Question

The point whose abscissa is equal to its ordinate and which is equidistant from A (5, 0) and B (0,3) is (a) (1,1)

Answer 1

Answer:  The required co-ordinates of the point are (1, 1).

Step-by-step explanation:  Given that a point with equal ordinate and abscissa is equidistant from the points A(5, 0) and B(0, 3).

We are to find the co-ordinates of the point.

Let (a, a) be the co-ordinates of the required point.

Then, according to the given information, we have

$\textup{distance between (a,a) and (5, 0)}=\textup{distance between (a, a) and (0, 3)}\\\\\Rightarrow \sqrt{(a-5)^2+(a-0)^2}=\sqrt{(a-0)^2+(a-3)^2}\\\\\Rightarrow a^2-10a+25+a^2=a^2+a^2-6a+9~~~~~~~~~[\textup{Squaring both sides}]\\\\\Rightarrow -10a+25=-6a+9\\\\\Rightarrow 10a+6a=25-9\\\\\Rightarrow 16a=16\\\\\Rightarrow a=1.$

Thus, the required co-ordinates of the point are (1, 1).

Answer 2

Answer:

Answer: The required co-ordinates of the point are (1, 1).

Step-by-step explanation: Given that a point with equal ordinate and abscissa is equidistant from the points A(5, 0) and B(0, 3).

We are to find the co-ordinates of the point.

Let (a, a) be the co-ordinates of the required point.

Then, according to the given information, we have

\begin{gathered}\textup{distance between (a,a) and (5, 0)}=\textup{distance between (a, a) and (0, 3)}\\\\\Rightarrow \sqrt{(a-5)^2+(a-0)^2}=\sqrt{(a-0)^2+(a-3)^2}\\\\\Rightarrow a^2-10a+25+a^2=a^2+a^2-6a+9~~~~~~~~~[\textup{Squaring both sides}]\\\\\Rightarrow -10a+25=-6a+9\\\\\Rightarrow 10a+6a=25-9\\\\\Rightarrow 16a=16\\\\\Rightarrow a=1.\end{gathered}

distance between (a,a) and (5, 0)=distance

distance between (a,a) and (5, 0)=distance between (a, a) and (0, 3)

⇒ (a−5) 2+(a−0) 2 = (a−0) 2+(a−3) 2

⇒a 2 −10a+25+a 2 =a 2 +a 2 −6a+9 [Squaring both sides]

⇒−10a+25=−6a+9

⇒10a+6a=25−9

⇒16a=16

⇒a=1.

Thus, the required co-ordinates of the point are (1, 1).

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