IIT-JEE 2008 Paper 2 Offline | Straight Lines and Pair of Straight Lines Question 6 | Mathematics

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider three points $$P = ( - \sin (\beta - \alpha ), - cos\beta ),Q = (cos(\beta - \alpha ),\sin \beta )$$ and $$R = (\cos (\beta - \alpha + \theta ),\sin (\beta - \theta ))$$ where $$0 < \alpha ,\beta ,\theta < {\pi \over 4}$$. Then :

P lies on the line segment RQ

Q lies on the line segment PR

R lies on the line segment QP

P, Q, R are non-collinear

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider the lines given by:

$${L_1}:x + 3y - 5 = 0$$

$${L_2}:3x - ky - 1 = 0$$

$${L_3}:5x + 2y - 12 = 0$$

Match the Statement/Expressions in Column I with the Statements/Expressions in Column II.

	Column I		Column II
(A)	L$$_1$$, L$$_2$$, L$$_3$$ are concurrent, if	(P)	$$K = - 9$$
(B)	One of L$$_1$$, L$$_2$$, L$$_3$$ is parallel to atleast one of the other two, if	(Q)	$$K = - {6 \over 5}$$
(C)	L$$_1$$, L$$_2$$, L$$_3$$ form a triangle, if	(R)	$$K = {5 \over 6}$$
(D)	L$$_1$$, L$$_2$$, L$$_3$$ do not form a triangle, if	(S)	$$K = 5$$

A - iv; B - ii; C - iii; D - i, ii

A - iv; B - i, ii; C - iii; D - i, ii, iv

A - iv; B - i; C - iii; D - i, ii

A - ii; B - i, iii; C - iii; D - i, ii, iv

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

Let $$O\left( {0,0} \right),P\left( {3,4} \right),Q\left( {6,0} \right)$$ be the vertices of the triangles $$OPQ$$. The point $$R$$ inside the triangle $$OPQ$$ is such that the triangles $$OPR$$, $$PQR$$, $$OQR$$ are of equal area. The coordinates of $$R$$ are

$$\left( {{4 \over 3},3} \right)$$

$$\left( {3,{2 \over 3}} \right)$$

$$\left( {3,{4 \over 3}} \right)$$

$$\left( {{4 \over 3},{2 \over 3}} \right)$$

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

The lines $${L_1}:y - x = 0$$ and $${L_2}:2x + y = 0$$ intersect the line $${L_3}:y + 2 = 0$$ at $$P$$ and $$Q$$ respectively. The bisector of the acute angle between $${L_1}$$ and $${L_2}$$ intersects $${L_3}$$ at $$R$$.

Statement-1: The ratio $$PR$$ : $$RQ$$ equals $$2\sqrt 2 :\sqrt 5 $$. because
Statement-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.

Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement- 1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is False.

Statement-1 is False, Statement-2 is True.

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