Question

If A(1,1), B(2,3), C(-1,1) are the points
of P is a point such that the area of the
quadrilateral PABC is 3 sq units then locus
of Pis
1) y2 +6y=0 2) y? - 6y = 0
3) x+y=0 4) x² - 6 y = 0​

Answer 1

Step-by-step explanation:

Given 26. A(1, 1), B (2,3), C (-1,1) are three points. If P is a point such that area of the  quadrilateral PABC is 3 square unit, then the locus of P​

Now area of (PABC) = area of triangle PAB + area of triangle PBC

                                 = l ½ l h        k           1                               l h         k        1

                                             1        1           1            +    l1/2 l       2          3       1

                                              2         3        1 l                          -   1           1         1   l

      3 = ½ [ (h(1 – 3) – k(1 – 2) – 1 (3 – 2) ] + l h (3 – 1) – k (2 + 1) + 1(2 + 3)

                         6 = l h(- 2) – k(- 1) + 1 l + l 2 h – 3 k + 5 l

            So 6 > = l -2h + k + 1 + 2h – 3k + 5 l   (since lal + lb l > = l a + b l

                  6 > = mod l – 2k + 6 l

                    3 > = mod l (- k + 3 ) l

                      3 > = (- k + 3)

Squaring we get

                   3^2 > = (- k + 3)^2

                   9 > = k^2 + 6k + 9

      So k^2 + 6k < = 0

          Or k^2 + 6k = 0

So (h,k) may be replaced by (x,y)

Therefore equation will be y^2 + 6y = 0