12. Another problem on coordinate geometry (GCSE)

수학 모음 (Maths collection) 전체보기

 

Question. In the diagram, $ABC$ is the line with equation $$ \begin{align} y=-\frac12x+5 \end{align} $$

 

It is also given that $AB=BC$ and that $D$ is the point with coordinates $(-13,0)$.

 

Find an equation of the line through $A$ and $D$.

 

Solution.

• $B$ is the $y$-intercept of $y=-\frac12x+5$: $$ \begin{align} B=(0,5) \end{align} $$
• $C$ is the $x$-intercept of $y=-\frac12x+5$: $$ \begin{align} 0&=-\frac12x+5 \\ x&=10 \\ \\ \Rightarrow\quad C&=(10,0) \end{align} $$
• Since $AB=BC$, by considering similar triangles for example, we find $$ \begin{align} A=(-10,10) \end{align} $$

 

To find the line $AD$, we need the gradient and a point on the line:

• Gradient: $$ \begin{align} \frac{\Delta y}{\Delta x} =\frac{10-0}{-10-(-13)} =\frac{10}{3} \end{align} $$ So we have $y=\frac{10}{3}x+c$.
• To find $c$: $$ \begin{align} {\rm Going\; through\; }A(-10,10)\;:&& 10&=\frac{10}{3}\times(-10)+c &&\Rightarrow& c&=10+\frac{100}{3}=\frac{130}{3} \\ {\rm Going\; through\; }D(-13,0)\;:&& 0&=\frac{10}{3}\times(-13)+c &&\Rightarrow& c&=\frac{130}{3} \quad\checkmark \end{align} $$
• Thus, we obtain: $$ \begin{align} AD\;:\qquad y=\frac{10}{3}x+\frac{130}{3} \quad\checkmark \end{align} $$