Question

The perimeter of triangle is 20 and the points (-2,-3) and (-2,3) locus of 3rd vertex?

Answer 1

Let A (3, 4) and B (–2, 3) be the given points and C (x, y) be the third vertex of the equilateral ΔABC. Then,

AB = BC = CA

Answer 2

Concept: The lengths of the line segments AB, BC, and CA are the side lengths. The distance formula will be used.

From point P(x1,y1) to point Q(x2,y2), the distance is:

Distance = $\sqrt{(x_{2}-x_{1})^{2} + (y_{2}- y_{1} )^{2} }$

Given: Perimeter of triangle= 20

Coordinates of two vertices=  (-2,-3) and (-2,3)

To find: Locus of third vertex

Solution:

Let the third vertex's coordinates be A (x,y)

Let the coordinates of two vertices be B (-2,-3) and C(-2,3)

The perimeter  of the triangle ABC= AB+AC+BC.... (i)

Using distance formula:

$AB= \sqrt{(x+2)^{2} +(y+3) ^{2} }$

$AC= \sqrt{(x+2)^{2} +(y-3)^{2}} \\BC= \sqrt{(-2+2)^{2}+ (3+3)^{2}} \\BC =\sqrt{36}=6$

Substituting the values in (i), we get

$\sqrt{(x+2)^{2} +(y+3)^{2}} + \sqrt{(x+2)^{2} +(y-3)^{2}} +6 =20$

$\sqrt{(x+2)^{2} +(y+3)^{2}} =14 - \sqrt{(x+2)^{2} +(y-3)^{2}}$

On squaring, we get

$(x+2)^{2} + (y+3)^{2} = 196 + (x+2)^{2} + (y-3)^{2} - 28\sqrt{(x+2)^{2} + (y-3)^{2}}$

$12y- 196= [- 28\sqrt{(x+2)^{2} + (y-3)^{2}}]$

$49- 3y = 7\sqrt{(x+2)^{2} + (y-3)^{2}}$

On squaring both sides,

$2401 +9y^{2} -294y=49(x+2)^{2} +49y^{2} +441-294y\\1960=49(x+2)^{2} +40y ^{2}$

$\frac{(x+2)^{2} }{40} + \frac{y^{2} }{49} = 1$

Therefore, the locus of the third vertex is $\frac{(x+2)^{2} }{40} + \frac{y^{2} }{49} = 1$

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