CBSE Class 10 Mathematics Standard 2026: question 23(b)

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Step-by-step solution

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1. Let the required point be $P(x,y)$. It is equidistant from $(6,4)$ and $(5,2)$, so by the distance formula:
   $(x-6{)}^2+(y-4{)}^2=(x-5{)}^2+(y-2{)}^2$
   $\Rightarrow x^2-12x+36+y^2-8y+16=x^2-10x+25+y^2-4y+4$
   $\Rightarrow 52-12x-8y=29-10x-4y$
   $\Rightarrow 2x+4y=23\text{...(i)}$

   1 mark

2. P lies on the line, so $x+y=5$, which gives $2x+2y=10\text{...(ii)}$.
   Subtracting (ii) from (i): $2y=13\Rightarrow y=\frac{13}{2}$.
   Then $x=5-\frac{13}{2}=-\frac{3}{2}$.

   ½ mark

3. Hence the required point is $\left(-\frac{3}{2},\frac{13}{2}\right)$.
   (Check: both squared distances equal $62.5$.)

   ½ mark

Answer: $\left(-\frac{3}{2},\frac{13}{2}\right)$

Revise the chapter

• Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)