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Q.1. The area enclosed within the curve  x  +  y  = 1 is ................... (1981  2 Marks)
Ans. 2 sq. units.
Sol.  x  +  y  = 1
The curve represents four lines
x + y = 1, x – y = 1, – x + y = 1, – x – y = 1
which enclose a square of side = distance between opp. sides x + y = 1 and x + y = – 1
Side
∴ Req. area = (side)^{2} = 2 sq. units.
Q.2. y = 10^{x }is the reflection of y = log10^{ x} in the line whose equation is ................ (1982  2 Marks)
Ans. y = x
Sol. As y = log_{10} x can be obtained by replacing x by y and y by x in y = 10^{x}
∴ The line of reflection is y = x.
Q.3. The set of lines ax+by+c = 0, where 3a + 2b + 4c = 0 is concurrent at the point ................... (1982  2 Marks)
Ans. (3/4, 1/2)
Sol. Given that 3a + 2b + 4c = 0
⇒ The set of lines ax + by + c = 0 passes through the point (3/4, 1/2).
Q.4. Given the points A (0, 4) and B (0, –4), the equation of the locus of the point P(x, y) such that  AP – BP  = 6 is ................... (1983  1 Mark)
Ans.
Sol. AP – BP  = 6
We know that locus of a point, difference of whose distances from two fixed points is constant, is hyperbola with the fixed points as focii and the difference of distances as length of transverse axis.
Thus, ae = 4 and 2a = 6 ⇒ a = 3, e = 4 /3
∴ Equation is
(foci being on yaxis, it is vertical hyperbola)
Q.5. If a, b and c are in A.P., then the straight line ax + by + c = 0 will always pass through a fixed point whose coordinates are ................... (1984  2 Marks)
Ans. (1, – 2)
Sol. If a, b, c are in A.P. then a + c = 2b ⇒ a – 2b + c = 0
⇒ ax + by + c = 0 passes through (1,– 2).
Q.6. The orthocentre of the triangle for med by the lines x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0 lies in quadrant number ................... . (1985  2 Marks)
Ans. first quadrant
Sol.
The equations of sides of triangle ABC are
AB : x + y = 1
BC : 2x + 3y = 6
CA : 4x – y = – 4
Solving these pairwise we get the vertices of D as follows A (– 3/5, 8/5) B (– 3, 4) C (– 3/7, 16/7)
Now AD is line ⊥^{ lar} to BC and passes through A. Any line perpendicular to BC is 3x – 2y + l = 0 As it passes through A(– 3/5, 8/5)
∴ Equation of altitude AD is 3x – 2y + 5 = 0 ...(1)
Any line perpendicular to side AC is x + 4y + µ = 0
As it passes through point B (– 3, 4)
∴ – 3 + 16 + µ = 0 ⇒ µ = – 13
∴ Equation of altitude BE is x + 4y –13 = 0 ...(2)
Now orthocentre is the point of intersection of equations (1) and (2) (AD and BE)
Solving (1) and (2), we get x = 3/7, y = 22/7
As both the coordinates are positive, orthocentre lies in first quadrant.
Q.7. Let the algebraic sum of the perpendicular distances from the points (2, 0), (0, 2) and (1, 1) to a variable straight line be zero; then the line passes through a fixed point whose cordinates are ................... . (1991  2 Marks)
Ans. (1, 1)
Sol. Let the variable line be ax + by + c = 0 .....… (1)
Then ⊥^{lar} distance of line from (0, 2) =
⊥^{lar} distance of line from (1, 1) =
ATQ p_{1 }+ p_{2} + p_{3} = 0
⇒ 3a + 3b + 3c = 0
⇒ a + b + c = 0 .....… (2)
From (1) and (2), we can say variable line (1) passes through the fixed point (1, 1).
Q.8. The vertices of a triangle are A (–1, –7), B (5, 1) and C (1, 4).
The equation of th e bisector of the an gle ∠ABC is ................... . (1993  2 Marks)
Ans. x  7y + 2 = 0
Sol. Let BD be the bisector of ∠ABC.
NOTE THIS STEP :
Then AD : DC = AB : BC
And
AB =
∴ By section formula
Therefore equation of BD is
⇒ 7y – 7 = x – 5 ⇒ x – 7y + 2 = 0
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