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  • The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

Answers (1)

Solution:
               

Here points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2
(x1, y1) = (3, 2)            (x2, y2) = (5, 1)
m1 = 1, m2 = 2
By section formula we have
P\left ( x,y \right )= \left [ \frac{1\times 5+2\times 3}{1+2},\frac{1\times 1+2\times 2}{1+2} \right ]
                 = \left [ \frac{5+6}{3},\frac{1+4}{3} \right ]
                 = \left [ \frac{11}{3},\frac{5}{3} \right ]
Line is 3x – 18y + k = 0     ......(1)
 Put\, point\, P\left ( \frac{11}{3},\frac{5}{3} \right ) in \left ( 1 \right )
 Put\, x= \frac{11}{3},y= \frac{5}{3}
3\left ( \frac{11}{3} \right )-18\left ( \frac{5}{3} \right )+k= 0

11 – 30 + k = 0
–19 + k = 0
      k = 19

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