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Solution 2.2:5a

From Förberedande kurs i matematik 1

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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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 +
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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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 +
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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>
 +
and
 +
<math>\left( 3 \right., \left. 0 \right)</math>
 +
satisfy the equation of the line:
 +
 +
<math>\left( x \right., \left. y \right)=\left( 2 \right., \left. 3 \right)</math>: LHS=
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<math>3</math>
 +
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=
 +
<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


y=kx+m


where k and m are constants which we shall determine.

Since the points 23  and 30  should lie on the line, they must also satisfy the equation of the line,


3=k2+m and 0=k3+m


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


3−0=k2+m−k3+m 


3=−k

Substituting this into the equation 0=k3+m then gives us a value for m,


m=−3k=−3−3=9 


The equation of the line is thus y=−3x+9.


NOTE: To be completely certain that we have calculated correctly, we check that the points 23  and 30  satisfy the equation of the line:

xy=23 : LHS= 3 and RHS= −32+9=3


xy=30 : LHS= 0 and LHS= −33+9=0


Image:S1_2_2_5_a.jpg

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