Question

The line joining the points (2, -1) and (5, -6) is bisected at p. if p lies on the line 2x+4y+k=0, find the value of k?

Answer 1

Step-by-step explanation:

$\huge\underline{\underline{\ Question}}$

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The line joining the points (2, -1) and (5,-6) is bisected at p. if p lies on the line 2x+4y+k=0, find the value of k?

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$\huge\underline{\underline{\ Answer}}$

Let...

(2,-1) = (x1, y1)

(5,-6) = (x2, y2)

By mid point formula...

$x = \frac{x1 + x2}{2} \: \: \: \: \: \: \: \: \: y = \frac{y1 + y2}{2} \\ x = \frac{5 + 2}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{ - 6 -1 }{2} \\ x = \frac{7}{2} \: and \: y = \frac{ - 7}{2}$

(x, y) = (7/2, -7/2)

It lies on line 2x +4y +k = 0

So it will satisfy its equation...

$\implies 2 (\frac{7}{2} ) + 4( \frac{ - 7}{2} ) + k = 0 \\ \implies 7 - 14 + k = 0 \\ \implies \boxed{k = 7}$