Question

Find the equation of the straight line passing through the point (2,1) and parallel to the join of the points (1,3) and (-3,1)​

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Answer 1

Answer:

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Answer 2

SOLUTION

TO DETERMINE

The equation of the straight line passing through the point (2,1) and parallel to the join of the points (1,3) and (-3,1)

EVALUATION

Here the equation of the line passing through the points (1,3) and (-3,1) is

$\displaystyle \: \sf{ \frac{y - 3}{x - 1} = \frac{3 - 1}{1 + 3} }$

$\displaystyle \: \sf{ \implies \frac{y - 3}{x - 1} = \frac{2}{4} }$

$\displaystyle \: \sf{ \implies \frac{y - 3}{x - 1} = \frac{1}{2} }$

$\displaystyle \: \sf{ \implies x - 1 = 2y - 6 }$

$\displaystyle \: \sf{ \implies x - 2y = - 5 }$

Now the equation of the line parallel to the line x - 2y = - 5 is x - 2y = k

Now the line x - 2y = k passes through the point (2,1)

∴ 2 - 2 = k

$\implies \sf{k = 0}$

Hence the required equation of the line is

$\sf{x - 2y = 0}$

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