}\).
Using point (1,5): \(y – 5 = \frac{1}{2}(x – 1)\) converts to \(x – 2y + 9 = 0\).