9th Class Math Chapter 7 Notes

Q1(i) — Distance: $A(6,7),B(0,-2)$

Find the distance between the points.

1. $d=\sqrt{(6-0{)}^2+(7-(-2){)}^2}=\sqrt{36+81}$.

Answer: $3\sqrt{13}$

Q1(ii) — Distance: $C(-5,-2),D(3,2)$

Find the distance.

1. $d=\sqrt{(-5-3{)}^2+(-2-2{)}^2}=\sqrt{64+16}$.

Answer: $4\sqrt{5}$

Q1(iii) — Distance: $L(0,3),M(-2,-4)$

Find the distance.

1. $d=\sqrt{(0-(-2){)}^2+(3-(-4){)}^2}=\sqrt{4+49}$.

Answer: $\sqrt{53}$

Q1(iv) — Distance: $P(-8,-7),Q(0,0)$

Find the distance.

1. $d=\sqrt{64+49}$.

Answer: $\sqrt{113}$

Q2(i) — Distance & midpoint: $A(3,1),B(-2,-4)$

Find the distance and the midpoint.

1. $d=\sqrt{(3-(-2){)}^2+(1-(-4){)}^2}=\sqrt{25+25}=\sqrt{50}$.
2. Midpoint $=\left(\frac{3-2}{2},\frac{1-4}{2}\right)$.

Answer: $d=5\sqrt{2}$; midpoint $=\left(\frac{1}{2},-\frac{3}{2}\right)$

Q2(ii) — Distance & midpoint: $A(-8,3),B(2,-1)$

Find the distance and the midpoint.

1. $d=\sqrt{(-8-2{)}^2+(3-(-1){)}^2}=\sqrt{100+16}$.
2. Midpoint $=\left(\frac{-8+2}{2},\frac{3-1}{2}\right)$.

Answer: $d=2\sqrt{29}$; midpoint $=(-3,1)$

Q2(iii) — Distance & midpoint: $A(-\sqrt{5},-\frac{1}{3}),B(-3\sqrt{5},5)$

Find the distance and the midpoint.

1. $d=\sqrt{(-\sqrt{5}+3\sqrt{5}{)}^2+(-\frac{1}{3}-5{)}^2}=\sqrt{(2\sqrt{5}{)}^2+(-\frac{16}{3}{)}^2}=\sqrt{20+\frac{256}{9}}=\sqrt{\frac{436}{9}}$.
2. Midpoint $=\left(\frac{-\sqrt{5}-3\sqrt{5}}{2},\frac{-\frac{1}{3}+5}{2}\right)$.

Answer: $d=\frac{2\sqrt{109}}{3}$; midpoint $=\left(-2\sqrt{5},\frac{7}{3}\right)$

Q3(i) — Is $(\sqrt{176},7)$ at distance 15 from the origin?

Check the distance from the origin.

1. $d=\sqrt{176+49}=\sqrt{225}$.

Answer: $d=15$ — YES, exactly 15 units

Q3(ii) — Is $(10,-10)$ at distance 15?

Check the distance from the origin.

1. $d=\sqrt{100+100}=\sqrt{200}=10\sqrt{2}$.

Answer: $d=10\sqrt{2}\approx 14.14$ — NOT 15

Q3(iii) — Is $(1,15)$ at distance 15?

Check the distance from the origin.

1. $d=\sqrt{1+225}=\sqrt{226}$.

Answer: $d=\sqrt{226}\approx 15.03$ — NOT exactly 15

Q4(i) — Show right triangle: $A(0,2),B(\sqrt{3},1),C(0,-2)$

Show these are vertices of a right triangle.

1. $AB=\sqrt{3+1}=2$; $BC=\sqrt{3+9}=2\sqrt{3}$; $AC=\sqrt{0+16}=4$.
2. $A{B}^2+B{C}^2=4+12=16=A{C}^2$ ✓ — right angle at $B$.

Answer: Right triangle, right angle at $B$

Q4(ii) — Show isosceles: $A(3,1),B(-2,-3),C(2,2)$

Show these are vertices of an isosceles triangle.

1. $AB=\sqrt{25+16}=\sqrt{41}$; $BC=\sqrt{16+25}=\sqrt{41}$; $AC=\sqrt{1+1}=\sqrt{2}$.
2. $AB=BC$ ✓ isosceles.

Answer: Isosceles, since $AB=BC=\sqrt{41}$

Q4(iii) — Show parallelogram: $A(5,2),B(-2,3),C(-3,-4),D(4,-5)$

Show these are vertices of a parallelogram.

1. Midpoint of $AC=\left(\frac{5-3}{2},\frac{2-4}{2}\right)=(1,-1)$.
2. Midpoint of $BD=\left(\frac{-2+4}{2},\frac{3-5}{2}\right)=(1,-1)$.
3. Diagonals share the same midpoint ✓ parallelogram.

Answer: Parallelogram, since both diagonals bisect at $(1,-1)$

Q5(i) — Find $h$: right angle at $A$ for $A(\sqrt{3},-1),B(0,2),C(h,-2)$

Find $h$ so the triangle has a right angle at $A$.

1. $\overrightarrow{AB}=(-\sqrt{3},3)$, $\overrightarrow{AC}=(h-\sqrt{3},-1)$.
2. Right angle at $A$ means $\overrightarrow{AB}\cdot \overrightarrow{AC}=0$: $-\sqrt{3}(h-\sqrt{3})-3=0\Rightarrow -\sqrt{3}h+3-3=0$.

Answer: $h=0$

Q5(ii) — Find $h$: collinear $A(-1,h),B(3,2),C(7,3)$

Find $h$ so the points are collinear.

1. Slope $AB=$ slope $BC$: $\frac{2-h}{4}=\frac{1}{4}$.

Answer: $h=1$

Q5(iii) — Find $h$: right angle at $A$ for $A(h,1),B(2,7),C(-6,-7)$

Find $h$ so the triangle has a right angle at $A$.

1. $\overrightarrow{AB}=(2-h,6)$, $\overrightarrow{AC}=(-6-h,-8)$.
2. $(2-h)(-6-h)+6(-8)=0\Rightarrow h^2+4h-12-48=0\Rightarrow h^2+4h-60=0$.
3. $(h+10)(h-6)=0$.

Answer: $h=-10$ or $h=6$

Q6 — Centre/radius: diameter ends $A(-5,-2),B(5,-4)$

Find the centre and radius of the circle.

1. Centre $=$ midpoint $=\left(\frac{-5+5}{2},\frac{-2-4}{2}\right)$.
2. Radius $=\frac{1}{2}\sqrt{(-5-5{)}^2+(-2+4{)}^2}=\frac{1}{2}\sqrt{104}$.

Answer: Centre $=(0,-3)$; radius $=\sqrt{26}$

Q7 — Midpoints of $A(9,3),B(-7,7),C(-3,-7),D(5,-5)$

Find the midpoints of the sides and show they form a parallelogram.

1. Midpoints: $AB\to (1,5)$, $BC\to (-5,0)$, $CD\to (1,-6)$, $DA\to (7,-1)$.
2. Midpoint of $(1,5)$ & $(1,-6)$ $=(1,-0.5)$; midpoint of $(-5,0)$ & $(7,-1)$ $=(1,-0.5)$ — same point.

Answer: The four midpoints form a parallelogram (diagonals bisect at $(1,-0.5)$) — an instance of Varignon's theorem

Q8(i) — Slope/inclination: $(-2,4),(5,11)$

Find the slope and inclination.

1. Slope $=\frac{11-4}{5-(-2)}=\frac{7}{7}=1$.
2. $\tan \theta =1\Rightarrow \theta =45^{\circ }$.

Answer: Slope $=1$; inclination $=45^{\circ }$

Q8(ii) — Slope/inclination: $(3,-2),(2,7)$

Find the slope and inclination.

1. Slope $=\frac{7-(-2)}{2-3}=\frac{9}{-1}=-9$.
2. Negative slope $\Rightarrow$ obtuse inclination: $\theta =180^{\circ }-\arctan (9)\approx 96.34^{\circ }$.

Answer: Slope $=-9$; inclination $\approx 96.34^{\circ }$

Q8(iii) — Slope/inclination: $(4,6),(4,8)$

Find the slope and inclination.

1. Equal $x$-coordinates $\Rightarrow$ vertical line.

Answer: Slope undefined; inclination $=90^{\circ }$

Q9(i) — Collinear: $A(-1,-3),B(1,5),C(2,9)$

Show these points lie on the same line.

1. Slope $AB=\frac{8}{2}=4$; slope $BC=\frac{4}{1}=4$ — equal.

Answer: Collinear (both slopes $=4$)

Q9(ii) — Collinear: $P(4,-5),Q(7,5),R(10,15)$

Show these points lie on the same line.

1. Slope $PQ=\frac{10}{3}$; slope $QR=\frac{10}{3}$ — equal.

Answer: Collinear (both slopes $=\frac{10}{3}$)

Q9(iii) — Collinear: $L(-4,6),M(3,8),N(10,10)$

Show these points lie on the same line.

1. Slope $LM=\frac{2}{7}$; slope $MN=\frac{2}{7}$ — equal.

Answer: Collinear (both slopes $=\frac{2}{7}$)

Q9(iv) — Collinear: $X(a,2b),Y(c,a+b),Z(2c-a,2a)$

Show these points lie on the same line.

1. Slope $XY=\frac{(a+b)-2b}{c-a}=\frac{a-b}{c-a}$.
2. Slope $YZ=\frac{2a-(a+b)}{(2c-a)-c}=\frac{a-b}{c-a}$ — equal.

Answer: Collinear (both slopes $=\frac{a-b}{c-a}$)

Q10(i) — Find $k$ (parallel): $A(7,3),B(k,-6)$ ∥ $C(-4,5),D(-6,4)$

Find $k$ so the two lines are parallel.

1. Slope $AB=\frac{-9}{k-7}$; slope $CD=\frac{-1}{-2}=\frac{1}{2}$.
2. $\frac{-9}{k-7}=\frac{1}{2}\Rightarrow -18=k-7$.

Answer: $k=-11$

Q10(ii) — Find $k$ (perpendicular): same lines

Find $k$ so the two lines are perpendicular.

1. $\frac{-9}{k-7}\times \frac{1}{2}=-1\Rightarrow \frac{-9}{k-7}=-2\Rightarrow 9=2(k-7)$.

Answer: $k=\frac{23}{2}$

Q11 — Show right triangle: $A(6,1),B(2,7),C(-6,-7)$

Use slopes to show this is a right triangle.

1. Slope $AB=\frac{6}{-4}=-\frac{3}{2}$; slope $AC=\frac{-8}{-12}=\frac{2}{3}$.
2. $\left(-\frac{3}{2}\right)\left(\frac{2}{3}\right)=-1$ — right angle at $A$.

Answer: Right triangle, right angle at $A$

Q12(i) — Test: $(1,-2),(2,4)$ and $(4,1),(-8,2)$

Determine if the two lines are parallel, perpendicular, or neither.

1. Slope${}_1=\frac{6}{1}=6$; slope${}_2=\frac{1}{-12}=-\frac{1}{12}$.
2. Product $=6\times \left(-\frac{1}{12}\right)=-\frac{1}{2}$ — not $-1$, and slopes unequal.

Answer: Neither parallel nor perpendicular

Q12(ii) — Test: $(-3,4),(6,2)$ and $(4,5),(-2,-7)$

Determine if the two lines are parallel, perpendicular, or neither.

1. Slope${}_1=\frac{-2}{9}$; slope${}_2=\frac{-12}{-6}=2$.
2. Product $=-\frac{2}{9}$ — not $-1$, slopes unequal.

Answer: Neither parallel nor perpendicular

Q13(i) — Equation: horizontal line through $(7,-9)$

Find the equation.

1. A horizontal line has the form $y=c$.

Answer: $y=-9$

Q13(ii) — Equation: vertical line through $(-5,3)$

Find the equation.

1. A vertical line has the form $x=c$.

Answer: $x=-5$

Q13(iii) — Equation: through $A(-6,5)$, slope $7$

Find the equation.

1. Point-slope: $y-5=7(x+6)$.

Answer: $y=7x+47$

Q13(iv) — Equation: through $(8,-3)$, slope $0$

Find the equation.

1. Slope $0$ means horizontal.

Answer: $y=-3$

Q13(v) — Equation: through $(-8,5)$, slope undefined

Find the equation.

1. Undefined slope means vertical.

Answer: $x=-8$

Q13(vi) — Equation: through $(-5,-3)$ and $(9,-1)$

Find the equation.

1. Slope $=\frac{-1+3}{9+5}=\frac{2}{14}=\frac{1}{7}$.
2. $y+3=\frac{1}{7}(x+5)\Rightarrow y=\frac{x}{7}-\frac{16}{7}$.

Answer: $y=\frac{1}{7}x-\frac{16}{7}$

Q13(vii) — Equation: $y$-intercept $-7$, slope $-5$

Find the equation.

1. $y=mx+c$ with $m=-5,c=-7$.

Answer: $y=-5x-7$

Q13(viii) — Equation: $x$-intercept $-3$, $y$-intercept $4$

Find the equation.

1. Intercept form $\frac{x}{-3}+\frac{y}{4}=1$; multiply by $-12$: $4x-3y=-12$.

Answer: $y=\frac{4}{3}x+4$

Q13(ix) — Equation: $x$-intercept $-9$, slope $-4$

Find the equation.

1. Passes through $(-9,0)$: $y-0=-4(x+9)$.

Answer: $y=-4x-36$

Q14(i) — Perpendicular bisector of $A(3,5),B(9,8)$

Find the equation of the perpendicular bisector.

1. Midpoint $=(6,6.5)$; slope $AB=\frac{3}{6}=\frac{1}{2}$, so perpendicular slope $=-2$.
2. $y-6.5=-2(x-6)$.

Answer: $y=-2x+18.5$

Q14(ii) — Through $(-4,-6)$, perpendicular to slope $-\frac{3}{2}$

Find the equation.

1. Perpendicular slope $=\frac{2}{3}$.
2. $y+6=\frac{2}{3}(x+4)$.

Answer: $y=\frac{2}{3}x-\frac{10}{3}$

Q14(iii) — Through $(11,-5)$, parallel to slope $-24$

Find the equation.

1. Same slope $-24$: $y+5=-24(x-11)$.

Answer: $y=-24x+259$

Q15(i) — Convert: $2x-4y+11=0$

Convert into all three forms.

1. Slope-intercept: $y=\frac{x}{2}+\frac{11}{4}$.
2. Two-intercept: $x$-int $=-\frac{11}{2}$, $y$-int $=\frac{11}{4}$.
3. Normal: $r=\sqrt{4+16}=2\sqrt{5}$; since $c=11>0$, divide by $-r$: $\cos \alpha =-\frac{1}{\sqrt{5}},\sin \alpha =\frac{2}{\sqrt{5}},p=\frac{11}{2\sqrt{5}}$.

Answer: $y=\frac{x}{2}+\frac{11}{4}$; intercepts $-\frac{11}{2},\frac{11}{4}$; normal form as above

Q15(ii) — Convert: $4x+7y-2=0$

Convert into all three forms.

1. Slope-intercept: $y=-\frac{4}{7}x+\frac{2}{7}$.
2. Two-intercept: $x$-int $=\frac{1}{2}$, $y$-int $=\frac{2}{7}$.
3. Normal: $r=\sqrt{16+49}=\sqrt{65}$; $c=-2<0$, divide by $+r$: $\cos \alpha =\frac{4}{\sqrt{65}},\sin \alpha =\frac{7}{\sqrt{65}},p=\frac{2}{\sqrt{65}}$.

Answer: $y=-\frac{4}{7}x+\frac{2}{7}$; intercepts $\frac{1}{2},\frac{2}{7}$; normal form as above

Q15(iii) — Convert: $15y-8x+3=0$

Convert into all three forms.

1. Slope-intercept: $y=\frac{8}{15}x-\frac{1}{5}$.
2. Two-intercept: $x$-int $=\frac{3}{8}$, $y$-int $=-\frac{1}{5}$.
3. Normal: $r=\sqrt{64+225}=17$; $c=3>0$, divide by $-r$: $\cos \alpha =\frac{8}{17},\sin \alpha =-\frac{15}{17},p=\frac{3}{17}$.

Answer: $y=\frac{8}{15}x-\frac{1}{5}$; intercepts $\frac{3}{8},-\frac{1}{5}$; normal form as above

Q16(i) — Check: $2x+y-3=0$; $4x+2y+5=0$

Determine the relationship.

1. Slopes: $-2$ and $-2$ (from $4x+2y+5=0\Rightarrow y=-2x-2.5$) — equal.

Answer: Parallel

Q16(ii) — Check: $3y=2x+5$; $3x+2y-8=0$

Determine the relationship.

1. Slopes: $\frac{2}{3}$ and $-\frac{3}{2}$; product $=-1$.

Answer: Perpendicular

Q16(iii) — Check: $4y+2x-1=0$; $x-2y-7=0$

Determine the relationship.

1. Slopes: $-\frac{1}{2}$ and $\frac{1}{2}$; product $=-\frac{1}{4}$, and slopes unequal.

Answer: Neither parallel nor perpendicular

Q17(i) — Through $(-4,7)$, parallel to $2x-7y+4=0$

Find the equation.

1. Given line's slope $=\frac{2}{7}$.
2. $y-7=\frac{2}{7}(x+4)$.

Answer: $y=\frac{2}{7}x+\frac{57}{7}$

Q17(ii) — Through $(5,-8)$, perpendicular to join of $A(-15,-8),B(10,7)$

Find the equation.

1. Slope $AB=\frac{15}{25}=\frac{3}{5}$, so perpendicular slope $=-\frac{5}{3}$.
2. $y+8=-\frac{5}{3}(x-5)$.

Answer: $y=-\frac{5}{3}x+\frac{1}{3}$

WP1 — Distance between two houses: $(2,6),(9,12)$ km

Find the straight-line distance.

1. $d=\sqrt{(9-2{)}^2+(12-6{)}^2}=\sqrt{49+36}$.

Answer: $\sqrt{85}$ km

WP2 — Trail midpoint: $(5,7)$ to $(15,3)$

Find the midpoint.

1. Midpoint $=\left(\frac{5+15}{2},\frac{7+3}{2}\right)$.

Answer: $(10,5)$

WP3 — Distance between buildings: $(10,8),(4,3)$ m

Find the distance.

1. $d=\sqrt{(10-4{)}^2+(8-3{)}^2}=\sqrt{36+25}$.

Answer: $\sqrt{61}$ m

WP4 — Delivery distance: $(7,2),(12,10)$ km

Find the distance.

1. $d=\sqrt{(12-7{)}^2+(10-2{)}^2}=\sqrt{25+64}$.

Answer: $\sqrt{89}$ km

WP5 — Race track midpoint: $(3,9),(9,3)$

Find the midpoint.

1. Midpoint $=\left(\frac{3+9}{2},\frac{9+3}{2}\right)$.

Answer: $(6,6)$

WP6 — Road midpoint: $A(3,4),B(7,10)$

Find the midpoint.

1. Midpoint $=\left(\frac{3+7}{2},\frac{4+10}{2}\right)$.

Answer: $(5,7)$

WP7 — Distance between ports (coordinate practice): $(12,65),(20,45)$

Treating the given lat/long numbers as plane coordinates, find the distance.

1. $d=\sqrt{(20-12{)}^2+(65-45{)}^2}=\sqrt{64+400}$.

Answer: $4\sqrt{29}$ (coordinate-distance practice; real geographic distance needs spherical trigonometry)

WP8 — Fence perimeter: rectangle $(0,0),(0,5),(8,5),(8,0)$

Find the perimeter.

1. Sides $=5$ and $8$; perimeter $=2(5+8)$.

Answer: $26$ m

WP9 — Airplane distance: $(40,100),(50,80)$

Treating the given lat/long numbers as plane coordinates, find the distance.

1. $d=\sqrt{(50-40{)}^2+(100-80{)}^2}=\sqrt{100+400}$.

Answer: $10\sqrt{5}$

WP10 — Surveyor's perimeter: $(3,1),(3,6),(8,6),(8,1)$

Find the perimeter.

1. This is a square with side $5$; perimeter $=4\times 5$.

Answer: $20$

WP11 — Garden fence: $A(0,0),B(5,0),C(5,3),D(0,3)$

Find the amount of fencing needed.

1. Sides $=5$ and $3$; perimeter $=2(5+3)$.

Answer: $16$

Q18(i) — Distance: $A(2,3),B(7,8)$

Find the distance.

1. $d=\sqrt{(7-2{)}^2+(8-3{)}^2}=\sqrt{25+25}$.

Answer: $5\sqrt{2}$

Q18(ii) — Midpoint: $(4,-2),(-6,3)$

Find the midpoint.

1. Midpoint $=\left(\frac{4-6}{2},\frac{-2+3}{2}\right)$.

Answer: $\left(-1,\frac{1}{2}\right)$

Q18(iii) — Gradient: $(1,2),(4,6)$

Find the gradient.

1. Slope $=\frac{6-2}{4-1}$.

Answer: $\frac{4}{3}$

Q18(iv) — Equation $y=mx+c$ through $(3,7),(5,11)$

Find the equation.

1. Slope $=\frac{11-7}{5-3}=2$.
2. $y-7=2(x-3)$.

Answer: $y=2x+1$

Q18(v) — Parallel gradient, given $\frac{2}{3}$

Find the gradient of the other (parallel) line.

1. Parallel lines have equal slopes.

Answer: $\frac{2}{3}$

WP12 — Airplane distance: $A(12,5)$ to $B(8,-4)$

Find the distance.

1. $d=\sqrt{(12-8{)}^2+(5+4{)}^2}=\sqrt{16+81}$.

Answer: $\sqrt{97}$

WP13 — Path midpoint: $(2,3)$ to $(10,7)$

Find the midpoint.

1. Midpoint $=\left(\frac{2+10}{2},\frac{3+7}{2}\right)$.

Answer: $(6,5)$

WP14 — Drone gradient and distance: $(2,3)$ to $(10,15)$

Find the gradient and the total distance.

1. Gradient $=\frac{15-3}{10-2}=\frac{12}{8}$.
2. Distance $=\sqrt{(10-2{)}^2+(15-3{)}^2}=\sqrt{64+144}$.

Answer: Gradient $=\frac{3}{2}$; distance $=4\sqrt{13}$

Q19(i) — Slope-intercept form

Write the equation in slope-intercept form.

1. Directly from the given gradient and $y$-intercept.

Answer: $y=-3x+2$

Q19(ii) — Point-slope form (using the point $(0,2)$ on the line)

Write the equation in point-slope form.

1. $(0,2)$ lies on $y=-3x+2$: $y-2=-3(x-0)$.

Answer: $y-2=-3x$

Q19(iii) — Two-point form (using $(0,2)$ and $(1,-1)$, both on the line)

Write the equation in two-point form.

1. Slope check: $\frac{-1-2}{1-0}=-3$ ✓ matches.
2. $\frac{y-2}{x-0}=-3$.

Answer: $y-2=-3x$ (same line, confirming consistency)

Q19(iv) — Intercepts form

Write the equation in intercepts form.

1. $x$-intercept: $0=-3x+2\Rightarrow x=\frac{2}{3}$; $y$-intercept $=2$.
2. $\frac{x}{2/3}+\frac{y}{2}=1$.

Answer: $3x+y=2$

Q19(v) — Symmetric form

Write the equation in symmetric form.

1. Direction vector $(1,-3)$, magnitude $\sqrt{10}$.
2. $\frac{x-0}{1/\sqrt{10}}=\frac{y-2}{-3/\sqrt{10}}=r$.

Answer: $\frac{x}{1/\sqrt{10}}=\frac{y-2}{-3/\sqrt{10}}=r$

Q19(vi) — Normal form

Write the equation in normal form.

1. From $3x+y-2=0$: $r=\sqrt{9+1}=\sqrt{10}$; $c=-2<0$, divide by $+r$.

Answer: $\cos \alpha =\frac{3}{\sqrt{10}},\sin \alpha =\frac{1}{\sqrt{10}},p=\frac{2}{\sqrt{10}}=\frac{\sqrt{10}}{5}$