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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 3 · Set 1 · Q23
(a) Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, -2) and (1,6) respectively. (i) Find the co-ordinates of…
Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are $\displaystyle (9, -2)$ and $\displaystyle (1,6)$ respectively.
(i)
Find the co-ordinates of point P.
(ii)
Find the length of the side of the square.
Find the coordinates of a point on the line $\displaystyle \mathrm{x}+\mathrm{y}=5$ which is equidistant from $\displaystyle (6,4)$ and $\displaystyle (5, 2)$.
Marking-scheme solution
(i)
Coordinates of P are $\displaystyle \left(\frac{9+1}{2}, \frac{-2+6}{2}\right)=(5,2)$
(ii)
$\displaystyle 2 \mathrm{AB}^{2}=\mathrm{BD}^{2}$
\[\begin{array}{l}
\Rightarrow 2 \mathrm{AB}^{2}=(9-1)^{2}+(-2-6)^{2} \\
\Rightarrow \mathrm{AB}=8
\end{array}
\]
Hence, the length of the side of square is $\displaystyle 8$ units.
Let the required point be ( $\displaystyle \mathrm{x}, \mathrm{y}$ ) which is equidistant from $\displaystyle (6, 4)$ and $\displaystyle (5,2)$
\[\therefore(6-x)^{2}+(4-y)^{2}=(5-x)^{2}+(2-y)^{2}
\]
\[\Rightarrow 2 x+4 y=23
\]
Since point (x, y) also lies on the line $\displaystyle \mathrm{x}+\mathrm{y}=5$
\[\therefore x+y=5
\]
Solving (i) and (ii), we get $\displaystyle \mathrm{x}=-\frac{3}{2}$ and $\displaystyle \mathrm{y}=\frac{13}{2}$
So, the coordinates of the required point are $\displaystyle \left(-\frac{3}{2}, \frac{13}{2}\right)$
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.