If $A\left( {6, - 3} \right),B\left( { - 3,5} \right),C\left( {4, - 2} \right),P\left( {{\rm{\alpha }},{\rm{\beta }}} \right),$ then the ratio of the areas of the triangles $PBC,ABC$ is

$\left| {{\rm{\alpha }} + {\rm{\beta }} - 2} \right|$

$\left| {{\rm{\alpha }} + {\rm{\beta }}} \right|$

$\left| {{\rm{\alpha }} - {\rm{\beta }}} \right|$

$\left| {{\rm{\alpha }} + {\rm{\beta }} + 2} \right|$

JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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A\left( {6, - 3} \right),B\left( { - 3,5} \right),C\left( {4, - 2} \right),P\left( {\mathrm{\alpha },\mathrm{\beta }} \right),

then the ratio of the areas of the triangles

PBC,ABC

View answer and explanation

Correct option: D

\left| {\mathrm{\alpha } + \mathrm{\beta } - 2} \right|

Option D is correct because the ratio of areas of triangles PBC and ABC is found by calculating the determinants and simplifying to |α + β - 2|.

Practice All Chapter MCQs →

\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3