(a) A perpendicular to a horizontal line is vertical: \(x=-1\).

(b) \(2=20-6x\), so \(6x=18\), \(x=3\). The intersection is \((3,2)\).

\(m=\frac{10-20}{5-3}=\frac{-10}{2}=\boxed{-5}\).

(a) L passes through \((0,1)\) and \((3,-1)\). Its gradient is \(\frac{-1-1}{3-0}=\boxed{-\frac23}\).

(b) Rearrange to \(y=2x-3\). Plot \((0,-3)\) and \((3,3)\), then join with a straight line.

(a) \(AB=\sqrt{(-3-5)^2+(6-2)^2}=\sqrt{80}=\boxed{4\sqrt5}\) (approximately \(8.94\)).

(b) AB has gradient \((6-2)/(-3-5)=-1/2\), so the perpendicular gradient is 2. Substituting \((-2,5)\) into \(y=2x+c\) gives \(5=-4+c\), hence \(c=9\). The equation is \(\boxed{y=2x+9}\).

(a) At \(x=0\), \(y=-3\), so \(\boxed{(0,-3)}\).

(b) AB has midpoint \(((1+5)/2,(7+15)/2)=(3,11)\) and gradient \((15-7)/(5-1)=2\). The perpendicular gradient is \(-1/2\). Through the midpoint, \(y-11=-\frac12(x-3)\), giving \(\boxed{y=-\frac12x+\frac{25}{2}}\).

Midpoint \(=((8+20)/2,(0+6)/2)=(14,3)\). Gradient of PQ \(=6/12=1/2\), so the perpendicular gradient is \(-2\). Through the midpoint: \(y-3=-2(x-14)\), giving \(\boxed{y=-2x+31}\).

(a) Put \(x=0\): \(y=-10\), giving \(\boxed{(0,-10)}\).

(b) \(19=2x-10\), so \(\boxed{x=14.5}\).

(c) L has gradient 2, so P has gradient \(-1/2\). Substitute (2,3): \(3=-\frac12(2)+c\), giving \(c=4\). Thus \(\boxed{y=-\frac12x+4}\).

The perpendicular gradient is 2. Through (3,5): \(y-5=2(x-3)\), giving \(\boxed{y=2x-1}\).

(a) \(BD=\sqrt{(-5+3)^2+(9-1)^2}=\sqrt{68}=2\sqrt{17}\). Therefore \(AC=(3/2)(2\sqrt{17})=\boxed{3\sqrt{17}}\).

(b) Midpoint \(=\boxed{(-4,5)}\).

(c) Kite diagonals are perpendicular. BD has gradient \(8/(-2)=-4\), so AC has gradient 1/4. Through (−4,5): \(y-5=\frac14(x+4)\), giving \(\boxed{y=\frac14x+6}\).

Gradient \(=(4-0)/(0-2)=-2\); the y-intercept is 4. Thus \(\boxed{y=-2x+4}\).

Midpoint \(=((6+2)/2,(1+7)/2)=(4,4)\). Gradient AB \(=(7-1)/(2-6)=-3/2\), so the perpendicular gradient is 2/3. Use the midpoint: \(y-4=\frac23(x-4)\), giving \(\boxed{y=\frac23x+\frac43}\).

\(3=2x-5\), so x=4. Thus \(\boxed{P=(4,3)}\).

\(m=(1-7)/(3-(-2))=\boxed{-6/5}\).

\(m=(6-2)/(8-0)=1/2\); the intercept is 2. Thus \(\boxed{y=\frac12x+2}\).

(a)(i) \(\boxed{\overrightarrow{PQ}=\begin{pmatrix}4\\-12\end{pmatrix}}\). (ii) \(OP=\sqrt{1^2+7^2}=\sqrt{50}\), \(OQ=\sqrt{5^2+(-5)^2}=\sqrt{50}\), so they are equal. (iii) Radius√50, circumference=\(2\pi\sqrt{50}=\boxed{44.4}\). (iv) Midpoint=\(\boxed{(3,1)}\). (v) Gradient PQ=−12/4=−3; perpendicular gradient=1/3. Through(3,1), \(y-1=(x-3)/3\), so \(\boxed{y=x/3}\). (b) \(\overrightarrow{OM}=\mathbf a+\frac25(\mathbf b-\mathbf a)=\boxed{\frac35\mathbf a+\frac25\mathbf b}\).

(a)(i) \(\boxed{((6+3)/2,(2-4)/2)=(4.5,-1)}\). (ii) \(AB=\sqrt{3^2+6^2}=\boxed{6.71}\). (b)(i) \(y=-\frac43x+4\), so gradient=\(\boxed{-4/3}\). (ii) At x=0, \(\boxed{(0,4)}\). (iii) Perpendicular gradient 3/4; 5=(3/4)6+c gives c=1/2. Thus \(\boxed{y=\frac34x+\frac12}\).

Midpoint M=(20,24). Gradient of AB=\((39-9)/(23-17)=5\), so perpendicular gradient=−1/5. Through M: \(y-24=-\frac15(x-20)\), giving \(\boxed{y=-\frac15x+28}\).

(a) \(y=4\).

(b) Gradient \((0-4)/(8-0)=-1/2\); intercept 4. Hence \(y=-x/2+4\).

(c)(i) Perpendicular gradient \(=-1/(-1/2)=2\). Using \((2,3)\): \(3=2(2)+c\Rightarrow c=-1\), so \(y=2x-1\).

(ii) At \(D\), \(0=2x-1\Rightarrow x=1/2\). \(CD=\sqrt{(2-1/2)^2+(3-0)^2}=\sqrt{11.25}=3.35\).

(a) \(((5+13)/2,(-1+15)/2)=(9,7)\). (b) \((15+1)/(13-5)=2\).

(c) Perpendicular gradient \(-1/2\). Through \((9,7)\): \(y-7=-\frac12(x-9)\Rightarrow y=-\frac12x+\frac{23}2\).

(a) \(\overrightarrow{AB}=(3,4)\). The perpendicular side vector leading to positive coordinates for \(C\) is \((4,-3)\). Hence \(C=(5,2),D=(2,-2)\).

(b)(i) \((({-1}+6)/2,(3+4)/2)=(2.5,3.5)\). (ii) \(PQ=\sqrt{7^2+1^2}=\sqrt{50}=7.07\).

(iii) Gradient \(=(4-3)/(6+1)=1/7\). (iv) Through \((2,0)\), \(y=(x-2)/7=x/7-2/7\).

(a) \(MN=\sqrt{(4+2)^2+(1+7)^2}=\sqrt{100}=10\). (b) Gradient \(=(1+7)/(4+2)=4/3\).

(c) Midpoint \(((4-2)/2,(1-7)/2)=(1,-3)\). Perpendicular gradient \(-3/4\). Through the midpoint: \(y+3=-\frac34(x-1)\), giving \(y=-\frac34x-\frac94\).

(a)(i) \(\begin{pmatrix}2\\5\end{pmatrix}\). (ii) \(\begin{pmatrix}-6\\4\end{pmatrix}\). (b) Gradient \((5-1)/(2-8)=-2/3\). Using \((8,1)\) gives \(y=-\frac23x+\frac{19}3\).

(c) Midpoint \((5,3)\), perpendicular gradient \(3/2\). \(y-3=\frac32(x-5)\Rightarrow y=\frac32x-\frac92\). (d) \(PQ=\frac{19}3-(-\frac92)=\frac{65}6\).

(a) \(AB^2=3^2+1^2=10,BC^2=1^2+(-3)^2=10,AC^2=4^2+(-2)^2=20\). \(AB=BC\), so isosceles.

(b)(i) Gradient \((0-2)/(4-0)=-1/2\), so \(y=-x/2+2\). (ii) Midpoint \((2,1)\), perpendicular gradient 2. \(y-1=2(x-2)\Rightarrow y=2x-3\).

(iii) The kite’s other vertex \(D\) lies on this perpendicular bisector: \(4w+1=2w-3\), giving \(w=-2\). Hence \(D=(-2,-7)\).

(a) \(m=(-3-5)/(-2+6)=-2\). Using \(A\), \(5=12+c\) gives \(y=-2x-7\). (b) Negative reciprocal gradient \(1/2\) and intercept 0 give \(y=x/2\).

(a) Gradient 5. (b) Put \(x=0\) to get \((0,7)\).

(a) \(FG=\sqrt{(3+5)^2+(2-4)^2}=\sqrt{68}=8.25\). (b) Midpoint \((-1,3)\). Gradient \((4-2)/(-5-3)=-1/4\); perpendicular gradient 4. \(y-3=4(x+1)\) gives \(y=4x+7\).

(c) Write \(H=(0,y)\). \(25+(y-4)^2=169\) gives \(y-4=\pm12\). Thus \((0,-8)\) and \((0,16)\).

(a)(i) \((-0.5,1)\). (ii) \(\binom7{-3}\). (iii) \(\sqrt{7^2+(-3)^2}=\sqrt{58}=7.62\). (iv) Gradient \((-5-7)/(1+2)=-4\); using A gives \(c=-1\). Thus \(y=-4x-1\). (v) Perpendicular gradient \(1/4\). \(4=5/4+c\) gives \(y=x/4+11/4\).

(b) \(8-5x=2x^2+6x-13\) gives \(2x^2+11x-21=(2x-3)(x+7)=0\). For \(x=1.5\), \(y=0.5\); for \(x=-7\), \(y=43\). Therefore P and Q are \((1.5,0.5),(-7,43)\) in either order.

Kite diagonals are perpendicular, so the other gradient is \(-1/4\). \(-5=(-1/4)(-2)+c\) gives \(c=-11/2\). Thus \(y=-x/4-11/2\).

(a) \(B=2M-A=(24-2,14-3)=(22,11)\). (b) Gradient \(m=(11-3)/(22-2)=2/5\). Perpendicular gradient \(-5/2\). Through M: \(y-7=-\frac52(x-12)\). Multiply by 2: \(2y-14=-5x+60\), hence \(2y+5x=74\). (c) \(2n+5(2)=74\) gives n = 32. (d) AN is vertical with length 29, and M is 10 units from it. Area \(\frac12(29)(10)=145\) square units.

(a) \(\frac{dy}{dx}=6x^2-7\) gives \(6(-2)^2-7=17\). (b)(i) \(AB=\sqrt{(-5-7)^2+(8-2)^2}=\sqrt{180}=13.4\). (ii) \(m_{AB}=6/(-12)=-1/2\). Perpendicular gradient is 2. \(y-3=2(x+1)\) gives \(y=2x+5\).

(iii) Gradient \(b/a=1\) gives a = b. Magnitude \(a^2+b^2=8\) gives \(2a^2=8\), and positive a = b = 2. In a parallelogram \(\overrightarrow{AD}=\overrightarrow{BC}=(-2,-2)\). Thus \(D=(7-2,2-2)=(5,0)\).

The given gradient is \(4/3\). Its negative reciprocal is \(-3/4\).

(a) \(m=(-1-2)/(4-0)=-3/4\). (b) \(y=-3x/4+2\). (c) Perpendicular gradient \(4/3\). \(-7=(4/3)(12)+c\) gives \(c=-23\). Thus \(y=4x/3-23\).

(a) \(AB=\sqrt{4^2+(-8)^2}=\sqrt{80}=8.94\). (b) Gradient \(-8/4=-2\). Using (5, 7): \(7=-2(5)+c\) gives \(c=17\). Hence \(y=-2x+17\).

L has gradient −5, so the perpendicular gradient is \(1/5\). The intercept is 6: \(y=x/5+6\).

\(m=(10+2)/(3-2)=12\). \(-2=12(2)+c\) gives \(c=-26\). Thus \(y=12x-26\).

(a) \(m=(31-16)/(8-3)=3\). \(16=3(3)+c\) gives \(c=7\), hence \(y=3x+7\). (b) \(-1/.5=-2\).

\(m_{AB}=\frac{5-(-7)}{3-1}=6\), so perpendicular gradient \(-1/6\). Using A, \(y-5=-(x-3)/6\), hence \(y=-\frac16x+\frac{11}2\).

(a) \(y=(3/8)x+5/2\), so (i) gradient3/8 and (ii) intercept(0,2.5).

(b)(i) \(PQ=\sqrt{12^2+(-10)^2}=\sqrt{244}=15.6\). (ii) Gradient PQ=−10/12=−5/6. \(y+1=-\frac56(x-6)\Rightarrow y=-\frac56x+4\).

(iii) Midpoint(3,3); perpendicular gradient6/5. \(y-3=\frac65(x-3)\Rightarrow y=\frac65x-\frac35\).

Rewrite \(y=-x/2+5/8\), gradient−1/2. The perpendicular gradient is its negative reciprocal,2.

Perpendicular gradient−3. Through(2,3), \(y-3=-3(x-2)\Rightarrow y=-3x+9\). Equate \(-3x+9=x/3+1\), giving \((10/3)x=8\Rightarrow x=12/5=2.4\). Then y=1.8. P=(2.4,1.8).

(a)5. (b) Set y=0: \(5x+12=0\Rightarrow x=-12/5\). Point(−12/5,0). (c) The negative reciprocal is k=−1/5.

(a) \(((2+7)/2,(3-5)/2)=(4.5,-1)\). (b) AB gradient \((-5-3)/(7-2)=-8/5\), perpendicular gradient5/8. \(y-3=(5/8)(x-2)\Rightarrow y=5x/8+7/4\).

(a) x=0 gives (0,−8). (b) The coefficient of x is 3.

(a) \(m=(9+3)/6=2\), y-intercept −3, so \(y=2x-3\). (b) Perpendicular gradient −1/2; intercept 2, so \(y=-x/2+2\).

(a) \(y=1.5x+2\), so gradient1.5 and intercept(0,2). (b)(i) L passes through(0,6),(3,0); gradient−2, so \(y=-2x+6\). (ii) Gradient1/2; \(y-3=(x-9)/2\Rightarrow y=.5x-1.5\). (c)(i) \(AB=\sqrt{12^2+4^2}=\sqrt{160}=12.6\). (ii) \(((8-4)/2,(5+1)/2)=(2,3)\).

(a) \(m=(9+3)/(1+2)=4\), hence \(y=4x+5\). (b) \(\theta=\tan^{-1}4=76.0^\circ\). (c)(i) Diagonals are perpendicular: gradient−1/4. \(y-2=-\frac14(x-3.5)\Rightarrow y=-\frac14x+\frac{23}8\). (ii) AC bisects BD, so \(D=2(-.5,3)-(3.5,2)=(-4.5,4)\).

\(y=\frac54x-\frac52,\quad y=-\frac45x+7\). The gradients multiply to \(\frac54(-\frac45)=-1\), so the lines are perpendicular.

\(\frac{6-(-5)}{3k-k}=2\Rightarrow11=4k\Rightarrow k=2.75\).

The given gradient is 5/2. Its negative reciprocal is \(-\frac25\).

(a) \(\tan\theta=\frac{5-0}{2-1}=5\Rightarrow\theta=78.690\ldots\approx78.7^\circ\). (b) The perpendicular gradient is −1/3. \(y-10=-\frac13(x-6)\Rightarrow y=-\frac13x+12\).

(a) \(m_{PQ}=\frac{24-9}{22-16}=\frac52\), so perpendicular gradient=\(-\frac25\). \(y-1=-\frac25(x-5)\Rightarrow y=-\frac25x+3\). (b) N is two-thirds of the way from P to Q: \((16,9)+\frac23(6,15)=(20,19)\).

\(m_{CD}=\frac{7-2}{9-(-1)}=\frac12\). Perpendicular gradient=−2. \(y-3=-2(x-1)\Rightarrow y=-2x+5\).

(a)A=(5,6). (b) l has gradient−4/5 and intercept 3, so y=−4 x/5+3. (c)B=(0,−2), giving y=−4 x/5−2. (d)(i) Perpendicular gradient 5/4;14=(5/4)8+c gives c=4, so y=5 x/4+4. (ii) \(AC=\sqrt{(8-5)^2+(14-6)^2}=\sqrt{73}\approx8.54\). (iii) \(((0+8)/2,(-2+14)/2)=(4,6)\).

(a) Points (0,3) and (1,1) give \(m=(1-3)/(1-0)=-2\), so \(y=-2x+3\). (b) Perpendicular gradient \(-1/(-2)=1/2\). \(y+1=\frac12(x-3)\Rightarrow y=\frac12x-\frac52\).

\(\frac{23-8}{k-x}=5\Rightarrow k-x=3\Rightarrow x=k-3\).

(a) \(m=(4-0)/(0+2)=2\), and intercept 4, so \(y=2x+4\). (b) Midpoint (−1,2), perpendicular gradient −1/2: \(y-2=-\frac12(x+1)\Rightarrow y=-\frac12x+\frac32\).

(a) \(((-3+2)/2,(8-2)/2)=(-0.5,3)\). (b) \(m=(-2-8)/(2+3)=-2\); substituting B gives c=2, so y=−2x+2.

(c) Same gradient, intercept 7: y=−2x+7. (d) Perpendicular gradient 1/2; \(y-5=\frac12(x-1)\Rightarrow x-2y+9=0\).

(a) (−2,3). (b) Opposite sides are parallel, so \(D=A+C-B=(-2,3)+(3,-2)-(-1,-1)=(2,2)\). Join C to D and D to A.

(a)(i) 5. (ii) y=−1.5x+9, so −1.5. (b) \(0=5x-4\Rightarrow(4/5,0)\). (c) Gradient −1/5; \(y-9=-\frac15(x-10)\Rightarrow y=-0.2x+11\).

(d) \(3x+2(5x-4)=18\Rightarrow13x=26\Rightarrow x=2,y=6\). (e) The triangle has y-axis base from −4 to 9, length 13, and perpendicular height 2. Area=\(13(2)/2=13\) square units.

The midpoint is \((\frac{4+10}{2},\frac{1+15}{2})=(7,8)\). The gradient of AB is \(\frac{15-1}{10-4}=\frac73\), so the perpendicular gradient is \(-\frac37\). Through the midpoint, \(y-8=-\frac37(x-7)\), giving \(y=-\frac37x+11\).

(a) L passes through \((0,-1)\) and \((2,3)\). Its gradient is \(\frac{3-(-1)}{2-0}=2\). (b) A parallel line has the same gradient. Through \((0,6)\), its equation is \(y=2x+6\).

(a) \(AB=\sqrt{(4+2)^2+(13+5)^2}=\sqrt{360}=18.9737\ldots\approx19.0\). (b) The gradient is \((13+5)/(4+2)=3\). Substitute A: \(-5=3(-2)+c\), giving \(c=1\). Thus \(y=3x+1\). (c) Parallel lines have equal gradients, giving \(y=3x-5\). (d) The midpoint is \((1,4)\) and the perpendicular gradient is \(-1/3\). Hence \(y-4=-\frac13(x-1)\Rightarrow y=-\frac13x+\frac{13}{3}\).

(a) \(B=(7,1)\). (b) \(m=\frac{1-6}{7-3}=-\frac54\). (c) The perpendicular gradient is \(4/5\). Since the y-intercept is 2, \(y=\frac45x+2\).

\(m_{AB}=\frac{1-3}{12-8}=-\frac12\). The perpendicular gradient is 2 and the intercept is zero, so \(y=2x\).

Its gradient is \(-1/3\). Use point-gradient form: \(y-4=-\frac13(x-7)\), so \(x+3y=19\).

(a) \(m=(11-3)/(4-0)=2\) and the y-intercept is 3, so \(y=2x+3\). (b) The negative reciprocal gradient is \(-1/2\).

(a) \(((3+5)/2,(4+8)/2)=(4,6)\). (b) \(AB=\sqrt{(5-3)^2+(8-4)^2}=\sqrt{20}\approx4.47\). (c) Gradient \((8-4)/(5-3)=2\). Using A gives \(4=2(3)+c\Rightarrow c=-2\), so y = 2x − 2. (d) Perpendicular gradient −½; through the origin, y = −x/2. Thus r = −3.

The midpoint is (9, 4). Using \(y-4=(2/3)(x-9)\) gives \(y=(2/3)x-2\).

Gradient \((13-9)/(-3-5)=-1/2\). Using (5, 9), intercept \(9+2.5=11.5\). Thus \(y=-0.5x+11.5\).

(a) At x = 0, y = 5, so P = (0, 5). (b) −1.

\(m=(5-(-1))/(3-0)=2\). The intercept is −1, giving \(y=2x-1\).

(a) \(((5-2)/2,(23+2)/2)=(1.5,12.5)\). (b) \(m=(23-2)/(5-(-2))=3\). Using B, \(2=3(-2)+c\Rightarrow c=8\), so \(y=3x+8\). (c) \(3(3)+8=17\), confirming the point.

(a) \(((-8-2)/2,(3-3)/2)=(-5,0)\). (b) \(c=6-4(2)=-2\). (c) The gradients are 5/4 and k/2: \(k/2=5/4\Rightarrow k=5/2\).

(a) Set y = 0: \(x_P=-4,\ x_Q=6;\ PQ=6-(-4)=10\). (b) \(y=-4x+5\). (c) \(y=-4x+c;\ 4=-4(5)+c\Rightarrow c=24;\ y=-4x+24\).

(a) A(7,0), B(5,3): \(((7+5)/2,(0+3)/2)=(6,1.5)\). (b) C(0,4), B(5,3): \(m=(3-4)/(5-0)=-1/5\). Therefore \(y=-x/5+4\).

\(m=(13-1)/(3-0)=4\) and the y-intercept is 1, so \(y=4x+1\).

(a) \(AB=\sqrt{(3-1)^2+(0-3)^2}=\sqrt{13}\approx3.61\). (b) Gradient \((7-3)/(9-1)=1/2\). Using A: \(3=\tfrac12(1)+c\Rightarrow c=5/2\), so \(y=\tfrac12x+\tfrac52\).

Gradient \((3-5)/(3-2)=-2\). (b) \(y=-2x+c\) and \(5=-4+c\), so y = −2x+9. (a) \(1=-2k+9\Rightarrow k=4\).

(a) \(2x+4=8\Rightarrow A=(2,4)\). (b) \(3x=18\Rightarrow B=(6,0)\). (c)(i) \(M=((2+6)/2,(4+0)/2)=(4,2)\). (ii) Parallel gradient −3, so \(y-2=-3(x-4)\Rightarrow y=-3x+14\).

Gradient \((2-8)/(3-0)=-2\), intercept 8. Thus \(y=-2x+8\).

(a) Draw equal-radius arcs from A and B, using a radius greater than half AB. Join their two intersections. The bisector passes through (4,2) with gradient 1/2: \(y=x/2\). (b) \(((1+7)/2,(8-4)/2)=(4,2)\). (c) Gradient \((-4-8)/(7-1)=-2\); substituting A gives intercept 10. Thus \(y=-2x+10\).

(a) Set x = 0: A = (0,7). (b)(i) Same gradient 2 and intercept 3: \(y=2x+3\). (ii) C = (2,5), giving midpoint \(((0+2)/2,(3+5)/2)=(1,4)\).