Question

(b) Find the points on the y-axis which are at a distance of 15 units from the
point (9,-3).
​

Answer 1

Answer:

Let the required point on the y-axis be P(0, y). Then,

PA=13⇒PA2=169

⇒(0+5)2+(y−7)2=169

⇒y2−14y+74=169⇒y2−14y−95=0

⇒y2−19y+5y−95=0⇒y(y−19)+5(y−19)=0

⇒(y−19)(y+5)=0⇒y−19=0 or y+5=0

⇒y=19 or y=−5

Hence, the required points on the y-axis are B(0, 19) and C(0, -5).

Answer 2

Answer:

The points on the y-axis which are at a distance of 15 units from the point (9,-3) are (0, 9) and (0, -15).

Step-by-step explanation:

Let the coordinates of the points on the y - axis which are at a distance of 15 units from (9, -3) be (0, $y$).

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

$15=\sqrt{(0-9)^{2}+[y-(-3)]^{2}}$

$15=\sqrt{(-9)^{2}+(y+3)^{2}}$

$15=\sqrt{81+(y^{2}+9+6y)}$ (∵ $(a+b)^{2}=a^{2}+2ab+b^{2}$)

$15=\sqrt{81+y^{2}+9+6y}$

$15=\sqrt{90+y^{2}+6y}$

Squaring on both sides,

$225=90+y^{2}+6y$

$225-90=y^{2}+6y$

$135=y^{2}+6y$

$y^{2}+6y-135=0$

$y^{2}-9y+15y-135=0$

$y(y-9)+15(y-9)=0$

$(y+15)(y-9)=0$

$y-9=0\\y=0+9\\y=9$

$y+15=0\\y=0-15\\y=-15$

∴ The points on the y-axis which are at a distance of 15 units from the point (9,-3) are (0, 9) and (0, -15).