If the coordinates of the points P and Q are (4,3) and (-1,7). Then find the abscissa of a point R on the line segment PQ such that PR/PQ=3/5
Asked by | 17th Dec, 2012, 09:25: PM
PR/PQ = 3/5
Let PR = 3x and PQ = 5x
therefore, PR : PQ = 3 : 5
now coordinates of R will be ( x1m2 + x2m1 / m1 + m2 , y1m2 + y2m1 / m1 + m2 )
(where x1 = 4 , x2 = -1 , y1 = -3 , y2 = 7 , m1 = 3 and m2 = 5 )
so coordinates of R become ( 4*5 + ( -1)3 / 3 + 5 , ( -3) 5 + 7*3 / 3 + 5 )
i.e., ( 20 - 3 / 8 , -15 + 21 / 8 )
= ( 17/8 , 3/4 )
therefore, abscissa of point R will b 3/4....
PR/PQ = 3/5
Let PR = 3x and PQ = 5x
therefore, PR : PQ = 3 : 5
now coordinates of R will be ( x1m2 + x2m1 / m1 + m2 , y1m2 + y2m1 / m1 + m2 )
(where x1 = 4 , x2 = -1 , y1 = -3 , y2 = 7 , m1 = 3 and m2 = 5 )
so coordinates of R become ( 4*5 + ( -1)3 / 3 + 5 , ( -3) 5 + 7*3 / 3 + 5 )
i.e., ( 20 - 3 / 8 , -15 + 21 / 8 )
= ( 17/8 , 3/4 )
therefore, abscissa of point R will b 3/4....
Answered by | 18th Dec, 2012, 10:25: AM
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