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Bisection and Lengths of Line Segments - Questions

Question 1
17637

The midpoint of \(PQ\) has the co - ordinates \({\rm{(}}4, - 1).\,\) \(P\) is the point \({\rm{(3}}, - 4),\,\,Q\) is the point? 

\(Q(5,2)\)

\begin{align}
&\text{Let } Q \text{ have the coordinates } \left(x_{2}, y_{2}\right)\\
&\begin{aligned}
\therefore \frac{3+x_{2}}{2} &=4,& \frac{-4+y_2}{2}&=-1 \\
3+x_{2} &=8 &-4+y_2&=-2 \\
x_{2} &=5 & y_{2}&=2
\end{aligned}\\
&\rightarrow Q(5,2)
\end{align}

Question 2
17638

The perpendicular bisector of the line segment joining \(A( - 1,2)\) and \(B(3,4)\) is?

\(2x + y - 5 = 0\)

\begin{align}
&m_{AB}=\frac{4-2}{3-(-1)}=\frac{2}{4}=\frac{1}{2} \\
&\text {mid point of } A B \rightarrow\left(\frac{-1+3}{2}, \frac{2+4}{2}\right) \rightarrow(1,3)\\
&\begin{aligned}
\text {Equation is : }& y-3=-2(x-1)\\
&y-3=-2x+2\\
&2x+y-5=0
\end{aligned}
\end{align}

Question 3
17646

The equation of the line with a gradient of \( - 2\) and passing through the midpoint of \(A( - 1,3)\) and \(B(5,1)\) is? 

\(2x + y - 6 = 0\)

\begin{align}
&\text {midpoint } \rightarrow\left(\frac{-1+5}{2}, \frac{3+1}{2}\right) \rightarrow (2,2) \\
&y-2=-2(x-2) \\
&y-2=-2x+4 \\
&2x+y-6=0
\end{align}

Question 4
17647

The distance between the points \(A(1,3)\) and \(B(p,1)\) is \(\sqrt {13} .\) The value of \(p\) is?

\(p = 4\,or\,p = - 2\)

\begin{align}
d&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\
&=\sqrt{(p-1)^{2}+(1-3)^{2}}\\
d&=\sqrt{13}\\
&\begin{aligned}
(p-1)^{2}+4&=13\\
(p-1)^{2}&=9 \\
p-1&=\pm 3\\
p&=1 \pm 3\\
&=4 \text { or }-2
\end{aligned}
\end{align}

Question 5
17649

From the adjacent figure determine the 

length of \({\rm{CD}}\) given that \({\rm{CD}} \bot {\rm{AB}}\)

\(\frac{{4\sqrt 5 }}{5}\)

\(\begin{align}
&\textbf{Method 1}\\
&m_{AB}=\frac{4-0}{0-2}=-2 \\
&m_{CD}=\frac{1}{2}\\
&CD \Rightarrow y=\frac{1}{2} x \\
&AB \Rightarrow y=-2 x+4\\
&\text {To locate } D:\\
&\quad \begin{aligned}
\frac{1}{2} x&=-2 x+4 \\
\frac{5}{2} x&=4 \\
x&=\frac{8}{5},\; y=\frac{4}{5} \\
\end{aligned}\\
&\begin{aligned}
\therefore CD&=\sqrt{\frac{64}{25}+\frac{16}{25}}\\
&=\frac{\sqrt{80}}{5}\\
&=\frac{4 \sqrt{5}}{5}
\end{aligned}
\end{align}\)

\(\begin{align}
&\textbf{Method 2}\\
&\text{Area of }\triangle ACB=\frac{1}{2} \times 2 \times 4=4\\
&\begin{aligned}
BA &=\sqrt{(0-2)^{2}+(4-0)^{2}} \\
&=\sqrt{20} \\
&=2 \sqrt{5}
\end{aligned}\\
&\begin{aligned}
\text{Let } CD&=h \\
\therefore 4 &=\frac{1}{2} h \times 2 \sqrt{5} \\
h&=\frac{4}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \rightarrow CD=\frac{4 \sqrt{5}}{5}
\end{aligned}
\end{align}\)