Question

A(1, 6) and B(3, 5), find the equation of
the locus of point P such that segment
AB subtends right angle at P. (ZAPB
= 90°)​

Answer 1

Answer:

x²+y²-4x-11y+33=0

Step-by-step explanation:

Given, two points A and B having co-ordinates (1,6) and (3,5) respectively.

Also given that, P is a variable point such that ∠APB =90°.

We have to find the locus of the variable point P.

Let us assume that, at any instance, the co-ordinates of P is (h,k).

Now, slope of line AP is given by

$\frac{k-6}{h-1}$

=m (say) ......(1)

{We know that the slope of a straight line passing through two points (x,y)and (X,Y) is given by $\frac{Y-y}{X-x}$ }

Again, slope of line BP is given by

$\frac{k-5}{h-3}$

=n(say) .......(2)

Now, we have product of the slopes of two perpendicular line is always -1.

Hence, m×n= -1

From equations (1) and (2), we have

$(\frac{k-6}{h-1}).(\frac{k-5}{h-3})=-1$

⇒$(k-6)(k-5)=-(h-1)(h-3)$

⇒k²-11k+30=-h²+4h-3

⇒h²+k²-4h-11k+33=0

Now, converting into current co-ordinates, the equation of locus of point P will be, x²+y²-4x-11y+33=0 (Answer)

Answer 2

Answer:

x^2 + y^2 - 4x - 11y + 33=0 is the answer.....