Points, straight lines and shapes in 2D

Unit 4 Points, straight lines and shapes in 2D

Key unit competence: Apply the meaning of coordinates in solving problems involving points, lines and shapes in two dimensions.

Key ideas

• Distance $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. Midpoint $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. Point dividing $AB$ in the ratio $m:n$: $\left(\frac{n{x}_1+m{x}_2}{m+n},\frac{n{y}_1+m{y}_2}{m+n}\right)$.
• Equations of a line: $y=mx+c$; $ax+by+c=0$ (gradient $-\frac{a}{b}$); through a point with direction vector $\vec{u}=(u_1,u_2)$: parametric $x=x_0+t{u}_1$, $y=y_0+t{u}_2$; symmetric $\frac{x-x_0}{u_1}=\frac{y-y_0}{u_2}$.
• Lines are parallel if gradients are equal, perpendicular if $m_1{m}_2=-1$. Angle between lines: $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$.
• Distance from a point $(x_0,y_0)$ to the line $ax+by+c=0$: $d=\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$.
• Circle with centre $(a,b)$ and radius $r$: $(x-a{)}^2+(y-b{)}^2=r^2$. Expanded: $x^2+y^2+2gx+2fy+c=0$, centre $(-g,-f)$, $r=\sqrt{g^2+f^2-c}$.
• A line is tangent to a circle when the distance from the centre to the line equals $r$; the tangent at a point is perpendicular to the radius there.

Formulas to know

Common misconceptions

• "The centre of $x^2+y^2-4x+6y-3=0$ is $(-4,6)$." No: centre $(2,-3)$, half the coefficients with opposite signs.
• "The distance from a point to a line is the distance to any point on the line." No: it is the perpendicular (shortest) distance.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find the equation of the perpendicular bisector of $A(2,3)$, $B(6,7)$.

   Show answer

   Midpoint $(4,5)$; $m_{AB}=1$, so $m=-1$: $y=-x+9$.

   Common mistake: Using the gradient of $AB$ itself.

2. Find the circles with centre $(3,k)$ that pass through $(1,0)$ and are tangent to the y-axis.

   Show answer

   Tangent to the y-axis: $r=3$. Through $(1,0)$: $4+k^2=9$, $k=\pm \sqrt{5}$. Circles $(x-3{)}^2+(y\mp \sqrt{5}{)}^2=9$.

   Common mistake: Assuming $k=0$.

3. Find the distance from $(2,-1)$ to the line $5x-12y+4=0$.

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   $\frac{|10+12+4|}{13}=2$.

   Common mistake: Forgetting the absolute value or the square root.

4. Show that $A(1,1)$, $B(5,4)$, $C(2,8)$ form a right-angled triangle.

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   $A{B}^2=25$, $A{C}^2=50$, $B{C}^2=25$: $A{B}^2+B{C}^2=A{C}^2$, right angle at $B$.

   Common mistake: Comparing gradients incorrectly.

5. Write the circle $x^2+y^2+6x-2y-15=0$ in standard form.

   Show answer

   $(x+3{)}^2+(y-1{)}^2=25$: centre $(-3,1)$, radius 5.

   Common mistake: Sign errors when completing the square.

6. Write parametric and symmetric equations of the line through $A(1,3)$ and $B(4,-1)$.

   Show answer

   Direction $(3,-4)$: $x=1+3t$, $y=3-4t$; $\frac{x-1}{3}=\frac{y-3}{-4}$.

   Common mistake: Using the coordinates of $B$ as the direction.

7. Find the points where $y=x+1$ meets $x^2+y^2=13$.

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   $x^2+(x+1{)}^2=13$, $x^2+x-6=0$: $(2,3)$ and $(-3,-2)$.

   Common mistake: Expanding $(x+1{)}^2$ as $x^2+1$.

8. Find the equation of the line through $(1,-2)$ perpendicular to $2x+y=7$.

   Show answer

   Gradient of the given line $-2$, so $m=\frac{1}{2}$: $y+2=\frac{1}{2}(x-1)$, i.e. $x-2y-5=0$.

   Common mistake: Using $-2$ again.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Write the symmetric equation of the line through $(1,-2)$ with direction $(4,3)$.	a. about $81.9^{\circ }$
   2. Find the points where $y=x+1$ meets $x^2+y^2=13$.	b. $(2,3)$ and $(-3,-2)$
   3. Find the centre and radius of $x^2+y^2+6x-8y=0$.	c. $\frac{x-1}{4}=\frac{y+2}{3}$
   4. Find the acute angle between $2x-y+1=0$ and $x+3y-2=0$.	d. Centre $(-3,4)$, radius 5

   Show answer

   1 → c ($\frac{x-1}{4}=\frac{y+2}{3}$) · 2 → b ($(2,3)$ and $(-3,-2)$) · 3 → d (Centre $(-3,4)$, radius 5) · 4 → a (about $81.9^{\circ }$)

   • Step 1. $\frac{x-x_0}{u_1}=\frac{y-y_0}{u_2}$.
     Step 2. $x_0=1$, $y_0=-2$, $u_1=4$, $u_2=3$.
   • Step 1. $x^2+(x+1{)}^2=13$, so $2x^2+2x-12=0$.
     Step 2. $x^2+x-6=0$: $x=2$ or $x=-3$; then $y=x+1$.
   • Step 1. $(x+3{)}^2-9+(y-4{)}^2-16=0$.
     Step 2. $(x+3{)}^2+(y-4{)}^2=25$.
   • Step 1. $m_1=2$, $m_2=-\frac{1}{3}$.
     Step 2. $\tan \theta =\left|\frac{2+\frac{1}{3}}{1-\frac{2}{3}}\right|=\frac{7/3}{1/3}=7$, so $\theta \approx 81.9^{\circ }$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $c=\pm 4$  ·  $(x-3{)}^2+(y-4{)}^2=16$  ·  $c=\pm 2$  ·  $2x-3y-7=0$
   1. A line passes through $(2,-1)$ with direction vector $(3,2)$. Its Cartesian equation is .
   2. For which $c$ is $y=x+c$ tangent to $x^2+y^2=8$? →
   3. Find the circle with centre $(3,4)$ that touches the $x$-axis. →
   Show answer

   1: $2x-3y-7=0$ · 2: $c=\pm 4$ · 3: $(x-3{)}^2+(y-4{)}^2=16$

   • Step 1. Normal vector $(2,-3)$ is perpendicular to $(3,2)$: $2x-3y+c=0$.
     Step 2. Through $(2,-1)$: $4+3+c=0$, so $c=-7$.
   • Step 1. Distance from $O$ to $x-y+c=0$: $\frac{|c|}{\sqrt{2}}$.
     Step 2. $\frac{|c|}{\sqrt{2}}=\sqrt{8}$, so $|c|=4$.
   • Step 1. The radius is the distance to the $x$-axis: 4.
     Step 2. $(x-3{)}^2+(y-4{)}^2=16$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. For which $k$ is $y=2x+k$ tangent to $x^2+y^2=5$?	a. $(2,5)$
   2. Find the point dividing $[AB]$ in the ratio $2:1$ from $A$, with $A(-2,1)$ and $B(4,7)$.	b. $x+y=7$
   3. Find the point on the $x$-axis equidistant from $A(1,3)$ and $B(5,1)$.	c. $(2,0)$
   4. Find the perpendicular bisector of $[AB]$ with $A(1,2)$ and $B(5,6)$.	d. $k=\pm 5$

   Show answer

   1 → d ($k=\pm 5$) · 2 → a ($(2,5)$) · 3 → c ($(2,0)$) · 4 → b ($x+y=7$)

   • Step 1. Distance from $O$ to $2x-y+k=0$: $\frac{|k|}{\sqrt{5}}$.
     Step 2. Tangent when $\frac{|k|}{\sqrt{5}}=\sqrt{5}$: $|k|=5$.
   • Step 1. $P=\left(\frac{1\times (-2)+2\times 4}{3},\frac{1\times 1+2\times 7}{3}\right)$.
     Step 2. $=(2,5)$. Check: $\overrightarrow{AP}=(4,4)=\frac{2}{3}\overrightarrow{AB}$. ✓
   • Step 1. $(x-1{)}^2+9=(x-5{)}^2+1$.
     Step 2. $-2x+10=-10x+26$, so $8x=16$ and $x=2$.
   • Step 1. Midpoint $(3,4)$; gradient of $AB$ is 1, so the bisector has gradient $-1$.
     Step 2. $y-4=-(x-3)$: $x+y=7$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $19$  ·  $5$  ·  $9.5$  ·  $k=\frac{2}{3}$
   1. For which $k$ are $kx+2y=1$ and $3x-y=4$ perpendicular? →
   2. Find the area of the triangle $A(1,1)$, $B(5,2)$, $C(2,6)$. →
   3. Find the length of the tangent from $P(7,1)$ to $x^2+y^2=25$. →
   Show answer

   1: $k=\frac{2}{3}$ · 2: $9.5$ · 3: $5$

   • Step 1. Gradients $-\frac{k}{2}$ and 3.
     Step 2. $-\frac{k}{2}\times 3=-1$, so $k=\frac{2}{3}$.
   • Step 1. $\frac{1}{2}\left|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\right|$.
     Step 2. $\frac{1}{2}|1(-4)+5(5)+2(-1)|=\frac{19}{2}=9.5$.
   • Step 1. $O{P}^2=49+1=50$.
     Step 2. Tangent length $=\sqrt{O{P}^2-r^2}=\sqrt{50-25}=5$.

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