Question

Find the value of y for which the distance between the points P (2,-3) and Q(10,y) is 10 units.​

Answer 1

Given :-

• Distance between P(2, -3) and Q(10,y) is 10units

To find :-

• Value of y

Formula to know:-

Distance formula:-

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

Solution:-

P = (2, -3)

${x_1 =2}$

${y_1 = -3}$

Q = (10,y )

${x_2 = 10}$

${y_2 = y}$

PQ = 10

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}= 10$

$\sqrt{(2 - 10) {}^{2} + (3 - y) {}^{2} } = 10$

Squaring on both sides

$( \sqrt{(2 - 10) {}^{2} + (3 - y) {}^{2} } ) {}^{2} = (10) {}^{2}$

$(2 - 10) {}^{2} + (3 - y) {}^{2} = 100$

$( - 8) {}^{2} + (3 - y) {}^{2} = 100$

$64 + (3 - y) {}^{2} = 100$

$(3 - y) {}^{2} = 100 - 64$

$(3 - y) {}^{2} = (6) {}^{2}$

$3 - y = ±6$

$3 - y = 6$

$y = - 3$

$3 - y = - 6$

$y = 9$

So, the value of y is -3 or 9

Know more :-

Mid point formula:-

$\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2}$

Centroid formula:-

$\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}$

Section formula Internal

division

$\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}$

Section formula External division

$\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}$