Question

A (2, 3) and B(-3, 4) are two given points. Find the equation of locus of P so that
the area of the triangle PAB is 8.5​

Answer 1

Answer:

Hope it will be helpful to you

Step-by-step explanation:

Let coordinate of P are (h,t)A(2,3),B(3,−4)

point P will lies on the both side of AB and the

Area of triangle PAB =

2

1

(h(3−4)+2(4−k)−3(k−3))

±8.5=

2

1

[h(−1)+8−2k−3k+9]

±17=−h+8−5x+9

taking

′

+

′

sign

17+h−17+5k=0

h+5k=0

∴ locus of P is

x+5y=0

taking -ve sign

−17+h+5k−17=0

h+5k=34

∴ locus of P is

x+5y=34

If you do not understand this I have gave you a pic also see it fro there

Answer 2

$\huge\fcolorbox{black}{lime}{AnsweR:}$

Given: The points A(2,3), B(-3,4) and area of triangle PAB is 8.5

To find The equation of locus of P.

Solution:

Now we have given a point P. Let coordinates of P be (h,t).

Now as per question, the point P will be on both the sides of the line AB, so:

Area of triangle PAB =  1/2 x (h(3-4) + 2(4-k) - 3(k-3))

•                ±8.5 =  1/2 x (h(3-4) + 2(4-k) - 3(k-3))

•                ±17 =  (h(-1) + 8 - 2k - 3k + 9)

•                ±17 =  -h + 8 - 5k + 9

•                ±17 =  -h + 17 - 5k

If we take positive sign, we get:

•                17 = -h + 17 - 5k

•                h + 5k = 0

So locus of P is x + 5y = 0 .

If we take negative sign, we get:

•                -17 = -h + 17 - 5k

•                h + 5k = 34

So locus of P is x + 5y = 34 .

Answer:

        So the equation of locus of point P is  x + 5y = 34 and x + 5y = 0.

$\large\bold{\bigstar\:SeKhEr\:HeRe\:\bigstar}$