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1

### AIEEE 2007

Let A $$\left( {h,k} \right)$$, B$$\left( {1,1} \right)$$ 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
A
$$\left\{ { - 1,3} \right\}$$
B
$$\left\{ { - 3, - 2} \right\}$$
C
$$\left\{ { 1,3} \right\}$$
D
$$\left\{ {0,2} \right\}$$

## Explanation

Given : The vertices of a right angled triangle $$A\left( {1,k} \right),$$

$$B\left( {1,1} \right)$$ and $$C\left( {2,1} \right)$$ and area of $$\Delta ABC = 1$$ square unit

We know that, area of night angled triangle

$$= {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$

$$\Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$$
2

### AIEEE 2006

If $$\left( {a,{a^2}} \right)$$ falls inside the angle made by the lines $$y = {x \over 2},$$ $$x > 0$$ and $$y = 3x,$$ $$x > 0,$$ then a belong to
A
$$\left( {0,{1 \over 2}} \right)$$
B
$$\left( {3,\infty } \right)$$
C
$$\left( {{1 \over 2},3} \right)$$
D
$$\left( {-3,-{1 \over 2}} \right)$$

## Explanation

Clearly for point $$P,$$

$${a^2} - 3a < 0$$ and $${a^2} - {a \over 2} > 0 \Rightarrow {1 \over 2} < a < 3$$
3

### AIEEE 2006

A straight line through the point $$A (3, 4)$$ is such that its intercept between the axes is bisected at $$A$$. Its equation is
A
$$x + y = 7$$
B
$$3x - 4y + 7 = 0$$
C
$$4x + 3y = 24$$
D
$$3x + 4y = 25$$

## Explanation

As is the mid point of $$PQ,$$ therefore

$${{a + 0} \over 2} = 3,{{0 + b} \over 2} = 4 \Rightarrow a = 6,b = 8$$

$$\therefore$$ Equation of line is $${x \over 6} + {y \over 8} = 1$$

or $$4x + 3y = 24$$
4

### AIEEE 2005

If a vertex of a triangle is $$(1, 1)$$ and the mid points of two sides through this vertex are $$(-1, 2)$$ and $$(3, 2)$$ then the centroid of the triangle is
A
$$\left( { - 1,{7 \over 3}} \right)$$
B
$$\left( {{{ - 1} \over 3},{7 \over 3}} \right)$$
C
$$\left( { 1,{7 \over 3}} \right)$$
D
$$\left( {{{ 1} \over 3},{7 \over 3}} \right)$$

## Explanation

Vertex of triangle is $$\left( {1,\,1} \right)$$ and midpoint of sides through -

this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$

$$\Rightarrow$$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$

$$\therefore$$ centroid is $${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$$

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