The vertices of a triangle are (2, 1), (5, 2) and (4, 4). The lengths of the perpendicular from these vertices on the opposite sides are

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Solution

The correct option is D
$L_{12} \equiv x - 3 y + 1 = 0 L_{23} \equiv 2 x + y - 12 = 0 L_{13} \equiv 3 x - 2 y - 4 = 0 T h e r e f o r e, t h e r e q u i r e d d i s t a n c e s a r e D_{3} = ∣ ∣ ∣ \frac{4 - 3 \times 4 + 1}{\sqrt{10}} ∣ ∣ ∣ = \frac{7}{\sqrt{10}} D_{1} = ∣ ∣ ∣ \frac{4 + 1 - 12}{\sqrt{5}} ∣ ∣ ∣ = \frac{7}{\sqrt{5}} D_{2} = ∣ ∣ ∣ \frac{3 \times 5 - 2 \times 2 - 4}{\sqrt{9 + 4}} ∣ ∣ ∣ = \frac{7}{\sqrt{13}}$

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The vertices of a triangle are (2, 1), (5, 2) and (4, 4). The lengths of the perpendicular from these vertices on the opposite sides are

The correct option is D L12≡x−3y+1=0L23≡2x+y−12=0L13≡3x−2y−4=0Therefore, the required distances areD3=∣∣∣4−3×4+1√10∣∣∣=7√10D1=∣∣∣4+1−12√5∣∣∣=7√5D2=∣∣∣3×5−2×2−4√9+4∣∣∣=7√13