Question

From the given right angled isosceles triangle PQR whose side QR is parallel to X axis, find the slope of the sides of the triangle.

Answer 1

Answer:

Let the slopes of PQ and PR be m and −

m

1

respectively.

Since △PQR is an isosceles triangle, ∠PQR=∠PRQ

⇒

∣

∣

∣

∣

∣

∣

1−2m

m+2

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∣

∣

1+

m

2

−

m

1

+2

∣

∣

∣

∣

∣

∣

∣

∣

∣

as slope of QR=2

⇒m+2=±(1−2m)⇒m=3 or −

3

1

So the equation of PQ and PR are

(y−1)=3(x−2) and y−1=(−

3

1

)(x−2)

Thus, joint equation representing PQ and PR is

(3(x−2)−(y−1))((x−2)+3(y−1))=0

⇒3(x−2)

2

−3(y−1)

2

+8(x−2)(y−1)=0

⇒3x

2

−3y

2

+8xy−20x−10y+25=0