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Find an equation of the line that satisfies the given conditions.

Through
(8, 8);
  perpendicular to the line passing through
(5, 4) and (1, 2)
 Jun 11, 2014

Best Answer 

 #1
avatar+90 
+5

Let the line passing(-8,-8) be L1 and another one be L2

 

Slope of L2 : $${\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$=$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

Since L1 perpendicular to L2

Slope of L1 x Slope of L2 = -1

Slope of L1 x $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ = -1

Slope of L1 = 2

 

Equation of L1 : $${\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}}$$=2

y+8=2x+16

2x-y+8=0

Therefore, equation of the line passing through (-8,-8) is 2x-y+8=0

 Jun 11, 2014
 #1
avatar+90 
+5
Best Answer

Let the line passing(-8,-8) be L1 and another one be L2

 

Slope of L2 : $${\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$=$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

Since L1 perpendicular to L2

Slope of L1 x Slope of L2 = -1

Slope of L1 x $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ = -1

Slope of L1 = 2

 

Equation of L1 : $${\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}}$$=2

y+8=2x+16

2x-y+8=0

Therefore, equation of the line passing through (-8,-8) is 2x-y+8=0

problem Jun 11, 2014

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