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The line segment joining the points A $(-5, 1)$ and B $(7, 6)$ is trisected at the points P and Q such that P is nearer to A. If P lies on the line $x + y = k$, then find the value of $k$.
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OR
(b) AP : PB = $1 : 2$ (
frac{1}{2})
Coordinates of P are $(\frac{1\times 7+2\times(-5)}{1+2}, \frac{1\times 6+2\times 1}{1+2})$ i.e. $(-1, \frac{8}{3})$ (1
frac{1}{2})
$\therefore$ P lies on $x + y = k$ therefore $k = -1 + \frac{8}{3} = \frac{5}{3}$ (1)
(b) AP : PB = $1 : 2$ (
frac{1}{2})
Coordinates of P are $(\frac{1\times 7+2\times(-5)}{1+2}, \frac{1\times 6+2\times 1}{1+2})$ i.e. $(-1, \frac{8}{3})$ (1
frac{1}{2})
$\therefore$ P lies on $x + y = k$ therefore $k = -1 + \frac{8}{3} = \frac{5}{3}$ (1)