Question

What is the value of k if the area of triangle whose vertices are P(k 0) Q(2 2) R(4 3) is (3) /(2) sq. unit?


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Answer 1

Answer:

Alternative method by coordinate geometry:

Take QR as base and PM perpendicular to base.

Equation of QR is

y-2= (3–2)/( 4–2) ( x-2)

Simplifying, x-2y +2=0

PM = | ( k-2.0 +2)/ SQRT( ( 1^2 + (-2)^2))|

Simplifying. PM= | (k+2)/ SQRT5)|

By distance formula,

QR= SQRT( ( 4–2^2 +( 3–2)^2))

Simplifying, QR= SQRT.5

So 1/2. QR. PM = 3/2

1/2. SQRT5. | k+2)/ SQRT 5| = 3/2

This gives . | k+2| = 3

So k+2 = 3 or -3

So k= 1. -5.