Solution 2.2:5d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 2.2:5d moved to Solution 2.2:5d: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
If two non-vertical lines are perpendicular to each other, their gradients
-
<center> [[Image:2_2_5d-1(2).gif]] </center>
+
<math>k_{1}</math>
-
{{NAVCONTENT_STOP}}
+
and
 +
<math>k_{2}</math>
 +
satisfy the relation
 +
 
 +
<math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a gradient that is given by
 +
 
 +
 
 +
<math>k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}</math>
 +
 
 +
 
 +
since the line
 +
<math>y=2x+5</math>
 +
has a gradient
 +
<math>k_{1}=2</math>
 +
(the coefficient in front of
 +
<math>x</math>
 +
).
 +
 
 +
The line we are looking for can thus be written in the form
 +
 
 +
 
 +
<math>y=-\frac{1}{2}x+m</math>
 +
 
 +
 
 +
with
 +
<math>m</math>
 +
as an unknown constant.
 +
 
 +
Because the point
 +
<math>\left( 2 \right.,\left. 4 \right)</math>
 +
should lie on the line,
 +
<math>\left( 2 \right.,\left. 4 \right)</math>
 +
must satisfy the equation of the line,
 +
 
 +
 
 +
<math>4=-\frac{1}{2}\centerdot 2+m</math>
 +
 
 +
 
 +
i.e.
 +
<math>m=5</math>. The equation of the line is .
 +
 
 +
 
 +
<math>y=-\frac{1}{2}x+5</math>
 +
 
 +
 
 +
 
 +
 
{{NAVCONTENT_START}}
{{NAVCONTENT_START}}
<center> [[Image:2_2_5d-2(2).gif]] </center>
<center> [[Image:2_2_5d-2(2).gif]] </center>
-
{{NAVCONTENT_STOP}}
 
-
[[Image:2_2_5_d.jpg|center|300px]]
 

Revision as of 09:50, 18 September 2008

If two non-vertical lines are perpendicular to each other, their gradients \displaystyle k_{1} and \displaystyle k_{2} satisfy the relation

\displaystyle k_{1}k_{2}=-1, and from this we have that the line we are looking for must have a gradient that is given by


\displaystyle k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}


since the line \displaystyle y=2x+5 has a gradient \displaystyle k_{1}=2 (the coefficient in front of \displaystyle x ).

The line we are looking for can thus be written in the form


\displaystyle y=-\frac{1}{2}x+m


with \displaystyle m as an unknown constant.

Because the point \displaystyle \left( 2 \right.,\left. 4 \right) should lie on the line, \displaystyle \left( 2 \right.,\left. 4 \right) must satisfy the equation of the line,


\displaystyle 4=-\frac{1}{2}\centerdot 2+m


i.e. \displaystyle m=5. The equation of the line is .


\displaystyle y=-\frac{1}{2}x+5



Image:2_2_5d-2(2).gif
Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.2:5d"