# The Point a Divides the Join Of P (−5, 1) And Q(3, 5) in the Ratio K:1. Find the Two Values Of K for Which the Area of δAbc Where B Is (1, 5) And C(7, −2) is Equal to 2 Units. - Mathematics

#### Question

The point A divides the join of P (−5, 1)  and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, −2) is equal to 2 units.

#### Solution

GIVEN: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

Coordinates of point B (1, 5) and C (7, −2)

TO FIND: The value of k

PROOF: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

So the coordinates of A are (3k+1)/(k+1),(5k+1)/(k+1)

We know area of triangle formed by three points (x1,y1)(x2,y2) and (x3,y3)s given by

Δ = 1/2([x1y2+x2y3+x3y1)-(x2y1+x3y2+x1y3)]

Now Area of ΔABC= 2 sq units.

Taking three points A [(3k+1)/(k+1),(5k+1)/(k+1)], B (1, 5) and C (7, −2)

2=1/2[{((3k-5)/(k+1))5-2+7((5k+1)/(k+1))}-((5k+1)/(k+1)) +35-2((3k-5)/(k-1))]

2=1/2[{((15k-25)/(k+1))-2+((35k+7)/(k+1))}-((5k-1)/(k+1))+35-((6k-10)/(k+1))]

2=1/2[((15k-25-2k+35k+7)/(k+1))-((5k+1+35k-6k+10)/(k+1))]

2=1/2[((48-20-(34k+46))/(k+1))]

2=1/2[((48k-20)-(34k+46))/(k+1))]

2=1/2[((14k-66)/(k+1))]

2=[((7k-33)/(k-1))]

+-=((7k-33)/(k+1))]

⇒ 7k -33= ±2(k+1)

⇒ 7k -33= 2(k+1), 7k-33= -2(k+1)

⇒7k -33=2k +2, 7k+2k= +33- 2

⇒ 7k -2k=33 +2, 7k+2k =+33- 2

⇒ 5k  =35,9k=31

⇒k =35/5,k=31/9

⇒ k=7,k=31/9

Hence k=7 or 31/9

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#### APPEARS IN

RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 6: Co-Ordinate Geometry
Exercise 6.5 | 24 | Page no. 54
RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 6: Co-Ordinate Geometry
Exercise 6.5 | 24 | Page no. 54