# A line passes through (6 ,5 ) and (8 ,3 ). A second line passes through (0 ,1 ). What is one other point that the second line may pass through if it is parallel to the first line?

Aug 2, 2016

The equation of the second line is: $\text{ } y = - x + 1$

#### Explanation:

If the two lines are parallel then they have the same gradient (slope).

For the first line:
Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(6 , 5\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(8 , 3\right)$

For the second line:
Let point 3 be ${P}_{3} \to \left({x}_{3} , {y}_{3}\right) = \left(0 , 1\right)$

For both lines:
Let the gradient (slope) be $m$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider the first line with } {P}_{1} \mathmr{and} {P}_{2}}$

gradient $= m = \left(\text{change in the y axis")/("change in the x axis}\right)$

$\implies m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{3 - 5}{8 - 6} = \frac{- 2}{2} = - 1$

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider the second line with } {P}_{3}}$

$\textcolor{b r o w n}{\text{as this line is parallel to the first it has the same gradient (m)}}$

The standard form equation for a straight line is:

$y = m x + c$

But $m = - 1$ giving

$y = \left(- 1\right) x + c \text{ "->" } y = - x + c$

We know that this line passes through the point ${P}_{3}$

So the value of the coordinates for ${P}_{3}$ must be true for this equation giving:

${P}_{3} \to \left({x}_{3} , {y}_{3}\right) \to \left(\textcolor{m a \ge n t a}{0} , \textcolor{g r e e n}{1}\right) \text{ " =>" "color(green)( 1)" "=" } - 1 \left(\textcolor{red}{0}\right) + c$

Thus $c = 1$ giving

color(brown)(y=mx+c)color(blue)(" "->" "y=-x+1