Question

Find the equation of the straight line passing through the points (4; 2) and (2; 4).

Answer 1

(x - x 1 ) / ( x 2 - x 1 ) = (y - y 1 ) / ( y 2 - y1 ) ;
( x - 4 ) / ( 2 - 4 ) = ( y - 2 ) / ( 4 - 2 ) ;
( x - 4 ) / -2 = ( y - 2 ) / 2 ;
( x - 4 ) / - 1 = y - 2 ;
- x + 4 = y - 2 ;
-x - y + 6 = 0 ;
x + y - 6 = 0 ;

Answer 2

Use point-slope form.

But first let's find the slope.

(4, 2), (2, 4)

m = y2-y1/x2-x1

m = (4 - 2)/(2 - 4)

m = 2/-2

m = -1

So this is the slope, now we can plug this into the equation along with any point.

y - y1 = m(x - x1)

Where y1 is the y-value of the point, x1 is the x-value, and 'm' is the slope.

Let's plug in point (4, 2):

y - 2 = -1(x - 4)

This can be one of your options.

Distribute -1:

y - 2 = -x + 4

Add 2 to both sides:

y = -x + 6

This can be another of your options.

Add 'x' to both sides:

x + y = 6

This can be another of your options.