Advertisement Remove all ads

Find the Value of K So that the Area of the Triangle with Vertices a (K+1, 1), B(4, -3) and C(7, -k) is 6 Square Units - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

Find the value of k so that the area of the triangle with vertices A (k+1, 1), B(4, -3) and C(7, -k) is 6 square units

Advertisement Remove all ads

Solution

`"Let" A(x_1,y_1) = A(k+1,1) , B(x_2,y_2)= B (4,-3) and C(x_3,y_3) = C(7,-k) now`

`"Area "(Δ ABC) = 1/2 [x_1 (y_2-y_3) + x_2 (y_3-y_1) +x_3(y_1-y_2)}`

`⇒ 6=1/2 [(k+1) (-3+k)+4(-k-1) +7(1+3)]`

`⇒6=1/2[k^2 -2k-3-4k-4+28]`

`⇒ k^2-6k+9=0`

`⇒(k-3)^2 = 0⇒k=3`

Hence , k=3

Concept: Area of a Triangle
  Is there an error in this question or solution?

APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 11
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×