Find a point on the y-axis which is equidista | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that any point on the $y$-axis is of the form $(0, y)$. Let the point $P(0, y)$ be equidistant from points $A(6, 5)$ and $B(-4, 3)$.Accord

Vedclass · Accepted Answer

$(0, 9)$

Vedclass · Answer

(A) We know that any point on the $y$-axis is of the form $(0, y)$. Let the point $P(0, y)$ be equidistant from points $A(6, 5)$ and $B(-4, 3)$.According to the distance formula,$PA = PB$,so $PA^2 = PB^2$.$(6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2$$36 + (25 - 10y + y^2) = 16 + (9 - 6y + y^2)$$61 - 10y + y^2 = 25 - 6y + y^2$Subtracting $y^2$ from both sides:$61 - 10y = 25 - 6y$$61 - 25 = 10y - 6y$$36 = 4y$$y = 9$Thus,the required point is $(0, 9)$.

Find a point on the $y$-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.

Similar Questions

If $P (0,0), Q (0,5)$ and $R (6,0),$ then $\Delta PQR$ is $\ldots \ldots \ldots$ triangle.

If $A(0,0)$,$B(0,12)$,and $C(5,0)$,then $BC = \ldots$

Name the type of triangle $PQR$ formed by the points $P(\sqrt{2}, \sqrt{2})$,$Q(-\sqrt{2}, -\sqrt{2})$,and $R(-\sqrt{6}, \sqrt{6})$.

The distance of the point $P(-6, 8)$ from the origin is

In $\Delta ABC,$ if $A(0, 0), B(4, 0)$ and $C(0, 3)$ then $\Delta ABC$ is a $\ldots \ldots \ldots$ triangle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module