Question 30494: Find the equation, in standard form, of the line that passes through the points (5,6) and (7, 3)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! Find the equation, in standard form, of the line that passes through the points (5,6) and (7, 3)
Given two points P(x1,y1) and Q(x2,y2) on a given line
the equation to the given line is
(y-y1)/(y1-y2) = (x-x1)/(x1-x2)----(1)
Therefore given two points P(5,6) and Q(7, 3),
the equation to the line passing through P and Q is
(here x1 = 5, y1 = 6; x2 = 7, y2 = 3 )
(y-6)/(6-3) = (x-5)/(5-7)
That is (y-6)/3 = (x-5)/(-2)
Multiplying by 6 on both the sides (since lcm of 3 and 2 is 6)
2(y-6) = -3(x-5)
2y-12 = -3x+15
3x+2y-12-15 = 0 (grouping like terms,changing sign while changing side)
3x+2y-27 =0 ----(*)
which is the equation to the line in the general form
If by standard form you mean the slope and the y-intercept form
retain the y-term on the left side and take the other terms to the right
and then divide by the coefficient 2 of y
then the slope and y-intercept form of the required line is
y = (-3/2)x +(27/2) ----(**)
Verification: Testing for P(5,6) and Q(7, 3)in (*)
LHS = 3x+2y-27
= 3X(5)+2X(6)-27 (putting x = 5 and y = 6 in (1) )
= 13+12-27
= 27-27
= 0
=RHS
P(5,6) is a point on (*)
Now putting x = 7 and y = 3 in (*)
that is checking if Q(7,3) is a point on (*)
LHS = 3x+2y-27
= 3X(7)+2X(3)-27
= 21+6-27
= 27-27
= 0
=RHS
This implies Q(7,3) is a point on (*)
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