Answer
a) By using Pythagorean theorem: $( 3,11)$ and $\left( 3,-13 \right)$.
b) By using distance formula: $(3,11)$ and $( 3,-13)$.
Work Step by Step
a)
The coordinate of point ${{p}_{1}}$ is $\left( -2,-1 \right)$.
The $x-$co-ordinate of points is $3$, and their distance from the point$\left( -2,-1 \right)$ is $13$.
Therefore, the coordinate of point ${{p}_{2}}$is $\left( 3,y \right)$, and the coordinate of point ${{p}_{3}}$is $\left( 3,-1 \right)$.
The horizontal distance from ${{p}_{1}}$ to ${{p}_{3}}$ is $=\,|{{x}_{2}}-{{x}_{1}}|\,$ .
$=\,|3-\left( -2 \right)|\,$
$=\,|5|$
$=5$
Therefore, the horizontal distance from ${{p}_{1}}$ to ${{p}_{3}}$ is $5$.
The vertical distance from ${{p}_{3}}$to ${{p}_{2}}$is $=\,|{{y}_{2}}-{{y}_{1}}|\,$.
$=\,|y-\left( -1 \right)|\,$.
Therefore, the vertical distance between ${{p}_{3}}$ and ${{p}_{2}}$is $\,|y-\left( -1 \right)|\,$.
The distance between the point ${{p}_{1}}$ and the point${{p}_{2}}$ is $13$.
The length of the hypotenuse of the right angle triangle is $13$.
By using Pythagorean Theorem,
$\Rightarrow {{\left( 13 \right)}^{2}}={{5}^{2}}+|y+1{{|}^{2}}$
By simplifying,
$169=25+|y+1{{|}^{2}}$
Subtract $25$on both sides,
$169-25=25-25+|y+1{{|}^{2}}$
$144=|y+1{{|}^{2}}$
\[\Rightarrow \,|y+1{{|}^{2}}=144\]
By using square root method,
\[\Rightarrow \,|y+1|=\sqrt{144}\]
\[\Rightarrow \,|y+1|=12\]
By using the absolute value, if $|a|\,=b$then \[a=b\,\]or \[a=-b\,\],
\[\Rightarrow y+1=12\] or \[y+1=-12\]
By simplifying for \[y\],
\[\Rightarrow y=11\] or \[y=-13\]
The \[y\]coordinate of points are \[11\] and \[-13\].
Therefore, the points are \[\left( 3,11 \right)\]and \[\left( 3,-13 \right)\].
(b)
The points whose $x-$co-ordinate is $3$ means the points represent line $x=3$ or \[\left( 3,y \right)\].
From the graph of part (a) , and by using the distance formula with \[{{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,-1 \right)\] , ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,y \right)$ and $d=13$.
$13=\sqrt{{{\left( 3-\left( -2 \right) \right)}^{2}}+{{\left( y-\left( -1 \right) \right)}^{2}}}$
By simplifying,
$13=\sqrt{{{\left( 3+2 \right)}^{2}}+{{\left( y+1 \right)}^{2}}}$
Solving for $y$,
$13=\sqrt{{{\left( 5 \right)}^{2}}+{{\left( y+1 \right)}^{2}}}$
Squaring on both sides,
${{\left( 13 \right)}^{2}}={{\left( \sqrt{25+{{\left( y+1 \right)}^{2}}} \right)}^{2}}$
$169=25+{{\left( y+1 \right)}^{2}}$
Subtract $25$on both sides,
$169-25=25-25+{{\left( y+1 \right)}^{2}}$
$144={{\left( y+1 \right)}^{2}}$
$\Rightarrow {{\left( y+1 \right)}^{2}}=144$
By square root method,
$\Rightarrow y+1=\pm \sqrt{144}$
$\Rightarrow y+1=\pm 12$
Subtract $1$ on both sides,
$\Rightarrow y+1-1=\pm 12-1$
$\Rightarrow y=12-1$ and $y=-12-1$
$\Rightarrow y=11$ and $y=-13$
Therefore, the points are $\left( 3,11 \right)$ and $\left( 3,-13 \right)$.
