Question

find type of triangle formed by the points P(2a,4a),Q(2a,6a),R(2a+a root 3,5a)

Answer 1

P(2a , 4a )
Q(2a, 6a)
R(2a + a√3, 5a)

now, use distance formula ,
$distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$now, PQ=\sqrt{(2a-2a)^2+(6a-4a)^2}\\PQ=2a$
similarly ,
$QR=\sqrt{(2a+a\sqrt{3}-2a)^2+(5a-6a)^2}\\QR=\sqrt{3a^2+a^2}=2a$
$RP =\sqrt{(2a-2a-a\sqrt{3})^2+(5a-4a)^2}\\RP=\sqrt{3a^2+a^2}=2a$
you can see that ,
PQ = QR = RP = 2a
all sides of triangle are same . so, this is an equilateral triangle.