NCERT Exemplar Straight Lines - (Maths) for JEE Main & Advanced PDF Download

Q.1. Find the equation of the straight line which passes through the point (1, - 2) and cuts off equal intercepts from axes.
Ans. Use the intercept form of a straight line:

where a and b are the intercepts on the axes.
Given that a = b.
∴   ...(i)
If eq. (i) passes through the point (1, - 2), substitute x = 1, y = -2:

Solving gives a = -1, so the line becomes x + y + 1 = 0.
Hence, the required equation is x + y + 1 = 0.
Q.2. Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, - 1).
Ans. First find the slope of the line joining (2, 3) and (3, -1):

So slope m = (-1 - 3)/(3 - 2) = -4.
Slope of the required line, being perpendicular, is m' = 1/4 (since m · m' = -1).
Equation of the line passing through (5, 2) with slope 1/4 is:

Simplify: 4(y - 2) = (x - 5) ⇒ 4y - 8 = x - 5 ⇒ x - 4y + 3 = 0.
Hence, the required equation is x - 4y + 3 = 0.
Q.3. Find the angle between the lines y = (2 - √3) (x + 5) and y = (2 + √3 ) (x - 7).
Ans. The two lines are:
y = (2 - √3)(x + 5)  ...(i)
y = (2 + √3)(x - 7)  ...(ii)
So their slopes are m₁ = 2 - √3 and m₂ = 2 + √3.
Let θ be the angle between them. Then

Compute tan θ = |(m₂ - m₁)/(1 + m₁m₂)| = |(2√3)/(1 + (2 - √3)(2 + √3))|.
Since (2 - √3)(2 + √3) = 4 - 3 = 1, the denominator is 1 + 1 = 2, so tan θ = √3.
Thus θ = 60°. The supplementary angle 120° also satisfies the tangent value (signs considered), so the angle between the lines is 60° (acute) or 120° (obtuse).
Hence, the required angle is 60° or 120°.
Q.4. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.
Ans. Let the x- and y-intercepts be a and b respectively. The intercept form is
...(i)
Given a + b = 14 ⇒ b = 14 - a, so the equation becomes:
...(ii)
Substitute point (3, 4) into (ii):

This yields the quadratic in a:

⇒ a² - 13a + 42 = 0
Factor: (a - 6)(a - 7) = 0 ⇒ a = 6 or 7.
Then b = 14 - a gives b = 8 or 7 respectively.
Hence the two lines are the ones with intercepts (6, 8) and (7, 7):
and .
Q.5. Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.
Ans. Let (x₁, y₁) lie on x + y = 4 ⇒ x₁ + y₁ = 4 ...(i).
Distance from (x₁, y₁) to 4x + 3y = 10 is |4x₁ + 3y₁ - 10|/5. This must equal 1, so |4x₁ + 3y₁ - 10| = 5.
Take +5: 4x₁ + 3y₁ = 15 ...(ii). Using y₁ = 4 - x₁ gives x₁ = 3, y₁ = 1 ⇒ point (3, 1).
Take -5: 4x₁ + 3y₁ = 5 ...(iii). Using y₁ = 4 - x₁ gives x₁ = -7, y₁ = 11 ⇒ point (-7, 11).
Hence, the required points are (3, 1) and (-7, 11).
Q.6. Show that the tangent of an angle between the lines

is

Ans.
Given the two lines in the stated form:
...(i)
and ...(ii)
Their slopes are, respectively,
m₁ = and m₂ = .
Let θ be the angle between the two lines. Then

Carrying out the algebra (substituting m₁, m₂ and simplifying) leads to the required expression:

Hence proved.
Q.7. Find the equation of lines passing through (1, 2) and making angle 30° with y-axis.
Ans. If a line makes 30° with the y-axis, it makes 60° with the x-axis. So slope m = tan 60° = √3.
Equation through (1, 2) with slope √3 is:
y - 2 = √3(x - 1)
⇒ y - √3x + √3 - 2 = 0.
Hence, the required equation is y - √3x + √3 - 2 = 0.
Q.8. Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.
Ans. Let the intersection point of 2x + y - 5 = 0 and x + 3y + 8 = 0 be P. The family of lines through P is
(2x + y - 5) + λ(x + 3y + 8) = 0. ...(iv)
This expands to (2 + λ)x + (1 + 3λ)y - 5 + 8λ = 0.
Slope of (iv) is -(2 + λ)/(1 + 3λ). Slope of 3x + 4y = 7 is -3/4.
Equate slopes: -(2 + λ)/(1 + 3λ) = -3/4 ⇒ (2 + λ)/(1 + 3λ) = 3/4.
Solve: 8 + 4λ = 3 + 9λ ⇒ 5λ = 5 ⇒ λ = 1. Put λ = 1 into (iv): (2x + y - 5) + (x + 3y + 8) = 0 ⇒ 3x + 4y + 3 = 0.
Hence, the required equation is 3x + 4y + 3 = 0.
Q.9. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x - 3y + 6 = 0 on the axes.
Ans. Write ax + by + 8 = 0 as ax + by = -8. Its intercepts are (-8/a, 0) and (0, -8/b).
For 2x - 3y + 6 = 0 ⇒ 2x - 3y = -6, so intercepts are (-6/2, 0) = (-3, 0) and (0, -6/(-3)) = (0, 2). Thus intercepts are -3 on x-axis and 2 on y-axis.
The question says the intercepts of ax + by + 8 = 0 are equal in magnitude but opposite in sign to these, so we require:
-8/a = 3 and -8/b = -2 ⇒ a = -8/3 and b = 4.
Hence, the required values are a = -8/3 and b = 4.
Q.10. If the intercept of a line between the coordinate axes is divided by the point (-5, 4) in the ratio 1 : 2, then find the equation of the line.
Ans. Let the intercepts be A(a, 0) and B(0, b). The point (-5, 4) divides AB internally in the ratio 1:2, so using section formula:
Coordinates of the point dividing A(a,0) and B(0,b) in ratio 1:2 are ((2a + 0)/3, (0 + b)/3) = (2a/3, b/3).
Equate to (-5, 4): 2a/3 = -5 ⇒ a = -15/2. And b/3 = 4 ⇒ b = 12.
So A = (-15/2, 0), B = (0, 12). Equation of line AB in intercept form is x/(-15/2) + y/12 = 1 ⇒ multiply by 60: -4x + 5y = 60 ⇒ 8x - 5y + 60 = 0 (after multiplying by -2).
Hence, the required equation is 8x - 5y + 60 = 0.
Q.11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
[Hint: Use normal form, here ω =30°.]
Ans.
Normal form: x cos θ + y sin θ = p, where p is the perpendicular from origin and θ is the angle the normal makes with x-axis.
The line makes 120° with the positive x-axis, so the normal makes θ = 30° (since normal is 90° from the line direction). Given p = 4.
Thus x cos 30° + y sin 30° = 4 ⇒ (√3/2)x + (1/2)y = 4 ⇒ √3 x + y = 8.
Hence, the required equation is √3 x + y = 8.
Q.12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).
Ans. Hypotenuse has slope -3/4. For an isosceles right triangle the other two sides make equal acute angles with the hypotenuse; their slopes satisfy the condition given by angle bisector relations. Let one side have slope m.
From the relation for equal angles with slope -3/4, solve the equation obtained (details use slope-angle formulas): we get m = 1/7 or m = -7.
Line through (2, 2) with slope 1/7: y - 2 = (1/7)(x - 2) ⇒ x - 7y + 12 = 0.
Line with slope -7: y - 2 = -7(x - 2) ⇒ 7x + y - 16 = 0.
Hence, the required equations are x - 7y + 12 = 0 and 7x + y - 16 = 0.
LONG ANSWER TYPE QUESTIONS

Q.13. If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, - 1), then find the length of the side of the triangle. 
[Hint: Find length of perpendicular (p) from (2, - 1) to the line and use p = l sin 60°, where l is the length of side of the triangle].
Ans. Base AB: x + y = 2. The perpendicular distance p from A(2, -1) to the line x + y - 2 = 0 is

Compute numerator: |2 + (-1) - 2| = |-1| = 1. Denominator √(1^2 + 1^2) = √2. So p = 1/√2.
In an equilateral triangle p = l sin 60° = l(√3/2). Hence l = p/(√3/2) = (1/√2)·(2/√3) = 2/(√6) = √(4/6) = √(2/3). Simplify to l = 2/(√6) (or rationalised if desired).
Thus the side length l = 2/√6.
Q.14. A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P. 
[Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed point P (x₁, y₁) is y - y₁ = m (x - x₁). Taking the algebraic sum of perpendicular distances equal to zero, we get y - 1 = m (x - 1). Thus (x₁, y₁) is (1, 1).]
Ans. Let the fixed point be P(x₁, y₁) and slope be m. Equation: y - y₁ = m(x - x₁).
Perpendicular distances from A(2, 0), B(0, 2), C(1, 1) to this line are d₁, d₂, d₃ respectively (expressed algebraically with signs).
Setting d₁ + d₂ + d₃ = 0 and simplifying yields mx₁ - y₁ - m + 1 = 0. This equation must hold for all m (since the line is variable through fixed P), so comparing coefficients gives x₁ = 1 and y₁ = 1.
Hence, P = (1, 1).
Q.15. In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance

 from the given point.
Ans.

Let the intersection B be (a, b) on x + y = 4, and A = (1, 2). Slope of the line AB is m = (b - 2)/(a - 1). Given AB equals the specified distance (as shown in ). Using the distance formula between A and B and the condition a + b = 4, solving the two equations yields the coordinates of B () and hence the slope m (), giving the required direction of the line through (1, 2).
Q.16. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.

This implies that

line passes through the fixed point (k, k).]
Ans. Intercept form: x/a + y/b = 1. Given 1/a + 1/b = 1/k (constant).
Rewrite the line as kx + ky = ab·(x/a + y/b) = ab. Eliminating a and b using the given relation shows that every such line passes through (k, k). In particular, substituting x = k, y = k satisfies the family, so the fixed point is (k, k).
Q.17. Find the equation of the line which passes through the point (- 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Ans. Let intercepts be A(a, 0) and B(0, b). Point (-4, 3) divides AB internally in ratio 5:3, so:
Using section formula for division internally, coordinates equalities give 3a/(5 + 3) = -4 ⇒ 3a/8 = -4 ⇒ a = -32/3. Similarly, 5·0 + 3b over 8 matches y = 3 ⇒ b = 8.
Form the intercept equation x/a + y/b = 1 with these a, b, simplify to get 9x - 20y + 96 = 0 (after clearing denominators and signs).
Hence, the required equation is 9x - 20y + 96 = 0.
Q.18. Find the equations of the lines through the point of intersection of the lines x - y + 1 = 0 and 2x - 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5.
Ans. Solve the pair: x - y + 1 = 0 and 2x - 3y + 5 = 0. Intersection is (2, 3).
Equation of any line through (2, 3): y - 3 = m(x - 2) ⇒ mx - y + 3 - 2m = 0.
Distance from (3, 2) to this line is |m·3 - 2 + 3 - 2m|/√(m² + 1) = 7/5. Simplify numerator to |m - (-1)| etc., square both sides and solve resulting quadratic in m:
24m² - 50m + 24 = 0 ⇒ (3m - 4)(4m - 3) = 0 ⇒ m = 4/3 or m = 3/4.
Lines are:
For m = 4/3: y - 3 = (4/3)(x - 2) ⇒ 4x - 3y + 1 = 0.
For m = 3/4: y - 3 = (3/4)(x - 2) ⇒ 3x - 4y + 6 = 0.
Hence, the required equations are 4x - 3y + 1 = 0 and 3x - 4y + 6 = 0.
Q.19. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point. 
[Hint: Given that

, which gives four sides of a square.]
Ans. Let P(x, y). Distances from axes are |x| and |y|. Given |x| + |y| = 1. This describes the diamond (a square rotated 45°) with vertices (1, 0), (0, 1), (-1, 0), (0, -1). The locus is the union of the four straight line segments satisfying x + y = 1, -x - y = 1, -x + y = 1 and x - y = 1 as appropriate in each quadrant.
Hence, the locus is the boundary of that square (the diamond).
Q.20. P₁, P₂ are points on either of the two lines

 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P₁, P₂ on the bisector of the angle between the given lines. 
[Hint: Lines are y = √3 x + 2 and y = - √3 x + 2 according as x ≥ 0 or x < 0. y-axis is the bisector of the angles between the lines. P₁, P₂ are the points on these lines at a distance of 5 units from the point of intersection of these lines which have a point on y-axis as common foot of perpendiculars from these points. The y-coordinate of the foot of the perpendicular is given by 2 + 5 cos30°.]
Ans. The two lines meet at Q(0, 2). Points P₁, P₂ lie 5 units from Q along the given lines at angles 60° and 120°. The bisector is the y-axis. The foot of perpendicular from either point to the y-axis has x = 0 and y = 2 + 5 cos 30° = 2 + 5·(√3/2) = 2 + (5√3)/2.
Thus the common foot has coordinates (0, 2 + (5√3)/2).
Q.21. If p is the length of perpendicular from the origin on the line

 and a², p², b² are in A.P, then show that a⁴ + b⁴ = 0.
Ans. Given line is in the form . Perpendicular from origin to this line has length p = |...|/√(...). Carrying out the algebra (square both sides, simplify) gives an expression (i). Since a², p², b² are in A.P., 2p² = a² + b². Substitute p² into (i) and simplify to obtain (a² + b²)² = 2a²b², which leads to a⁴ + b⁴ = 0. Hence proved.
OBJECTIVE ANSWERS TYPE QUESTIONS

Q.22. A line cutting off intercept - 3 from the y-axis and the tangent at angle to the x-axis is 3/5, its equation is
(a) 5y - 3x + 15 = 0 
(b) 3y - 5x + 15 = 0 
(c) 5y - 3x - 15 = 0 
(d) None of these
Ans. (a)
Solution.
Line passes through (0, -3). Given tan θ = 3/5 ⇒ slope m = 3/5. Using point-slope form y + 3 = (3/5)(x - 0) ⇒ 5(y + 3) = 3x ⇒ 5y + 15 = 3x ⇒ 5y - 3x + 15 = 0. Hence option (a).
Q.23. Slope of a line which cuts off intercepts of equal lengths on the axes is
(a) - 1
(b) - 0
(c) 2
(d) √3
Ans. (a)
Solution.
Intercept form with a = b gives x/a + y/a = 1 ⇒ x + y = a ⇒ y = -x + a. So slope = -1. Hence (a).
Q.24. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is 
(a) x - y = 5 
(b) x + y = 5 
(c) x + y = 1 
(d) x - y = 1
Ans. (b)
Solution.
Line through (3, 2) perpendicular to y = x (slope 1) has slope -1. Equation: y - 2 = -1(x - 3) ⇒ y = -x + 5 ⇒ x + y = 5. Hence (b).
Q.25. The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is 
(a) y - x + 1  = 0 
(b) y - x - 1 = 0 
(c) y - x + 2  = 0 
(d) y - x - 2  = 0
Ans. (b)
Solution.
Line x + y + 1 = 0 has slope -1. A perpendicular line has slope 1, so general form x - y + k = 0. Passing through (1, 2): 1 - 2 + k = 0 ⇒ k = 1 ⇒ x - y + 1 = 0 ⇒ y - x - 1 = 0. Hence (b).
Q.26. The tangent of angle between the lines whose intercepts on the axes are a, - b and b, - a, respectively, is
(a)

(b)

(c)

(d) None of these
Ans. (c)
Solution.
First line: x/a + y/(-b) = 1 ⇒ bx - ay = ab. Slope m₁ = b/a. Second line: x/b + y/(-a) = 1 ⇒ ay - bx = ab ⇒ slope m₂ = a/b. Then tan θ = |(m₁ - m₂)/(1 + m₁m₂)| = |(b/a - a/b)/(1 + 1)| = |(b² - a²)/(2ab)| = |(b - a)(b + a)/(2ab)| which matches option (c) as given. Hence (c).
Q.27. If the line

passes through the points (2, -3) and (4, -5), then (a, b) is 
(a) (1, 1) 
(b) (- 1, 1) 
(c) (1, - 1) 
(d) (- 1, -1)
Ans. (d)
Solution.
Line through (2, -3) and (4, -5) has slope (-5 + 3)/(4 - 2) = -2/2 = -1. Using point-slope: y + 3 = -1(x - 2) ⇒ x + y + 1 = 0 ⇒ comparing with given form yields a = -1, b = -1. Hence (d).
Q.28. The distance of the point of intersection of the lines 2x - 3y + 5 = 0 and 3x + 4y = 0 from the line 5x - 2y = 0 is
(a)

 
(b)

(c)

(d) None of these
Ans. (a)
Solution.
Solve 3x + 4y = 0 ⇒ y = -3x/4. Substitute in 2x - 3y + 5 = 0 to find intersection x = -20/17, y = 15/17. Distance from point (-20/17, 15/17) to 5x - 2y = 0 is |5(-20/17) - 2(15/17)|/√(5² + (-2)²) = |(-100 - 30)/17|/√29 = (130/17)/√29 = (130)/(17√29) which reduces to the value in option (a). Hence (a).
Q.29. The equations of the lines which pass through the point (3, -2) and are inclined at 60° to the line √3x + y = 1 is 
(a) y + 2 = 0,  √3x - y - 2 - 3√3 = 0 
(b) x - 2 = 0,  √3x - y + 2 + 3√3 = 0 
(c) √3x - y - 2 - 3 √3 = 0 
(d) None of these
Ans. (a)
Solution.
Line √3x + y = 1 ⇒ slope m₁ = -√3. Let m be required slope. Use formula for tan of angle between lines with given acute angle 60°. Solving gives m = √3 or m = 0. Lines through (3, -2): with m = √3 ⇒ √3x - y - 2 - 3√3 = 0; with m = 0 ⇒ y + 2 = 0. Hence option (a).
Q.30. The equations of the lines passing through the point (1, 0) and at a distance

from the origin, are
(a) √3x +  y - √3 = 0, √3x -  y - √3 = 0 
(b) √3x +  y + √3 = 0, √3x -  y  + √3 = 0 
(c) x + √3y - √3 = 0, x - √3y - √3 = 0 
(d) None of these.
Ans. (a)
Solution. 
Line through (1, 0): y = m(x - 1) ⇒ mx - y - m = 0. Distance from origin is |-m|/√(m² + 1) = given value (from ); solving gives m = ±√3. Thus equations: √3x + y - √3 = 0 and √3x - y - √3 = 0. Hence (a).
Q.31. The distance between the lines y = mx + c₁ and y = mx + c₂ is
(a)
(b)
(c)
(d) 0
Ans. (b)
Solution.
Parallel lines with same slope m have perpendicular distance |c₂ - c₁|/√(1 + m²), which is option (b).
Q.32. The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by
(a)

(b)

(c)

(d)

Ans. (b)
Solution.
Line y = 3x + 4 ⇒ 3x - y + 4 = 0. Perpendicular from (2, 3) has slope -1/3 and passes through (2, 3): equation x + 3y = 11. Solve the two equations to get intersection point (X, Y) = (-1/5, 14/5) which matches option (b). Hence (b).
Q.33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be 
(a) 2x + 3y = 12 
(b) 3x + 2y = 12 
(c) 4x - 3y = 6 
(d) 5x - 2y = 10
Ans. (a)
Solution.
If intercepts are (a, 0) and (0, b), midpoint is (a/2, b/2) = (3, 2) ⇒ a = 6, b = 4. Intercept form gives x/6 + y/4 = 1 ⇒ multiply by 12: 2x + 3y = 12. Hence (a).
Q.34. Equation of the line passing through (1, 2) and parallel to the line y = 3x - 1 is 
(a) y + 2 = x  + 1 
(b) y + 2 = 3 (x  + 1) 
(c) y - 2 = 3 (x  - 1) 
(d) y - 2 = x - 1
Ans. (c)
Solution.
Parallel lines have same slope 3. Through (1, 2): y - 2 = 3(x - 1). Hence (c).
Q.35. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are 
(a) y = x, y + x = 1 
(b) y = x, x + y = 2
(c) 2y = x, y + x = 1/3
(d) y = 2x,  y + 2x = 1
Ans. (a)
Solution.
Square vertices: (0,0),(1,0),(1,1),(0,1). Diagonals are y = x and x + y = 1. Hence (a).
Q.36. For specifying a straight line, how many geometrical parameters should be known? 
(a) 1 
(b) 2 
(c) 4 
(d) 3
Ans. (b)
Solution.
Typical forms (slope-intercept, intercept form, one-point form, normal form) all involve two parameters, so two geometric parameters suffice. Hence (b).
Q.37. The point (4, 1) undergoes the following two successive transformations : 
(i) Reflection about the line y = x 
(ii) Translation through a distance 2 units along the positive x-axis 
Then the final coordinates of the point are
(a) (4, 3) 
(b) (3, 4) 
(c) (1, 4)
(d)

Ans. (b)
Solution.
Reflection of (4, 1) in y = x swaps coordinates to (1, 4). Then translate 2 units along +x to get (3, 4). Hence (b).
Q.38. A point equidistant from the lines 4x + 3y + 10 = 0,  5x - 12y + 26 = 0 and 7x + 24y - 50 = 0 is 
(a) (1, -1) 
(b) (1, 1) 
(c) (0, 0) 
(d) (0, 1)
Ans. 
Solution.
Let point be (x, y). Equate distances to first two lines: |4x + 3y + 10|/5 = |5x - 12y + 26|/13 etc. Solving the system, one finds (0, 0) satisfies equal distances to all three lines. Hence (c).
Q.39. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
(a) 1/3
(b) 2/3
(c) 1 
(d) 4/3
Ans.
Solution.
A line perpendicular to 3x + y = 3 has form x - 3y = λ. Passing through (2, 2): 2 - 6 = λ ⇒ λ = -4 ⇒ x - 3y = -4 ⇒ y = (x + 4)/3. So y-intercept is 4/3. Hence (d).
Q.40. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y - 5 = 0 is 
(a) 1 : 2 
(b) 3 : 7 
(c) 2 : 3 
(d) 2 : 5
Ans. (b)
Solution.
Distances from 3x + 4y + 2 = 0 to 3x + 4y + 5 = 0 and to 3x + 4y - 5 = 0 are |5 - 2|/5 = 3/5 and |-5 - 2|/5 = 7/5 respectively. Ratio = 3 : 7. Hence (b).
Q.41. One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is 
(a) (-1, -1) 
(b) (2, 2) 
(c) (-2, -2) 
(d) (2, -2)
[Hint: Let ABC be the equilateral triangle with vertex A (h, k) and let D (α, β) be the point on BC. Then

. Also α + β - 2 = 0 and

Ans. (c)
Solution.
Let vertex be (x₁, y₁) and midpoint D(a, b) of BC. Centroid at origin means (x₁ + 2a)/3 = 0, (y₁ + 2b)/3 = 0 ⇒ x₁ = -2a, y₁ = -2b. D lies on x + y - 2 = 0 ⇒ a + b = 2. Also AD ⟂ BC, so slopes satisfy perpendicularity; this gives a = b = 1. Hence vertex is (-2, -2). So (c).
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Q.42. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____.
Ans. Given ax + by + c = 0 and a, b, c in A.P. ⇒ a + c = 2b ⇒ a - 2b + c = 0. Comparing with ax + by + c = 0 shows that (x, y) = (1, -2) satisfies the equation for all such triples. So the line always passes through (1, -2).
Q.43. The line which cuts off equal intercept from the axes and pass through the point (1, -2) is ____.
Ans. Intercept form: x/a + y/a = 1 ⇒ x + y = a. Substitute (1, -2): 1 - 2 = a ⇒ a = -1. So line is x + y = -1 ⇒ x + y + 1 = 0.
Q.44. Equations of the lines through the point (3, 2) and making an angle of 45° with the line x - 2y = 3 are ____.
Ans. Let required line be y - 2 = m(x - 3). Given line slope is 1/2. Use formula for angle 45° between lines:
Solve for m obtaining m = 3 and m = -1/3. Thus lines:
For m = 3: 3x - y - 7 = 0. For m = -1/3: x + 3y - 9 = 0.
Hence the two equations are 3x - y - 7 = 0 and x + 3y - 9 = 0.
Q.45. The points (3, 4) and (2, - 6) are situated on the ____ of the line 3x - 4y - 8 = 0.
Ans. Evaluate 3x - 4y - 8 at (3, 4): 9 - 16 - 8 = -15 (< 0).="" at="" (2,="" -6):="" 6="" +="" 24="" -="" 8="22" (=""> 0). Signs are opposite, so the points lie on opposite sides of the line. Hence answer: opposite.
Q.46. A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x - 12y = 3. The equation of its locus is ____.
Ans. Let moving point be (x, y). Distance squared to (3, -2) is (x - 3)² + (y + 2)². Distance to the line 5x - 12y - 3 = 0 is |5x - 12y - 3|/13. Given equality (numerical): (x - 3)² + (y + 2)² = |5x - 12y - 3|/13. Removing absolute value properly and squaring both sides, simplifying gives the locus:
13x² + 13y² - 83x + 64y + 172 = 0.
Hence the locus is 13x² + 13y² - 83x + 64y + 172 = 0.
Q.47. Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is ____.
Ans. Line intersects x-axis at a = p/ sin θ and y-axis at b = p/ cos θ. Midpoint C(h, k) has h = a/2 = p/(2 sin θ), k = b/2 = p/(2 cos θ). Eliminating θ by squaring and adding:
h² + k² = p²(1/(4 sin²θ) + 1/(4 cos²θ)) leads to
4h²k² = p²(h² + k²) after simplification. Thus the locus is 4x²y² = p²(x² + y²).
STATE TRUE OR FALSE

Q.48. If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
Ans. True. An equilateral triangle with integer coordinates would force pairwise squared distances to be integers with certain irrational ratios; no non-degenerate equilateral triangle has all three vertices with integer coordinates.
Q.49. The points A (- 2, 1), B (0, 5), C (- 1, 2) are collinear.
Ans. False. Compute area of triangle using determinant formula; area ≠ 0, so points are not collinear.
Q.50. Equation of the line passing through the point (a cos³θ, a sin³θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ - y sin θ = a sin 2θ.
Ans. False. A line perpendicular to x sec θ + y cosec θ = a can be written x cosec θ - y sec θ = k. Substitute the given point and simplify to obtain x cos θ - y sin θ = a(cos²θ - sin²θ) = a cos 2θ, not a sin 2θ. Hence the statement is False.
Q.51. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y - 10 = 0 and 2x + y + 5 = 0.
Ans. True. Solve the pair to find intersection (-5/3, 25/3) (as worked out), and verify substituting into 5x + 4y gives 0; the equality holds, so True.
Q.52. The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are y - 3 = (2 ± √3 ) (x - 2).
Ans. True. If the opposite side has slope -1, the other two sides through (2, 3) make ±60° with that side leading to slopes 2 ± √3, so equations are y - 3 = (2 ± √3)(x - 2). Hence True.
Q.53. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y - 1 = 0 and 7x - 3y - 35 = 0 is equidistant from the points (0, 0) and (8, 34). 
Ans. True. The intersection of the two lines is (2, -7). Line joining (3, 5) and (2, -7) is 12x - y - 31 = 0. Compute perpendicular distances from (0, 0) and (8, 34) to this line; both distances are equal, so the statement is True.
Q.54. The line

moves in such a way that

where c is a constant.
The locus of the foot of the perpendicular from the origin on the given line is x² + y² = c².
Ans. True. If the line moves so that its normal distance from origin is constant c, then the foot of perpendicular from origin lies on the circle x² + y² = c². Squaring and adding the forms of line and its perpendicular confirms this, so True.
Q.55. The lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent if a, b, c are in G.P.
Ans. False. Solving pairwise and imposing concurrency yields 2b = a + c, which means a, b, c are in arithmetic progression (A.P.), not geometric progression (G.P.). Hence False.
Q.56. Line joining the points (3, - 4) and (- 2, 6) is perpendicular to the line joining the points (-3, 6) and (9, -18).
Ans. False. Slope of first line is (6 + 4)/(-2 - 3) = 10/-5 = -2. Slope of second line is (-18 - 6)/(9 + 3) = -24/12 = -2. Slopes are equal, so the lines are parallel, not perpendicular. Hence False.
Match the questions given under Column C₁ with their appropriate answers given under the Column C₂ 
Q.57. 

Ans.
(a) For the line x + 5y = 13, let P(x₁, y₁) satisfy x₁ + 5y₁ = 13. Distance from 12x - 5y + 26 = 0 to P is 2. Solve |12x₁ - 5y₁ + 26|/√(12² + (-5)²) = 2, using x₁ = 13 - 5y₁, leads to x₁ = 1 or -3 and corresponding y values 12 and 4. So points are (1, 12) and (-3, 4). Hence (a) ↔ (iii).
(b) For x + y = 4, set distance to 4x + 3y - 10 = 0 equal to 2 and solve with x + y = 4. This yields points (3, 1) and (-7, 11). Hence (b) ↔ (i).
(c) Given AP = PQ = QB along line from A(-2, 5) to B(3, 1). Use section and midpoint formulas to obtain P(0, 11/3) and Q(1/2, 17/6). Hence (c) ↔ (ii).
Q.58. The value of the λ, if the lines (2x + 3y + 4) + λ (6x - y + 12) = 0 are

Ans.
(a) For the line to be parallel to the y-axis its y-coefficient must be 0: 3 - λ = 0 ⇒ λ = 3. Hence (a) ↔ (iv).
(b) For perpendicularity to 7x + y - 4 = 0 (slope -7), equate slopes from combined line and solve gives λ = -17/41. Hence (b) ↔ (iii).
(c) For the line to pass through (1, 2): (2 + 6λ)(1) + (3 - λ)(2) + 4 + 12λ = 0 ⇒ λ = -3/4. Hence (c) ↔ (i).
(d) For the line to be parallel to the x-axis its x-coefficient must be 0: 2 + 6λ = 0 ⇒ λ = -1/3. Hence (d) ↔ (ii).
Q.59. The equation of the line through the intersection of the lines 2x - 3y = 0 and 4x - 5y = 2 and

Ans.
(a) Lines: 2x - 3y = 0 and 4x - 5y = 2. Family through intersection: (2x - 3y) + k(4x - 5y - 2) = 0. Passing through (2, 1) gives k = -1 and equation x - y - 1 = 0. Hence (a) ↔ (iii).
(b) Choosing k so line is perpendicular to x + 2y + 1 = 0 (slope -1/2) gives k = -2/3, leading to 2x - y = 4. Hence (b) ↔ (i).
(c) For parallelism to 3x - 4y + 5 = 0, solve to get k = 1, giving 3x - 4y - 1 = 0. Hence (c) ↔ (iv).
(d) For the required slope -1 (equally inclined to axes), choose k to get x + y - 5 = 0. Hence (d) ↔ (ii).