Question

3. If Y ordinate of a point is 4 units more than its abscissa and this point is equidistant from point(-2, 3) and (6,-3) then find the coordinates of the point.​

Answer 1

Answer:

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 =

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 =

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y)

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y) 2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y) 2 ⇒y=2

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)(12−0) 2 +(−3−y) 2 = (13−0) 2 +(2−y) 2 ⇒ 144+(y+3) 2 = 169+(2−y) 2 ⇒y=2∴ The required point on Y-axis is (0, 2). ( hint: since x cordinates are equal to '0' on y-aixs)