![Tony B]()
Given:" "3y+7x=4
Write as color(blue)(y=-7/3x+4/3" "larr" first line")....(1)
Thus the gradient of the line normal to this (perpendicular) is:
(-1)xx1/m" "->" "(-1)xx(-3/7)=+3/7 giving:
" "color(blue)(y=3/7x+c" "larr" second line")......(2)
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color(brown)("Determine the value of "c)
The second line passes through the point P_1->(x,y)->(2,4)
so by substitution:
color(red)(y=3/7x+c)color(green)(" "->" "4=3/7(2)+c)
c" "=" "4-6/7" "=" " 3 1/7 -> 22/7 giving:
" "cancel(color(blue)(y=3/7x+19/7" "larr" second line")...(3))
" "color(blue)(y=3/7x+color(red)(22/7)" "larr" second line corrected")...(3)" "
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color(brown)("Determine point of intersection " P_2) (mid point)
Relating equation (3) to equation (1) through y
" "3/7x+22/7" "=" "y" "=" "-7/3x+4/3
" "=>3/7x+7/3x" "=" "4/3-22/7
" "(9+49)/21x" "=" "(28-66)/7
" "58x" "=" "-38
" "x=-38/58" "->" "-19/29
" "P_2->(x,y)=(-19/29 , y )
substitute into equation (3) to find P_2(y)
" "y=3/7x+22/7" "->" " y=3/7(-19/29)+ 22/7
" "y=2 25/29 -> 81/29
color(blue)(" "P_2->(x,y)=(-19/29 , 81/29 ))
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color(brown)("Determine point "P_3)
P_1(x)" to "P_2(x)" " =" " P_2(x)-P_1(x)
" "=" "2-(-19/22) = 2 19/22
,................................................................................................
" "P_3(x)" "=" "P_1(x)-2(2 19/22)
" "P_3(x)" "=" "2-2(2 19/22)" "=" "-3 8/11
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Substitute for x in equation (3) to find P_3(y)
P_3(y)=3/7x+22/7" "->" "3/7(-3 8/11)+22/7
P_3(y)=1 6/11
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P_3(x,y)=(-3 8/11,1 6/11)
