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put this solution on YOUR website!Use the intercept method to find the equation of a straight line passing through (-3,0) and (0,2)
The intercept form of equation to a straight line is given by
x/(a) +y/(b) = 1 -----(1)
where a and b are the x-intercept and the y-intercept respectively
The given points on the line are A(-3,0) and B(0,2)
Observe that the y-coordinate of A is zero and hence A is a point on the x-axis
and the x-coordinate of B is zero and hence B is a point on the y-axis
A(-3,0) and B(0,2)means the line traverses through the first and third quadrant via the second quadrant meeting the x-axis at A which is 3 units to the left of the origin and meeting the y-axis at B which is 2 units above the origin
Therefore a = (-3) and b= 2
Therefore (1) becomes
x/(-3) +y/(2) = 1
-x/3 +y/2 = 1
Multiplying through out by 6(which is the lcm of 3 and 2)
(-x)X2 +yX(3) = 1X6
-2x+3y = 6
That is 2x-3y = -6 (multiplying by (-1) through out)
That is 2x-3y+6 = 0 ----(*)
Verification: We check if A(-3,0) and B(0,2) hold in (*)
Putting A in (*) that is putting x=-3 and y=0 in (*)
LHS = 2x-3y+6 = 2X(-3)-3X(0)+6 = -6-0+6 = 0 =RHS
Therefore A(-3,0) is a point on our line
Putting B in (*) that is putting x=0 and y=2 in (*)
LHS = 2x-3y+6 = 2X(0)-3X(2)+6 = 0-6+6 = 0 =RHS
Therefore B(0,2) is a point on our line
Therefore the equation: 2x-3y+6 = 0 is correct
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