Question

The relation between x and y such that the distance of the
point (x,y) from the point (2, 3) is twice of its distance
from the point (1,5), is

A 3x2 + 3y2 - 4x - 34y + 91 = 0
B
3x2 + 4y2 + 12x + 40y + 109 - 0
с
2x2 + 3y2 + 11x + 36y - 119 = 0
D
3x2 + 2y2 + 4x + 44y - 109 = 0​

Answer 1

Answer :-

Option - a

Topic :-

Co - ordinate Geometry

Given :-

The point (x, y) is from the point (2, 3) is twice the distance from point (1, 5)

To find:-

• The relation between x and y

Solution :-

Let ,

• P =(x, y)
• A = (2, 3)
• B = (1 , 5 )

According to the question ,

Distance between P and A is twice the distance between P and B So,

PA = 2 PB

By using distance formula

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\sqrt{(x-2)^2+(y-3)^2} = 2 \sqrt{(x-1)^2+(y-5)^2}$

Squaring on both sides

$\bigg(\sqrt{(x-2)^2+(y-3)^2} \bigg)^2= \bigg(2 \sqrt{(x-1)^2+(y-5)^2}\bigg)^2$

${(x-2)^2+(y-3)^2} = 4 {(x-1)^2+(y-5)^2}$

Simplifying the equation :-

Expanding the equation by algebraic identity (a-b)²= a²-2ab+b²

$x^2 - 4x +4+y^2-6y+9 = 4[x^2-2x+1+y^2-10y+25]$

$x^2-4x +y^2-6y +4+9 = 4[x^2-2x+y^2-10y+1+25]$

$x^2-4x +y^2-6y+ 13 = 4[x^2-2x+y^2-10y+26]$

$x^2-4x +y^2-6y +13 = 4x^2-8x+4y^2-40y+104$

Transposing R.H.S equation to L.H.S

$x^2-4x +y^2-6y +13 -[ 4x^2-8x+4y^2-40y+104] =0$

$x^2-4x +y^2-6y +13 -4x^2+8x-4y^2+40y-104=0$

$x^2-4x^2+y^2-4y^2-4x+8x-6y+40y+13-104=0$

$-3x^2-3y^2+4x+34y-91=0$

Take common -

$3x^2+3y^2-4x -34y +91=0$

Hence option "a" is the correct

Used formulae:-

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ - Distance formula

( a- b)² = a²- 2ab + b²- Algebraic identity

Answer 2

$\sf◉Answer◉$

TO FIND -:

• Find the equation of the locus of the point whose distance from y-axis is half the distance from origin .

S O L U T I O N :

• Let us assume that , such a point R exists with the coordinates

( h, k ) .

• Now , the distance of the point from the y axis is half the distance from the origin .

Coordinates of that point on the Y axis -

$\sf⇝ [ 0, k ]$

Here ,

$\sf PR = ½ PQ$

• $\sf ⇝ 2PR = PQ$

• $\sf⇝ 4 [ PR ]² = [ PQ ]²$

• $\sf ⇝ 4 h² = h² + k²$

• $\sf⇝ 3h² = x²$

Replacing the coordinates by x and y -

• $\sf ⇝ 3x² = y²$

• $\sf ⇝ 3x² - y² = 0$

$\sf\color{purple} ☆ Thus , \:Option \:2 \:is\: the\: correct \:answer☆$

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