Question

find the distance from point c to point r in the graph, given that the equation of line de is \\(\frac{x}{200} + \frac{y}{60} = 1\\).
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$$\begin{tikzpicture}scale=0.5 \\draw-> (0,0) -- (210,0) noderight {$x$}; \\draw-> (0,0) -- (0,70) nodeabove {$y$}; \ odeabove left at (0,60) {$d$}; \ oderight at (200,0) {$e$}; \ odeabove at (32,63) {$a (32,63)$}; \ odeabove at (52,64) {$c (52,64)$}; \ odebelow at (72,-12) {$b (72,-12 )$}; \ odebelow left at (some point on de near a) {$r$}; \\draw (0,60) -- (200,0); \\draw (32,63) -- (72,-12); \\fill (32,63) circle (2pt); \\fill (52,64) circle (2pt); \\fill (72,-12) circle (2pt); \\fill (0,60) circle (2pt); \\fill (200,0) circle (2pt); \\fill (some point on de near a) circle (2pt); \\end{tikzpicture}$$

Explanation:

Step1: Find equation of line AB

First, find the slope of line AB using points \( A(32, 63) \) and \( B(72, -12) \). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
So, \( m=\frac{-12 - 63}{72 - 32}=\frac{-75}{40}=-\frac{15}{8} \).
Using point - slope form \( y - y_1=m(x - x_1) \) with point \( A(32, 63) \):
\( y - 63=-\frac{15}{8}(x - 32) \)
\( y - 63=-\frac{15}{8}x+60 \)
\( y=-\frac{15}{8}x + 123 \)

Step2: Find equation of line DE

The equation of line DE is \( \frac{x}{200}+\frac{y}{60}=1 \). We can rewrite it in slope - intercept form \( y=mx + b \).
Multiply through by 60 to get \( \frac{60x}{200}+y = 60 \), so \( y=-\frac{3}{10}x + 60 \)

Step3: Find coordinates of point R (intersection of AB and DE)

Set the equations of AB and DE equal to each other:
\( -\frac{15}{8}x + 123=-\frac{3}{10}x + 60 \)
Multiply through by 40 (the least common multiple of 8 and 10) to clear the fractions:
\( - 75x+4920=-12x + 2400 \)
\( -75x + 12x=2400 - 4920 \)
\( -63x=-2520 \)
\( x = 40 \)
Substitute \( x = 40 \) into the equation of DE: \( y=-\frac{3}{10}(40)+60=-12 + 60 = 48 \). So, point \( R \) is \( (40, 48) \)

Step4: Find distance between point \( C(52, 64) \) and point \( R(40, 48) \)

Use the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \).
Here, \( x_1 = 52,y_1 = 64,x_2 = 40,y_2 = 48 \).
\( d=\sqrt{(40 - 52)^2+(48 - 64)^2}=\sqrt{(-12)^2+(-16)^2}=\sqrt{144 + 256}=\sqrt{400}=20 \)

Answer:

The distance from point \( C \) to point \( R \) is \( 20 \)