Question

1.
Find the value of a, for which point P(a/3,2) is the mid-point of the line sgement joining
the points (-5, 4) and R (-1,0).​

Answer 1

Answer:

$a = -9$

Step-by-step explanation:

P(a/3,2) is the mid point of QR, where Q(-5,4) and R(-1,0)

∴ $PQ^2 = PR^2$

∴ $[\frac{a}{3} - (-5)]^2 + (2 - 4)^2 = [\frac{a}{3} - (-1)]^2 + (2 - 0)^2$

$=>[\frac{a}{3} - (-5)]^2 + 4 = [\frac{a}{3} - (-1)]^2 + 4$

$=>[\frac{a}{3} - (-5)]^2 = [\frac{a}{3} - (-1)]^2$

$=>[\frac{a}{3} - (-5)] =$ ± $[\frac{a}{3} - (-1)]$

$=>[\frac{a}{3} - (-5)] =$ $-[\frac{a}{3} - (-1)]$  [Positive value results in an inequality]

$=> \frac{a}{3} + 5 = -\frac{a}{3} - 1$

$=> \frac{2a}{3} = - 6$

∴ $a = -9$