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Solution 2.2:5d

From Förberedande kurs i matematik 1

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m (Lösning 2.2:5d moved to Solution 2.2:5d: Robot: moved page)
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If two non-vertical lines are perpendicular to each other, their gradients
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<center> [[Image:2_2_5d-1(2).gif]] </center>
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<math>k_{1}</math>
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and
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<math>k_{2}</math>
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satisfy the relation
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<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
 +
 
 +
 
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<math>k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}</math>
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 +
 
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since the line
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<math>y=2x+5</math>
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has a gradient
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<math>k_{1}=2</math>
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(the coefficient in front of
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<math>x</math>
 +
).
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The line we are looking for can thus be written in the form
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<math>y=-\frac{1}{2}x+m</math>
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 +
 
 +
with
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<math>m</math>
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as an unknown constant.
 +
 
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Because the point
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<math>\left( 2 \right.,\left. 4 \right)</math>
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should lie on the line,
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<math>\left( 2 \right.,\left. 4 \right)</math>
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must satisfy the equation of the line,
 +
 
 +
 
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<math>4=-\frac{1}{2}\centerdot 2+m</math>
 +
 
 +
 
 +
i.e.
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<math>m=5</math>. The equation of the line is .
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 +
 
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<math>y=-\frac{1}{2}x+5</math>
 +
 
 +
 
 +
 
 +
 
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<center> [[Image:2_2_5d-2(2).gif]] </center>
<center> [[Image:2_2_5d-2(2).gif]] </center>
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[[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 k1 and k2 satisfy the relation

k1k2=−1, and from this we have that the line we are looking for must have a gradient that is given by


k2=−1k1=−21


since the line y=2x+5 has a gradient k1=2 (the coefficient in front of x ).

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


y=−21x+m


with m as an unknown constant.

Because the point 24  should lie on the line, 24  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
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