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# The Vertices

#### The Equation of The Line Parallel to BC

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## The vertices of a ?OBC are O (0, 0), B(-3,-1),C(-1,-3). Find the equation of the line parallel to BC and intersecting the sides OB and OC and whose perpendicular distance from the origin is 12.

Answer: Considering the given statement, a diagram such as the below mentioned one can be drawn:Here, we need to calculate the equation of the line PQ. Now, from the equation of the line, it is evident that the equation of BC line will be

y+1 = [(-3+1)/ (-1+3)] * (x+3)

Y+1= (-2/2) * (x+3)

Y+1= (-1) (x+3)

Y+1 = -x-3

Y= -x-4 ------------------------equation number of (1)

Now, we need to find the equation of the line that is parallel to BC, hence the slope of line BC will be equal to slope of line PQ.Hence the equation of line PQ is y = -x + C as the slope of line BC is (-1)

Now, we need to calculate the value of C.

Hence, y+x – c = 0-------------------------------------------------------- equation number of (2)

Now, the distance of the line from (0,0) is ?{-c / (√1+1)?

As per the question, ?{-c / (√1+1)? = 1/2

Hence c/ √2 = 1/ 2

Hence, c = √2/2

Hence, c = 1/ √2

y+x- (1/ √2)

or, √2x + √2y = 1

Hence the equation of PQ is √2x + √2y = 1

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