Tiêu đề 10. STRAIGHT LINES.pdf ✅

(a.) a(–3, 0)  1 ,1 3       (b.) a (–, 3)  1 ,1 3       (c.) a 1 3, 3     −   (d.) a  1 , 3        a (a) Since origin and the point(a2 , a + 1) lie on the same side of both the lines, therefore we have 3a2 –(a + 1) + 1 > 0 i.e. a(3a – 1) > 0 gives a (– , 0)  1 , 3        and a 2 + 2(a + 1) – 5 < 0 i.e. a 2 + 2a – 3 < 0 i.e. (a – 1)(a + 3) < 0 gives a (–3, 1) Intersection of the above inequalities gives a (–3, 0)  1 ,1 3       . #QID# 82490 (4.) The point A(2, 1) is translated parallel to the line x – y = 3 by 4 units. If the new point lies in the third quadrant, then the coordinates of the new point are- (a.) 2 1 , 3 3   −     (b.) 1 1 , 2 2   −     (c.) (− + 2 1, – 2) (d.) (2 2 2,1– 2 2 − ) d (d) 2, 1 A (3, 0) (0, 3) (h, k)B AB = 4 4 Slope of AB = 1 i.e. k 1 h 2 − − = 1 i.e. h – k = 1 and AB = 4 i.e.(h – 2)2 +(k – 1)2 = 16 i.e.(k – 1)2 +(k – 1)2 = 16 given k = 1 – 8
and h = 2 – 8 .[k = 1 + 8 is not acceptable  B lies in the third quadrant] #QID# 82491 (5.) If the line y = 3 x cuts the curve x3 + y3 + 3xy + 5x2 + 3y2 + 4x + 5y – 1 = 0 at the points A, B, C, then OA · OB · OC is equal to- (a.) 4 13 (3 3 – 1) (b.) 4 3 3 1+ (c.) 2 + 1 3 (d.) 3 3 + 1 2 a (a) The abscissa of the intersection points of the given line and the given curve is given by the equation (3 3 + 1)x3 +(3 3 + 14)x2 +(5 3 + 4) x – 1 = 0 If x1, x2, x3 be the roots of the above equation, then A (x1, 3 x1), B (x2, 3 x2) and C (x3, 3 x3) Hence, we have OA · OB · OC = 2x1 · 2x2 · 2x3 = 8x1x2x3 = 8 × 1 3 3 1+ = 8 × 3 3 1 26 − = 4 13       (3 3 – 1). #QID# 82492 (6.) For how many integral values of m do the lines y + mx – 1 = 0 and 3x + 4y = 9 intersect in points having integral coordinates – (a.) 0 (b.) 1 (c.) 2 (d.) Infinite c (c) We have 3x + 4y = 9 i.e x = 9 4y 3 − = 3 – 4y 3 Thus, points lying on the above line and having integral coordinates are given by P (3 – 4k, 3k) k  I If P also lies on y + mx – 1 = 0, then we have 3k + m(3 – 4k) – 1 = 0 gives m = 3k 1 4k 3 − − , k  I For m to be an integer, we have |4k – 3|  |3k – 1| i.e. –(3k – 1) (4k – 3) (3k – 1) i.e. 4 7  k  2 There are only two integral values of k lying in the above integral, viz. k = 1, 2, Hence, there are only two integral values of m. #QID# 82493