Solution 2.2:5a

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m (Lösning 2.2:5a moved to Solution 2.2:5a: Robot: moved page)
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Let's write down the equation for a straight line as
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<math>y=kx+m</math>
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where
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<math>k</math>
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and
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<math>m</math>
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are constants which we shall determine.
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Since the points
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<math>\left( 2 \right., \left. 3 \right)</math>
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and
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<math>\left( 3 \right., \left. 0 \right)</math>
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should lie on the line, they must also satisfy the equation of the line,
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<math>3=k\centerdot 2+m</math>
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and
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<math>0=k\centerdot 3+m</math>
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If we take the difference between the equations,
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<math>m</math>
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disappears and we can work out the gradient
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<math>k</math>,
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<math>3-0=k\centerdot 2+m-\left( k\centerdot 3+m \right)</math>
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<math>3=-k</math>
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Substituting this into the equation
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<math>0=k\centerdot 3+m</math>
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then gives us a value for
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<math>m</math>,
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<math>m=-3k=-3\centerdot \left( -3 \right)=9</math>
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The equation of the line is thus
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<math>y=-3x+9</math>.
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NOTE: To be completely certain that we have calculated correctly, we check that the points
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<math>\left( 2 \right., \left. 3 \right)</math>
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and
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<math>\left( 3 \right., \left. 0 \right)</math>
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satisfy the equation of the line:
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<math>\left( x \right., \left. y \right)=\left( 2 \right., \left. 3 \right)</math>: LHS=
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<math>3</math>
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and RHS=
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<math>-3\centerdot 2+9=3</math>
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<math>\left( x \right., \left. y \right)=\left( 3 \right., \left. 0 \right)</math>: LHS=
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<math>0</math>
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and LHS=
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<math>-3\centerdot 3+9=0</math>
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Revision as of 09:02, 18 September 2008

Let's write down the equation for a straight line as


\displaystyle y=kx+m


where \displaystyle k and \displaystyle m are constants which we shall determine.

Since the points \displaystyle \left( 2 \right., \left. 3 \right) and \displaystyle \left( 3 \right., \left. 0 \right) should lie on the line, they must also satisfy the equation of the line,


\displaystyle 3=k\centerdot 2+m and \displaystyle 0=k\centerdot 3+m


If we take the difference between the equations, \displaystyle m disappears and we can work out the gradient \displaystyle k,


\displaystyle 3-0=k\centerdot 2+m-\left( k\centerdot 3+m \right)


\displaystyle 3=-k

Substituting this into the equation \displaystyle 0=k\centerdot 3+m then gives us a value for \displaystyle m,


\displaystyle m=-3k=-3\centerdot \left( -3 \right)=9


The equation of the line is thus \displaystyle y=-3x+9.


NOTE: To be completely certain that we have calculated correctly, we check that the points \displaystyle \left( 2 \right., \left. 3 \right) and \displaystyle \left( 3 \right., \left. 0 \right) satisfy the equation of the line:

\displaystyle \left( x \right., \left. y \right)=\left( 2 \right., \left. 3 \right): LHS= \displaystyle 3 and RHS= \displaystyle -3\centerdot 2+9=3


\displaystyle \left( x \right., \left. y \right)=\left( 3 \right., \left. 0 \right): LHS= \displaystyle 0 and LHS= \displaystyle -3\centerdot 3+9=0


Image:S1_2_2_5_a.jpg

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