The coordinates pf p are (-3,8) and the coordinates of Q are (9,-2) (ii) Find the equation of the line parallel to PQ that passes through the point (6,-1)
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Given
- Coordinates of p -> (-3,8)
- Coordinates of q -> (9,-2)
- Coordinates of point r (say) -> (6,-1)
To Find
Equation of line parallel to PQ passing through r (6,-1)
Solution
Let the equation of required line is y = mx + c [General form of line]
Here, as we know, m represents the slope and c the intercept on y-axis.
Also, Slope of line joining p and q would be given by difference of y Coordinates of two points upon difference of x Coordinates i.e.
Slope of PQ = [ (8) - (-2) ] ÷ [ (-3) - (9) ] = (10) ÷ (-12) = (-5/6)
Since, slope of parallel lines are equal, m = (-5/6)
As Line y = mx + c passes through r, r's coordinates satisfies it -> 6 = (-5/6)×(-1) + c
=> c = 6 - (-5/6) = 6 + 5/6 = 41/6
Therefore, required equation of line is
y = (-5/6)x + 41/6
Multiplying by -6 on both sides
-6y = 5x - 41
Rearranging the equation
5x + 6y = 41 is the required equation of line. [ANSWER]
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