(a)

\begin{align} \text{Gradient of } AB & = {7 - 5 \over 2 - (-3)} \\ & = {2 \over 5} \\ \\ \tan \theta & = {2 \over 5} \\ \theta & = \tan^{-1} \left(2 \over 5\right) \\ & = 21.801 \\ & \approx 21.8^\circ \text{ (1 d.p.)} \end{align}

(b)

\begin{align} \text{Gradient of } AB & = {6 - (-3) \over 4 - 4} \\ & = {9 \over 0} \text{ (Undefined)} \\ \\ \implies & AB \text{ is a vertical line} \\ \\ \therefore \text{Required angle} & = 90^\circ \end{align}

(c)

\begin{align} \text{Gradient of } AB & = {-4 - (-4) \over 7 - 5} \\ & = {0 \over 2} \\ & = 0 \\ \\ \implies & AB \text{ is a horizontal line} \\ \\ \therefore \text{Required angle} & = 0^\circ \end{align}

(a)

\begin{align} \text{Gradient of } AB & = {2 - 0 \over 4 - 1} \\ & = {2 \over 3} \\ \\ \text{Gradient of } CD & = {-1 - (-3) \over 3 - 0} \\ & = {2 \over 3} \\ \\ \text{Since gradient of } AB & = \text{gradient of } CD, \phantom{0} AB \phantom{.} // \phantom{.} CD \end{align}

(b)

\begin{align} \text{Gradient of } AB & = {7 - 5 \over 2 - 1} \\ & = 2 \\ \\ \text{Gradient of } CD & = {3 - 4 \over 1 - 0} \\ & = -1 \\ \\ \text{Since gradient of } AB \ne \text{gradient of } & CD, \phantom{0} AB \text{ is not parallel to } CD \end{align}

\begin{align} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \text{Gradient of } PQ & = {2 - 0 \over -1 - 5} \\ & = -{1 \over 3} \\ \\ \text{Gradient of } QR & = {0 - 4 \over 5 - 7} \\ & = 2 \\ \\ \text{Gradient of } RS & = {4 - 6 \over 7 - 1} \\ & = -{1 \over 3} \\ \\ \text{Gradient of } SP & = {6 - 2 \over 1 - (-1)} \\ & = 2 \\ \\ \\ \text{Gradient of } PQ & = \text{Gradient of } RS \\ \\ \therefore PQ \phantom{0} & // \phantom{0} RS \\ \\ \text{Gradient of } QR & = \text{Gradient of } SP \\ \\ \therefore QR \phantom{0} & // \phantom{0} SP \\ \\ \text{Since quadrilateral } PQRS \text{ has 2 pairs} & \text{ of parallel sides, } \text{it is a parallelogram.} \end{align}

Since all 3 points lie on the same line, gradient of AB = gradient of BC = gradient of AC

\begin{align} \text{Gradient of } AB & = {1 - (-5) \over 0 - 2} \\ & = -3 \\ \\ \text{Gradient of } AC & = {1 - c \over 0 - (-1)} \\ -3 & = {1 - c \over 1} \\ -3 & = 1 - c \\ c & = 1 + 3 \\ c & = 4 \end{align}

(a)

\begin{align} y & = 4x - 1 \\ \implies \text{Gradient of line} & = 4 \\ \\ y & = mx + c \\ y & = 4x + c \\ \\ \text{Using } (-1, 3), & \text{ when } x = -1 \text{ and } y = 3, \\ 3 & = 4(-1) + c \\ 3 & = -4 + c \\ 3 + 4 & = c \\ 7 & = c \\ \\ \therefore y & = 4x + 7 \end{align}

(b)

\begin{align} 2x + y & = 3 \\ y & = -2x + 3 \\ \implies \text{Gradient of line} & = - 2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Since } y- \text{inter} & \text{cept is } 1, \\ y & = -2x + 1 \\ y + 2x & = 1 \end{align}

(c)

\begin{align} x + 3y & = 12 \\ 3y & = -x + 12 \\ y & = -{1 \over 3}x + 4 \\ \implies \text{Gradient of line} & = -{1 \over 3} \\ \\ y & = mx + c \\ y & = -{1 \over 3}x + c \\ \\ \text{Using } (-2, 0), & \text{ when } x = -2 \text{ and } y = 0, \\ 0 & = -{1 \over 3}(-2) + c \\ 0 & = {2 \over 3} + c \\ -{2 \over 3} & = c \\ \\ y & = -{1 \over 3}x - {2 \over 3} \\ 3y & = -x - 2 \\ x + 3y + 2 & = 0 \end{align}

(d)

\begin{align} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \text{Gradient of } BC & = {4 - (-2) \over 6 - 3} \\ & = 2 \\ \\ \implies \text{Gradient of line} & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } (3, 1), & \text{ when } x = 3 \text{ and } y = 1, \\ 1 & = 2(3) + c \\ 1 & = 6 + c \\ 1 - 6 & = c \\ -5 & = c \\ \\ y & = 2x - 5 \end{align}

(a)

\begin{align} y & = 4x - 3 \\ \implies \text{Gradient of line} & = 4 \\ \\ y & = mx + c \\ y & = 4x + c \\ \\ \text{Using } (1, 2), & \text{ when } x = 1 \text{ and } y = 2, \\ 2 & = 4(1) + c \\ 2 & = 4 + c \\ 2 - 4 & = c \\ -2 & = c \\ \\ \therefore y & = 4x - 2 \end{align}

(b)

\begin{align} y & = 3x + 4 \\ \implies \text{Gradient of line} & = 3 \\ \\ y & = mx + c \\ y & = 3x + c \\ \\ \text{Since }y\text{-inter} & \text{cept} = 3, \\ y & = 3x + 3 \end{align}

\begin{align} ax + by + c & = 0 \\ by & = -ax - c \\ y & = {-ax - c \over b} \\ y & = {-ax \over b} - {c \over b} \\ y & = -{a \over b}x - {c \over b} \\ \\ \implies \text{Gradient of 1st line} & = -{a \over b} \\ \\ \\ dx + ey + f & = 0 \\ ey & = -dx - f \\ y & = {-dx - f \over e} \\ y & = {-dx \over e} - {f \over e} \\ y & = -{d \over e}x - {f \over e} \\ \\ \implies \text{Gradient of 2nd line} & = -{d \over e} \\ \\ \\ \text{Since both } & \text{lines are parallel,} \\ \\ -{a \over b} & = -{d \over e} \\ {a \over b} & = {d \over e} \phantom{0} \text{ (Shown)} \end{align}

\begin{align} \text{Gradient of } PQ & = {3k - (k - 2) \over k^2 - k} \\ & = {3k - k + 2 \over k^2 - k} \\ & = {2k + 2 \over k^2 - k} \\ & = {2(k + 1) \over k(k - 1)} \\ \text{Gradient of } RS & = {(k + 2) - 1 \over k - 1} \\ & = {k + 1 \over k - 1} \\ \\ \text{Since } PQ \phantom{.} & // \phantom{.} RS, \\ \\ \text{Gradient of } PQ & = \text{Gradient of } RS \\ {2(k + 1) \over k(k - 1)} & = {k + 1 \over k - 1} \\ {2(k + 1) \over k(k - 1)} - {k + 1 \over k - 1} & = 0 \\ {2(k + 1) - k(k + 1) \over k(k - 1)} & = 0 \\ {2k + 2 - k^2 - k \over k(k - 1)} & = 0 \\ {-k^2 + k + 2 \over k(k - 1)} & = 0 \\ \\ \therefore -k^2 + k + 2 & = 0 \\ (-k + 2)(k + 1) & = 0 \\ \\ -k + 2 = 0 \phantom{-00} &\text{or} \phantom{0000} k + 1 = 0 \\ -k = -2 \phantom{00} & \phantom{or0000+1} k = -1 \\ k = 2 \phantom{-00} \end{align}

\begin{align} \text{Gradient of } PQ & = {2a - a \over (a - b) - (a + b)} \\ & = {a \over a - b - a - b} \\ & = {a \over - 2b} \\ & = -{a \over 2b} \\ \\ \text{Gradient of } PR & = {a - c \over (a + b) - b} \\ & = {a - c \over a} \end{align}

Since the three points are collinear (i.e. they lie on the same straight line), the lines PQ and PR have the same gradient
\begin{align} \require{cancel} \text{Gradient of } PQ & = \text{Gradient of } PR \\ -{a \over 2b} & = {a - c \over a} \\ -a(a) & = 2b(a - c) \\ -a^2 & = 2ab - 2bc \\ 2bc & = a^2 + 2ab \\ \\ c & = {a^2 + 2ab \over 2b} \\ & = {a^2 \over 2b} + {\cancel{2}a\cancel{b} \over \cancel{2} \cancel{b}} \\ & = {a^2 \over 2b} + a \end{align}

(i)

\begin{align} 4x + 3y - 5 & = 0 \\ 3y & = -4x + 5 \\ y & = {-4x + 5 \over 3} \\ y & = -{4x \over 3} + {5 \over 3} \\ y & = -{4 \over 3}x + {5 \over 3} \\ \implies \text{Gradient of line} & = -{4 \over 3} \\ \\ \text{Since } AP \text{ is parallel} & \text{ to } 4x + 3y - 5 = 0, \\ \text{Gradient of } AP & = -{4 \over 3} \\ \\ \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ -{4 \over 3} & = {3 - (-5) \over -2 - p} \\ -{4 \over 3} & = {8 \over - 2 - p} \\ -4(-2 - p) & = 3(8) \\ 8 + 4p & = 24 \\ 4p & = 24 - 8 \\ & = 16 \\ p & = {16 \over 4} \\ & = 4 \end{align}

(ii) Coordinate of P is (4, - 5). From the previous part, the gradient of AP is -⁴/₃

\begin{align} y & = mx + c \\ y & = -{4 \over 3}x + c \\ \\ \text{Using } P(4, -5), & \text{ when } x = 4 \text{ and } y = -5, \\ -5 & = -{4 \over 3}(4) + c \\ -5 & = -{16 \over 3} + c \\ -5 + {16 \over 3} & = c \\ {1 \over 3} & = c \\ \\ y & = -{4 \over 3}x + {1 \over 3} \\ 3y & = -4x + 1 \\ 4x + 3y - 1 & = 0 \end{align}

(i)

\begin{align} x + y + 2 & = 0 \\ y & = - x - 2 \phantom{000} \text{ --- (1)} \\ \\ 3x - 2y + 1 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ 3x - 2(-x - 2) + 1 & = 0 \\ 3x + 2x + 4 + 1 & = 0 \\ 5x + 5 & = 0 \\ 5x & = -5 \\ x & = -{5 \over 5} \\ & = -1 \\ \\ \text{Substitute } & x = -1 \text{ into } (1), \\ y & = -(-1) - 2 \\ & = -1 \\ \\ \therefore & \phantom{.} A (-1, - 1) \end{align}

(ii)

\begin{align} 3x + 2y - 6 & = 0 \\ 2y & = -3x + 6 \\ y & = {-3x + 6 \over 2} \\ & = {-3x \over 2} + {6 \over 2} \\ & = -{3 \over 2}x + 3 \\ \implies \text{Gradient of line} & = -{3 \over 2} \\ \\ \text{The line passing through } & A(-1, -1) \text{ has a gradient of } -{3 \over 2}. \\ \\ y & = mx + c \\ y & = -{3 \over 2}x + c \\ \\ \text{Using } A(-1, -1), & \text{ when } x = -1 \text{ and } y = -1, \\ -1 & = -{3 \over 2}(-1) + c \\ -1 & = {3 \over 2} + c \\ -{5 \over 2} & = c \\ \\ y & = -{3 \over 2}x - {5 \over 2} \\ 2y & = -3x - 5 \\ 3x + 2y + 5 & = 0 \end{align}

(i)

\begin{align} y = 2x & \phantom{.} + 3 \\ \implies \text{Gradient of line} & = 2 \end{align}

Since the line through A(3, - 1) is parallel to the line y = 2x + 3, the gradient of the line through A is 2
\begin{align} y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } A(3, -1), & \text{ when } x = 3 \text{ and } y = -1, \\ -1 & = 2(3) + c \\ -1 & = 6 + c \\ -7 & = c \\ \\ y & = 2x - 7 \end{align}

(ii)

\begin{align} y & = 2x - 7 \phantom{000} \text{ --- (1)} \\ \\ y & = 3x - 11 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (1) into (2)}, \\ 2x - 7 & = 3x - 11 \\ 2x - 3x & = - 11 + 7 \\ -x & = -4 \\ x & = 4 \\ \\ \text{Substitute} & x = 4 \text{ into (1)}, \\ y & = 2(4) - 7 \\ & = 1 \\ \\ \therefore & \phantom{.} P(4, 1) \end{align}

(i) Since AB is parallel to the x-axis, it is a horizontal line. Thus both A and B has the same y-coordinate 2

\begin{align} \text{Let the coordinates of } & B \text{ be } (b, 2). \\ \\ \text{Distance between two points} & = 6 - b \\ 3 & = 6 - b \\ 3 - 6 & = - b \\ -3 & = -b \\ \\ b & = 3 \\ \\ \therefore & \phantom{/} B(3, 2) \end{align}

(ii)

\begin{align} 2y + x & = 0 \\ 2y & = -x \\ y & = -{1 \over 2}x \\ y & = -{1 \over 2}x + 0 \\ \implies \text{Gradient of line} & = -{1 \over 2} \\ \\ \text{Since } BC \text{ is pa} & \text{rallel to the line,} \\ \text{Gradient of } BC & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } (3, 2), & \text{ when } x = 3 \text{ and } y = 2, \\ 2 & = -{1 \over 2}(3) + c \\ 2 & = -{3 \over 2} + c \\ {7 \over 2} & = c \\ \\ y & = -{1 \over 2}x + {7 \over 2} \\ 2y & = - x + 7 \\ 2y + x & = 7 \end{align}

(iii) Since $AD$ and $BC$ are parallel, the gradient of both lines are $-{1 \over 2}$

\begin{align} y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } (6 , 2), & \text{ when } x = 6 \text{ and } y = 2, \\ 2 & = -{1 \over 2}(6) + c \\ 2 & = -3 + c \\ 5 & = c \\ \\ y & = -{1 \over 2}x + 5 \\ 2y & = -x + 10 \\ 2y + x & = 10 \phantom{00000} [\text{Equation of } AD] \end{align}

D is the y-intercept of the line AD
\begin{align} \text{Let } & x = 0, \\ 2y + (0) & = 10 \\ 2y & = 10 \\ y & = {10 \over 2} \\ & = 5 \\ \\ \therefore & \phantom{.} D(0, 5) \end{align}

(iv)

\begin{align} 2y - x & = 0 \\ 2y & = x \\ y & = {1 \over 2}x \\ \implies \text{Gradient of line} & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } (0, 5), & \text{ when } x = 0 \text{ and } y = 5, \\ 5 & = {1 \over 2}(0) + c \\ 5 & = c \\ \\ y & = {1 \over 2}x + 5 \\ 2y & = x + 10 \\ 2y - x & = 10 \phantom{00000} [\text{Equation of } CD] \end{align}

Use the equation of BC from part (ii) and the equation of CD to find the coordinates of C:
\begin{align} 2y + x & = 7 \\ x & = 7 - 2y \phantom{000} \text{ --- (1)} \\ \\ 2y - x & = 10 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (1) into (2)}, \\ 2y - (7 - 2y) & = 10 \\ 2y - 7 + 2y & = 10 \\ 2y + 2y & = 10 + 7 \\ 4y & = 17 \\ y & = {17 \over 4} \\ \\ \text{Substitute } & y = {17 \over 4} \text{ into (1)}, \\ x & = 7 - 2\left(17 \over 4\right) \\ & = -{3 \over 2} \\ \\ \therefore & \phantom{.} C \left(-{3 \over 2}, {17 \over 4}\right) \end{align}