Question

If A,B and P are the points (-4,3)(0,-2) and (alpha, beeta) respectively and P is equidistant from A and B, show that 8alpha-10beeta+21=0​

Answer 1

SOLUTION:-

Given,

Three points A(-4,3), B(0,-2) and P(a,b) such that P is the equidistant from A &B.

Now use the formula which states that the distance any two points is given by;

$= > \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } .$

So,

$pa = \sqrt{( - 4 - a) {}^{2} + (3 - b) {}^{2} } ............(1)$

Also,

$pb = \sqrt{(0 - a) {}^{2} + ( - 2 - b) {}^{2} } .............(2)$

Now, as PA= PB equating (1) & (2), we get

$= > \sqrt{( - 4 - a) {}^{2} + (3 - b) {}^{2} } = \sqrt{(0 - a) {}^{2} + ( - 2 - b) {}^{2} }$

Squaring both the sides;

(-4-a)² +(3-b)² = (0- a)² + (-2-b)²

Using the identity ( a+b)².

(-4,a)² + (3-b)²= (0-a)² + (-2-b)²

=)[(-4)² + (-a)² + 2(-4)(-a)] + [(3)² + (-b)²+ 2(3)(-b)] = [(0)² +(-a)² + 2(0)(-a)] +[(-2)² + (-b)² + 2(-2)(-b)]

=) 16+a²+8a+9+b² -6b=a² + 4+ b²+ 4b

=) 16 + a² + 8a +9+b² - 6b -a² -4-b²-4b=0

=) 25 +8a -6b -4-4b= 0

=) 8a -10b+21= 0 [Proved]

Hope it helps ☺️