Question

Number 1d please help me analytical geometry

Answer 1

For a) is just the distance formula

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ x}}\quad ,&{{ 1}})\quad % (c,d) B&({{ -4}}\quad ,&{{ 1}}) \end{array}\qquad % distance value \begin{array}{llll} d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2} \end{array}$
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for b)  is also the distance formula, just different coordinates and distance

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ -7}}\quad ,&{{ y}})\quad % (c,d) B&({{ -3}}\quad ,&{{ 4}}) \end{array}\ \ \begin{array}{llll} d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ 4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2} \end{array}$
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for c)  well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is $\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ -3}}\quad ,&{{ 0}})\quad % (c,d) B&({{ 5}}\quad ,&{{ -2}}) \end{array}\qquad % distance value \begin{array}{llll} d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\ d=\boxed{?} \end{array}$

now.. whatever that is, is  = BC, so  the distance for BC is

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) B&({{ 5}}\quad ,&{{ -2}})\quad % (c,d) C&({{ -13}}\quad ,&{{ y}}) \end{array}\qquad % distance value \begin{array}{llll} d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\ d=BC\\\\ BC=\boxed{?} \end{array}$

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
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now d)   we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

$\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) M&({{-2}}\quad ,&{{ 1}})\quad % (c,d) N&({{ x}}\quad ,&{{ 1}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P$

$\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies \begin{cases} \cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\ \cfrac{{{ y_2}} + {{ y_1}}}{2}=4 \end{cases}$

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"