Let A(h,k),B(1,1)\; and\; C(2,1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1 square unit, then the set of values which ' k' can take is given by

  • Option 1)

    {–1, 3}

  • Option 2)

    {–3, –2}

  • Option 3)

    {1, 3}

  • Option 4)

    {0, 2}

 

Answers (1)

As we learnt in 

 

Distance formula -

The distance between the point A\left ( x_{1},y_{1} \right )\: and \: B\left ( x_{2},y_{2} \right )

is \sqrt{\left ( x_{1} -x_{2}\right )^{2}+\left ( y_{1} -y_{2}\right )^{2}}

- wherein

 AC^{2}=AB^{2}+BC^{2}

(h-2)^{2}+(k-1)^{2}=(h+1)^{2}+(k-1)^{2}+1^{2}

4-4h=1-2h+1

2h=2\:\:\Rightarrow h=1

Also,  \frac{1}{2}AB.BC=1

\frac{1}{2}\times AB \times 1=1

AB=2=\sqrt{(h-1)^{2}+(k-1)^{2}}

(k-1)^{2}=2^{2}

k-1=2\:\:\:\:or\:\:\:\:k=-1=-2

k=3\:\:\:\:or\:\:\:\:k=-1


Option 1)

{–1, 3}

This option is correct.

Option 2)

{–3, –2}

This option is incorrect.

Option 3)

{1, 3}

This option is incorrect.

Option 4)

{0, 2}

This option is incorrect.

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