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Mathematics · 2024 · 2 marks
CBSE 2024 · Region 2 · Set 1 · Q25
In what ratio is the line segment joining the points $\displaystyle (3, -5)$ and $\displaystyle (-1, 6)$ divided by the line $\displaystyle \mathrm{y}=x$ ?$\displaystyle \mathrm{A}(3,0), \mathrm{B}(6,4)$ and $\displaystyle \mathrm{C}(-1,3)$ are vertices of a triangle ABC . Find length of its median BE.
In what ratio is the line segment joining the points $\displaystyle (3, -5)$ and $\displaystyle (-1, 6)$ divided by the line $\displaystyle \mathrm{y}=x$ ?
$\displaystyle \mathrm{A}(3,0), \mathrm{B}(6,4)$ and $\displaystyle \mathrm{C}(-1,3)$ are vertices of a triangle ABC . Find length of its median BE.
Marking-scheme solution
Let the required ratio be K:$\displaystyle 1$
Coordinates of point P are \(\displaystyle \left(\frac{-\mathrm{K}+3}{\mathrm{~K}+1}, \frac{6 \mathrm{~K}-5}{\mathrm{~K}+1}\right)\)
Point P lies on line \(\displaystyle \mathrm{y}=\mathrm{x} \Rightarrow \frac{-\mathrm{K}+3}{\mathrm{~K}+1}=\frac{6 \mathrm{~K}-5}{\mathrm{~K}+1}\)
Solving, we get \(\displaystyle \mathrm{K}=\frac{8}{7}\)
∴ Required ratio is $\displaystyle 8$: $\displaystyle 7$
Mid-point of AC is \(\displaystyle E\left(1, \frac{3}{2}\right)\)
Length of median BE
\[=\sqrt{(6-1)^{2}+\left(4-\frac{3}{2}\right)^{2}}=\sqrt{\frac{125}{4}} \text { or } \frac{5 \sqrt{5}}{2}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.