Analytical Geometry Part 2
Question 3
Shortest distance = straight perpendicular line that connets ‘A‘ to a point on line ‘GH‘
‘MGH=38+1642+30=5472=34‘
‘M⊥GH=4−3‘
‘y⊥GH−32=4−3(x+16)‘
‘y⊥GH=4−3x+20(1)‘
‘yGH+30=34(x+16)‘
‘yGH=34x−326(2)‘
\begin{cases}
y_{\perp GH} = \dfrac{-3}{4}x + 20 & \text{(1)} \\
\\
y_{GH} = \dfrac{4}{3}x - \dfrac{26}{3} & \text{(2}) \\
\end{cases}
Sub ‘(1)‘ into ‘(2)‘
‘4−3x+20=34x−326‘
‘−9x+(12)20=16x−4(26)‘
‘25x=344‘
‘x=25344(3)‘
Sub ‘(3)‘ into ‘(1)‘
‘y=4−3(25344)+20‘
‘y=25−258+20‘
‘y=25−258+25500‘
‘y=25242‘
Distance ‘=(−16−25344)2+(32−25242)2=37.2‘
‘∴‘ The shortest length
pipe is ‘37.2‘ units.
Question 4
Let ‘(x,y)‘ be the center of
the circle, and ‘r‘ be the radius
of the circle.
\begin{cases}
(x-4)^2 + (y-8)^2 = r^2 & \text{(1)} \\
(x-5)^2 + (y-1)^2 = r^2 & \text{(2)} \\
(x+2)^2 + y^2 = r^2 & \text{(3)} \\
\end{cases}
Sub ‘(1)‘ into ‘(2)‘
‘x2−8x+16+y2−16y+64=x2−10x+25+y2−2y+1‘
‘−8x−16y+80=−10x−2y+26‘
‘2x−14y=−54‘
‘x−7y=−27(4)‘
Sub ‘(2)‘ into ‘(3)‘
‘x2−10x+25+y2−2y+1=x2+4x+4+y2‘
‘−10x−2y+26=4x+4‘
‘14x+2y=22‘
‘7x+y=11‘
‘y=11−7x(5)‘
Sub ‘(5)‘ into ‘(4)‘
‘x−7(11−7x)=−27‘
‘x−77+49x=27‘
‘50x=50‘
‘x=1(6)‘
Sub ‘(6)‘ into ‘(5)‘
‘y=11−7(1)‘
‘y=4(7)‘
Sub ‘(6),(7)‘ into ‘(3)‘
‘(1+2)2+42=r2‘
‘r2=16+9‘
‘r2=25‘
‘∴(x−1)2+(y−4)2=25‘ is the equation of the circle.