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The line segment joining the points (2,1) and (5,-8) is trisected at the pointsP & Q. If the point lies on the line 2x-y+k=0, find the value of k.Please give the answer and show the detailed steps.Please give fast as this is important for an exam!!

Answer»

When a line segment JOINING the POINTS (x₁,y₁) and (x₂,y₂) is trisected at points P and Q, the coordinates of the points P and Q are given by:

P\ is\ [ \frac{2x_1+x_2}{3} , \frac{2y_1+y_2}{3} ]\ or\\ \\ P\ is\ [ \frac{2*2+5}{3} , \frac{2*1-8}{3} ]=[ \frac{9}{3} , \frac{-6}{3} ] = (3,-2)\\ \\ Q\ is\ [ \frac{x_1+2x_2}{3} , \frac{y_1+2y_2}{3} ]\ or\\ \\ Q\ is\ [ \frac{2+2*5}{3} , \frac{1+2*(-8)}{3} ]=[ \frac{12}{3} , \frac{-15}{3} ]=(4,-5)

If P(3,-2) lies on 2x-y+K=0;
2×3 - (-2) + k = 0
⇒ 6 + 2 + k = 0
⇒ k + 8 =0
⇒ k = -8

If Q(4,-5) lies on 2x-y+k=0;
2×4 - (-5) + k = 0
⇒ 8 + 5 + k = 0
⇒ k + 13 =0
⇒ k = -13


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