# A line segment is bisected by line with the equation  3 y + 7 x = 4 . If one end of the line segment is at (2 ,4 ), where is the other end?

Jun 22, 2016

${P}_{3} \left(x , y\right) = \left(- 3 \frac{8}{11} , 1 \frac{6}{11}\right)$

#### Explanation:

Given:$\text{ } 3 y + 7 x = 4$

Write as $\textcolor{b l u e}{y = - \frac{7}{3} x + \frac{4}{3} \text{ "larr" first line}} \ldots . \left(1\right)$
Thus the gradient of the line normal to this (perpendicular) is:

$\left(- 1\right) \times \frac{1}{m} \text{ "->" } \left(- 1\right) \times \left(- \frac{3}{7}\right) = + \frac{3}{7}$ giving:

" "color(blue)(y=3/7x+c" "larr" second line")......(2)
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$\textcolor{b r o w n}{\text{Determine the value of } c}$

The second line passes through the point ${P}_{1} \to \left(x , y\right) \to \left(2 , 4\right)$
so by substitution:

$\textcolor{red}{y = \frac{3}{7} x + c} \textcolor{g r e e n}{\text{ "->" } 4 = \frac{3}{7} \left(2\right) + c}$

$c \text{ "=" "4-6/7" "=" } 3 \frac{1}{7} \to \frac{22}{7}$ giving:

" "cancel(color(blue)(y=3/7x+19/7" "larr" second line")...(3))

$\text{ "color(blue)(y=3/7x+color(red)(22/7)" "larr" second line corrected")...(3)" }$

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$\textcolor{b r o w n}{\text{Determine point of intersection } {P}_{2}}$ (mid point)

Relating equation (3) to equation (1) through $y$

$\text{ "3/7x+22/7" "=" "y" "=" } - \frac{7}{3} x + \frac{4}{3}$

$\text{ "=>3/7x+7/3x" "=" } \frac{4}{3} - \frac{22}{7}$

$\text{ "(9+49)/21x" "=" } \frac{28 - 66}{7}$

$\text{ "58x" "=" } - 38$

$\text{ "x=-38/58" "->" } - \frac{19}{29}$

$\text{ } {P}_{2} \to \left(x , y\right) = \left(- \frac{19}{29} , y\right)$

substitute into equation (3) to find ${P}_{2} \left(y\right)$

$\text{ "y=3/7x+22/7" "->" } y = \frac{3}{7} \left(- \frac{19}{29}\right) + \frac{22}{7}$

$\text{ } y = 2 \frac{25}{29} \to \frac{81}{29}$

$\textcolor{b l u e}{\text{ } {P}_{2} \to \left(x , y\right) = \left(- \frac{19}{29} , \frac{81}{29}\right)}$
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$\textcolor{b r o w n}{\text{Determine point } {P}_{3}}$

${P}_{1} \left(x\right) \text{ to "P_2(x)" " =" } {P}_{2} \left(x\right) - {P}_{1} \left(x\right)$

$\text{ "=" } 2 - \left(- \frac{19}{22}\right) = 2 \frac{19}{22}$
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$\text{ "P_3(x)" "=" } {P}_{1} \left(x\right) - 2 \left(2 \frac{19}{22}\right)$

$\text{ "P_3(x)" "=" "2-2(2 19/22)" "=" } - 3 \frac{8}{11}$
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Substitute for $x$ in equation (3) to find ${P}_{3} \left(y\right)$

${P}_{3} \left(y\right) = \frac{3}{7} x + \frac{22}{7} \text{ "->" } \frac{3}{7} \left(- 3 \frac{8}{11}\right) + \frac{22}{7}$

${P}_{3} \left(y\right) = 1 \frac{6}{11}$
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${P}_{3} \left(x , y\right) = \left(- 3 \frac{8}{11} , 1 \frac{6}{11}\right)$