Question

Distance of any point p(3,2) from any point Q is root 2 and ordinate of Q is half of its absissca. Then find the coordinate of Q.

Answer 1

$<h2><font color = red > hey there$



$<font color = black > here we go......$



$pq = \sqrt{2}$



$pq {}^{2} = 2$



$(2a - 3) {}^{2} + (a - 2) {}^{2} = 2$



$4a {}^{2} + 9 - 12a + a {}^{2} + 4 - 4a = 2$



$5a {}^{2} - 16a + 11 = 20$



$5a {}^{2} - 11a - 5a + 11$



$a(5a - 11) - 1(5a - 11)$



$(5a - 11)(a - 1)$



$a = 1 \: or \: \frac{11}{5}$


Hope it helps you

Thanks for asking....


@ Prabhudutt

Answer 2

The coordinates of Q are $(2,1)\text{ and } (\frac{22}{5},\frac{11}{5})$

Step-by-step explanation:

Let the ordinate of Q is y then the abscissa of Q would be 2y.

Hence, the coordinate of Q is (2y, y)

Coordinate of P is (3,2)

Now, the distance formula is given by

$d^2=(x_2-x_1)^2+(y_2-y_1)^2$

Thus, the distance between P and Q is given by

$(\sqrt2)^2=(2y-3)^2+(y-2)^2$

Solve the equation for y

$2=4y^2+9-12y+y^2+4-4y\\\\5y^2-16y+11=0\\\\5y^2-11y-5y+11=0\\\\y(5y-11)-1(5y-11)=0\\\\(5y-11)(y-1)=0\\\\y=\frac{11}{5},1$

Hence, the values of absissca of Q are

$2\times1,2\times\frac{11}{5}\\\\=2,\frac{22}{5}$

Thus, the coordinates of Q are

$(2,1)\text{ and } (\frac{22}{5},\frac{11}{5})$

#Learn More:

Find the distance between the points p(3,2) and Q(-1,5)

https://brainly.in/question/13376025