QJEEM19Mad872: JEE Main 2019 Mathematics - Class 12 JEEM19Mad872 Solution: The vertices B and C of a ABC lie on the line, such that BC = 5 …

Question

Verified Solution • 2025–26 Session moderate

The vertices B and C of a

\Delta

ABC lie on the line,

{{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}

such that BC = 5 units.

Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :

Accepted Answer

Area of

\Delta

ABC =

{1 \over 2} \times BC \times AD

Given BC = 5 so we need perpendicular distance of A from line BC.

Let a point D on BC = (3

\lambda

- 2, 1, 4

\lambda

)

Direction ratio of AD

3

\lambda

- 3, 2, 4

\lambda

- 2

As AD perpendicular to the BC, so

DR of AD. DR of BC = 0

(3\lambda - 3)3 + 2(0) + (4\lambda - 2)4 = 0

9\lambda - 9 + 16\lambda - 8 = 0 \Rightarrow \lambda = {{17} \over {25}}

Hence,

D = \left( {{1 \over {25}},1,{{68} \over {25}}} \right)

\left| {\overline {AD} } \right| = \sqrt {{{\left( {{1 \over {25}} - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {{{68} \over {25}} - 2} \right)}^2}}

\Rightarrow \sqrt {{{\left( {{{ - 24} \over {25}}} \right)}^2} + 4 + {{\left( {{{18} \over {25}}} \right)}^2}}

\Rightarrow \sqrt {{{{{\left( {24} \right)}^2} + 4{{\left( {25} \right)}^2} + {{\left( {18} \right)}^2}} \over {{{25}^2}}}}

\Rightarrow \sqrt {{{576 + 2500 + 324} \over {{{25}^2}}}}

\Rightarrow \sqrt {{{3400} \over {{{25}^2}}}} \Rightarrow {{\sqrt {34} .10} \over {25}} = {{2\sqrt {34} } \over 5}

Area of triangle =

{1 \over 2} \times \left| {\overline {BC} } \right| \times \left| {\overline {AD} } \right|

=

{1 \over 2} \times 5 \times {{2\sqrt {34} } \over 5} = \sqrt {34}

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