Question

if A (3,y) is equidistant from points p(8,-3) and Q(7,6) find the value of y and find the distance AQ​

Answer 1

Answer:

The answer is given to the picture.

Answer 2

Solution :-

Given :-

A ( 3 , y ) is equidistant from points P ( 8 , -3 ) and Q ( 7 , 6 ).

That is,

AP = AQ

$\boxed{\bf Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\longrightarrow \sf AP = AQ \\\\\longrightarrow \sqrt{(8-3)^2+(-3-y)^2} = \sqrt{(7-3)^2+(6-y)^2} \\\\\longrightarrow \sqrt{5^2+(-3)^2 +y^2 - 2 \times -3 \times y } = \sqrt{4^2+6^2+y^2-2\times 6 \times y} \\\\\longrightarrow \sqrt{25+9+y^2+6y} = \sqrt{16+36+ y^2-12y} \\\\\longrightarrow \sqrt{34+y^2+6y} = \sqrt{52+y^2-12y } \\\\\longrightarrow 34+y^2+6y = 52+y^2-12y \\\\\longrightarrow 6y+12y = 52-34 \\\\\longrightarrow 18y = 18 \\\\\longrightarrow \bf y = 1$

Value of y = 1

$\bullet \sf \ AQ = \sqrt{(7-3)^2+(6-1)^2}$

         $= \sf \sqrt{4^2+5^2} \\\\=\sqrt{16+25} \\\\= \bf \sqrt{41} \ units$