A line passes through A(6,5) and B(4,2) find equation of a line parallel to AB and passes through (7,10)

2y = 3x-1

Step-by-step explanation:

Here, we want to find the equation of the line

We start by calculating the slope of the segment AB

mathematically, that will be

m = (y2-y1)/(x2-x-)

m = (2-5)/(4-6) = 3/2

Since the two lines are parallel, their slopes must be equal

We simply insert this value into the point-slope form

y-y1 = m(x-x1)

y-10 = 3/2(x-7)

Multiply through by 2

2y-20 = 3x - 21

2y = 3x-21 + 20

2y = 3x -1

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)