Question

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​

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

Reference link will be

https://brainly.in/question/1874150