Question 1

The locus of the point of intersection of the lines x=a ( 1-t^2 1+t^2 ) and y= 2 a t 1+t^2 represent (t being a parameter)

Accepted Answer

Given x=a ( 1-t^2 1+t^2 ) and y= 2 a t 1+t^2 aligned & Let t= \ & x=a ( 1- ^2 1+ ^2 ) \ & and y= 2 a 1+ ^2 \ & x=a 2 and y=a 2 \ & 2 = x a and 2 = y a aligned Squaring both sides, we get ^2 2 = x^2 a^2 ....(i) and ^2 2 = y^2 a^2 ....(ii) Adding Eqs. (i) and (ii), we get aligned ^2 2 + ^2 2 & = x^2 a^2 + y^2 a^2 \ x^2+y^2 & =a^2 aligned Locus is a circle having centre at origin and radius a.

Question 2

If the straight line 2 x+3 y-1=0, x+2 y-1=0 and a x+b y-1=0 form a triangle with origin as orthocentre, then (a, b) is equal to

Accepted Answer

Here, point A is the intersection of line A B and A C so equation of line passing through A. (x+2 y-1)+ (2 x+3 y-1)=0 .....(i) This line passes through the orthocentre (0,0), then -1- =0 =-1 On substituting =-1 in Eq. (i), we get x+y= 0 as the equation of A D. Since A D B C, therefore -1 - a b =-1 a+b=0 ...(ii) Similarly, by applying the condition that B E is perpendicular to C A, we get a+2 b=8 ....(iii) Now, solving Eqs. (ii) and (iii), we get a=-8, b=8

Question 3

The distance from the origin to the image of (1, 1) with respect to the line x+y+5=0 is

Accepted Answer

As we know that the image of (1,1) with respect to line x+y+5=0 is x-1 1 = y-1 1 = 2(1+1+5) 1+1 x-1=-7, y-1=-7 x=-6, y=-6 Image of point (1,1) is (-6,-6) Now, distance from origin This is the required transformed equation. D= (0+6)^2+(0+6)^2 D= 72 =6 2

Question 4

If (-4,5) is one vertex and 7 x-y+8=0 is one diagonal of a square, then the equation of second diagonal is

Accepted Answer

One vertex of square is (-4,5) and equation of one diagonal is 7 x-y+8=0 Diagonal of a square are perpendicular and bisect each other Let the equation of the other diagonal be y=m x+ c where m is the slope of the line and c is the y -intercept. Since this line passes through (-4,5) Since this line is at right angle to the line 7 x-y+8=0 or y=7 x+8, having slope =7, 7 m =-1 or m = -1 7 Putting this value of m in equation (i) we get 5=-4 ( -1 7 )+c or 5= 4 7 +c or c=5- 4 7 = 31 7 Hence equation of

Question 5

The coordinates of the foot of the perpendicular from the point (2,3) on the line x+y-11=0 are

Accepted Answer

Let (h, k) be the coordinates of the foot of the perpendicular from the point (2,3) on the i x+y-11=0. Then, the slope of the perpendicular line is k-3 h-2 . Again the slope of the given line x+y-11=0 is -1 Using the condition of perpendicularity of lines, we have Since (h, k) lies on the given line, we have, Solving (i) and (ii), we get h=5 and k=6. Thus (5,6) are the required coordinates of the foot of the perpendicular.

Question 6

What is the angle between the two straight lines y=(2- 3 ) x+5 and y=(2+ 3 ) x-7 ?

Accepted Answer

The given lines are: y=(2- 3 ) x+5 and y=(2+ 3 ) x-7 Therefore, slope of first line =m_1=2- 3 and slope of second line =m_2=2+ 3 = | m_2-m_1 1+m_1 m_2 |= | 2+ 3 -2+ 3 1+(4-3) | = | 2 3 2 |= 3 = 3 = 3 =60^

Question 7

If the coordinates of the points A and B be (3,3) and (7,6), then the length of the portion of the line AB intercepted between the axes is

Accepted Answer

Equation of line A B is y-3= 6-3 7-3 (x-3) 3 x-4 y+3=0 x -1 + y 3 / 4 =1 Hence required length is (-1)^2+ ( 3 4 )^2 = 5 4 .

Question 8

The distance between the parallel lines 3 x-4 y+7=0 and 3 x-4 y+5=0 is a b . Value of a+b is

Accepted Answer

Given parallel lines are 3 x-4 y+7=0 and 3 x-4 y+5=0 Required distance = |7-5| (3)^ 2 +(-4)^ 2 = 2 5 a=2, b=5

Question 9

The points (0,7,10),(-1,6,6) and (-4,9,6) form

Accepted Answer

Let P (0,7,10), Q (-1,6,6) and R (-4,9,6) be the vertices of a triangle Here, PQ = 1+1+16 =3 2 array l QR = 9+9+0 =3 2 \ PR = 16+4+16 =6 \ Now, PQ ^ 2 + QR ^ 2 =(3 2 )^ 2 +(3 2 )^ 2 =36=( PR )^ 2 array Therefore, PQR is a right angled triangle at Q . Also, OQ = QR . Hence, PQR is a right angled isosceles triangle.

Straight Lines — Previous Year Questions & Solutions