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5 months, 3 weeks ago

Can someone explain If an equilateral triangle, having vertexat the (2,-1), has a side along the line, x+y=2, then the area (in sq. units) of this triangle is :

If an equilateral triangle, having vertex at the (2,-1), has a side along the line, x+y=2, then the area (in sq. units) of this triangle is :

• Option 1)

• Option 2)

6

• Option 3)

• Option 4)

60 Views
A arun shelke

ACCORDING TO QUESTION ,WE JUST HAVE TO FIND THE PERPENDICULAR DISTANCE OF POINT (2,-1) FORM THE LINE

x + y = 2,THIS PERPENDICULAR DISTANCE IS EQUAL TO THE HEIGHT OF THE TRIANGLE FORMED,AND THEN WE CAN CALCULATE THE SIDE OF TRIANGLE,AND THEN BY APPLING FORMULA OF AREA OF TRIANGLE  {(1/2)*b*h}

SO,

DISTANCE OF POINT FRON THE LINE (h) = $\left |x_1+y_1-c \right |/\left | \sqrt{x_1^{2}+y_1^{2}} \right |$

AND ALSO WE KNOW THAT HEIGHT OF THE TRIANGLE = $(\sqrt{3}*a)/2$

NOW, h = $1/\sqrt{2 }$ AND ALSO h = $(\sqrt{3}*a)/2$

FROM HERE WE GET THE VALUE OF a(BASE OF TRIANGLE(b)) = $\sqrt{2/3}$

NOW BY FORMULA {(1/2)*b*h} = (1/2) *  $\sqrt{2/3}$ *  $1/\sqrt{2 }$ = $1/(\sqrt{3})*2$ sq.units.

BUT UNFORTUNATELY NO OPTION IS MATCHING OUR ANSWER.SO, THE ANSWER IS 1/2$\sqrt{3}$

S solutionqc
Answered 5 months, 3 weeks ago

Option 1)

Option 2)

6

Option 3)

Option 4)

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