COORDINATE GEOMETRY · HL

Area, Intersection & the Circle

Area of a triangle, intersection of lines, points on a line, and the circle.

Section 1 of 5

Area of a triangle

$A = \dfrac{1}{2}\left|x_1 y_2 - x_2 y_1\right|$

One vertex must be $(0,0)$ — slide every vertex by the same amount to move one to the origin.

(i) Area of $(-1,2)$, $(3,5)$, $(1,-4)$

$(-1,2) \xrightarrow{+1,\,-2} (0,0)$

$(3,5) \xrightarrow{+1,\,-2} (4,3) = x_1,y_1$

$(1,-4) \xrightarrow{+1,\,-2} (2,-6) = x_2,y_2$

$= \dfrac{1}{2}\left|4(-6) - 2(3)\right| = \dfrac{1}{2}\left|-24 - 6\right| = \dfrac{1}{2}\left|-30\right|$

$= 15 \text{ sq units}$

(ii) Area of $(3,1)$, $(5,7)$, $(6,4)$

$(3,1) \xrightarrow{-3,\,-1} (0,0)$

$(5,7) \to (2,6) = x_1,y_1 \qquad (6,4) \to (3,3) = x_2,y_2$

$= \dfrac{1}{2}\left|2(3) - 3(6)\right| = \dfrac{1}{2}\left|6 - 18\right| = \dfrac{1}{2}\left|-12\right|$

$= 6 \text{ sq units}$

(iii) $(7,-3)$, $(4,-1)$, $(-1,2)$

$(7,-3) \to (0,0) \qquad (4,-1) \to (-3,2) = x_1,y_1 \qquad (-1,2) \to (-8,5) = x_2,y_2$

$= \dfrac{1}{2}\left|-3(5) + 8(2)\right| = \dfrac{1}{2}\left|-15 + 16\right|$

$= \dfrac{1}{2} \text{ sq unit}$

(iv) $(-3,2)$, $(-1,5)$, $(3,-6)$

$(-3,2) \to (0,0) \qquad (-1,5) \to (2,3) = x_1,y_1 \qquad (3,-6) \to (6,-8) = x_2,y_2$

$= \dfrac{1}{2}\left|2(-8) - 6(3)\right| = \dfrac{1}{2}(34)$

$= 17 \text{ sq units}$

(v) Triangle $ABC$: $A(1,3)$, $B(-2,5)$, $C(4,9)$

$(1,3) \to (0,0) \qquad (-2,5) \to (-3,2) = x_1,y_1 \qquad (4,9) \to (3,6) = x_2,y_2$

$= \dfrac{1}{2}\left|-3(6) - 3(2)\right| = \dfrac{1}{2}\left|-24\right|$

$= 12 \text{ sq units}$

(vi) $(2,-1)$, $(-3,5)$, $(4,2)$

$(2,-1) \to (0,0) \qquad (-3,5) \to (-5,6) = x_1,y_1 \qquad (4,2) \to (2,3) = x_2,y_2$

$= \dfrac{1}{2}\left|-5(3) - 2(6)\right| = \dfrac{1}{2}\left|-27\right|$

$= \dfrac{27}{2} \text{ sq units}$

(vii) $(2,-1)$, $(3,2)$, $(-4,5)$

$(2,-1) \to (0,0) \qquad (3,2) \to (1,3) = x_1,y_1 \qquad (-4,5) \to (-6,6) = x_2,y_2$

$= \dfrac{1}{2}\left|1(6) + 6(3)\right| = \dfrac{1}{2}\left|24\right|$

$= 12 \text{ sq units}$

Section 2 of 5

Intersection of two lines

Intersection $=$ simultaneous equations.

(i) Where do $x + y = 5$ and $x - y = 1$ intersect?

$\begin{aligned} x + y &= 5 \\ x - y &= 1 \end{aligned}$ (add)

$2x = 6 \quad \to \quad x = 3$

Sub $x = 3$ into $x + y = 5$: $3 + y = 5$

$y = 2 \quad \to \quad (3, 2)$

(ii) Where do $2x + y = 3$ and $x + 3y = 4$ intersect?

$6x + 3y = 9$ (first $\times 3$)

$-x - 3y = -4$ (second $\times -1$)

$5x = 5 \quad \to \quad x = 1$

Sub $x = 1$ into $2x + y = 3$: $2 + y = 3$

$y = 1 \quad \to \quad (1, 1)$

Section 3 of 5

Midpoint in reverse — find the endpoint

Whatever takes you from $A$ to the midpoint $P$ takes you again from $P$ to $B$.

(i) $P(-1,3)$ is the midpoint of $A(4,-2)$ and $B$. Find $B$.

$A(4,-2) \xrightarrow{-5,\,+5} P(-1,3) \xrightarrow{-5,\,+5} B$

$B(-6, 8)$

(ii) $P(-2,6)$ is the midpoint of $A(4,-2)$ and $B$. Find $B$.

$A(4,-2) \xrightarrow{-6,\,+8} P(-2,6) \xrightarrow{-6,\,+8} B$

$B(-8, 14)$

Section 4 of 5

Point on a line & finding $k$

Key idea

1.Point on line $\Rightarrow$ sub in.

(i) Verify $(2,3)$ is on $x + 5y = 17$

$2 + 5(3) = 17$

$2 + 15 = 17$

$17 = 17$ True — point on line.

(ii) Is $(2,5)$ on $3x + y = 20$?

$3(2) + 5 = 20$

$6 + 5 = 20$

$11 = 20$ not true — point is NOT on line.

(iii) $(2,5)$ is on $3x + y = k$. Find $k$.

$3(2) + 5 = k$

$6 + 5 = k$

$k = 11$

(iv) $(2,-1)$ is on $3x - 5y = k$. Find $k$.

$3(2) - 5(-1) = k$

$6 + 5 = k$

$k = 11$

Section 5 of 5

The circle

$(x - h)^2 + (y - k)^2 = r^2 \qquad \text{Centre } (h, k)$

(i) Find centre and radius of $(x-3)^2 + (y+5)^2 = 25$

Centre $(3, -5) \qquad r = 5$

(ii) Is $(3,5)$ inside, on or outside $x^2 + y^2 = 35$?

$3^2 + 5^2$

$9 + 25$

$34 < 35 \quad \Rightarrow \quad$ inside.

SUM

The lot in one box

Area, intersection & circle toolkit

1.Area $= \dfrac{1}{2}\left|x_1 y_2 - x_2 y_1\right|$ — slide one vertex to $(0,0)$ first.

2.Intersection $=$ solve the two equations simultaneously.

3.Midpoint in reverse: repeat the step from $A \to P$ again to reach $B$.

4.Point on a line $\Rightarrow$ sub in.

5.Circle $(x-h)^2 + (y-k)^2 = r^2$: centre $(h,k)$, radius $r$.

End of lesson

Area, Intersection & the Circle — HL · Mathslive.ie