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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 5 · Set 1 · Q22
(a) Vertices of a right triangle ABC with ∠ B =90^° are A (3,4), B (1,1) and C(-8, 7). Find the value of tan A.
(a)
Vertices of a right triangle ABC with $\displaystyle \angle \mathrm{B}=90^{\circ}$ are $\displaystyle \mathrm{A}(3,4), \mathrm{B}(1,1)$ and C$\displaystyle (-8, 7)$. Find the value of $\displaystyle \tan \mathrm{A}$.
OR
(b)
Using distance formula, prove that the points A$\displaystyle (2, 3)$, B$\displaystyle (-7, 0)$ and C$\displaystyle (-1, 2)$ are collinear.
Marking-scheme solution
$\displaystyle \mathrm{BC}=\sqrt{(-8-1)^{2}+(7-1)^{2}}=\sqrt{117}=3 \sqrt{13}$
$\displaystyle \mathrm{AB}=\sqrt{(3-1)^{2}+(4-1)^{2}}=\sqrt{13}$
$\displaystyle \tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 \sqrt{13}}{\sqrt{13}}=3$
Using distance formula, prove that the points A$\displaystyle (2, 3)$, B$\displaystyle (-7, 0)$ and $\displaystyle \mathrm{C}(-1,2)$ are collinear.
\[\begin{array}{l}
A B=\sqrt{(-7-2)^{2}+(0-3)^{2}}=\sqrt{90}=3 \sqrt{10} \\
B C=\sqrt{(-1+7)^{2}+(2-0)^{2}}=\sqrt{40}=2 \sqrt{10} \\
A C=\sqrt{(-1-2)^{2}+(2-3)^{2}}=\sqrt{10}=\sqrt{10} \\
A C+B C=A B \\
\therefore A, B, C \text { are collinear. }
\end{array}
\]
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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.