Solution:

Let the required line through the point (1, 2) be inclined at an angle θ to the axis of x.

Then its equation is [x − 1] / [cosθ] = [y − 2] / [sinθ] = r …..(i)

where r is the distance of any point (x, y) on the line from the point (1, 2).

The coordinates of any point on the line (i) are (1 + r cosθ, 2 + r sinθ).

If this point is at a distance √6 / 3 form (1, 2), then r = √6 / 3.

Therefore, the point is (1 + [√6 / 3] cosθ, 2 + [√6 / 3] sinθ).

But this point lies on the line x + y = 4.

√6 / 3 (cosθ + sinθ) = 1 or

sinθ + cosθ = 3 / √6

[1 / √2] sinθ + [1 / √2] cosθ = √3 / 2, {Dividing both sides by √2}

sin (θ + 45^o) = sin60^o or sin 120^o

θ = 15^o or 75^o

Solution:

Required locus of the point (x, y) is the curve|x| + |y| = 1.

If the point lies in the first quadrant, then x > 0, y > 0 and so |x| + |y| = 1

⇒ x + y = 1, which is straight line AB.

If the point (x, y) lies in the second quadrant then x < 0, y > 0 and so |x| + |y| = 1

⇒ −x + y = 1.

Similarly, for the third and fourth quadrant, the equations are −x −y = 1 and x − y = 1.

Hence, the required locus is the curve consisting of the sides of the square.

Solution:

Let P (x₁, y₁), then the equation of the line passing through P and whose gradient is m, is y − y₁ = m (x − x₁).

Now according to the condition

[{−2m + (mx₁ − y₁)} / {√1 + m²}]+ [{2 + (mx₁ − y₁)} / {√1 + m²} + {1 − m + (mx₁ − y₁)} / {√1 + m²}=0

3 − 3m + 3mx₁ − 3y₁ = 0

⇒ y₁ − 1 = m (x₁ − 1)

Since it is a variable line, so hold for every value of m.

Therefore, y₁ = 1, x₁ = 1

⇒ P(1,1)

Let ABCD be a rectangle. Given A (1, 3) and C (5, 1). Equation B and D lie on y=2x+c We know that the intersecting point of diagonal of a rectangle is the same or at the midpoint. So, midpoint of AC is (3, 2). Hence, y = 2x + c passes through (3, 2). Therefore, c = −4.

[x + 5] / [cosθ] = [y + 4] / [sinθ] = r₁ / AB = r₂ / AC = r₃ / AD

(r₁ cosθ − 5, r₁ sinθ − 4) lies on x + 3y + 2 = 0

r₁ = 15 / [cosθ + 3 sinθ]

Similarly, 10 / AC = 2 cosθ + sinθ and 6 / AD = cosθ − sinθ

Putting in the given relation, we get (2 cosθ + 3 sinθ)² = 0

tan θ = −2 / 3

⇒ y + 4 = [−2 / 3] (x + 5)

2x + 3y + 22 = 0

Here O is the point (0, 0).

The line 2x + 3y = 12 meets the y-axis at B and so B is the point (0, 4).

The equation of any line perpendicular to the line 2x + 3y = 12 and passes through (5, 5) is 3x − 2y = 5 ……(i)

The line (i) meets the x-axis at C and so coordinates of C are (5 / 3, 0).

Similarly, the coordinates of E are (3, 2) by solving the line AB and (i).

Thus, O (0, 0), C (5 / 3, 0), E (3, 2) and B (0, 4).

Now the area of figure OCEB = area of ΔOCE + area of ΔOEB = [23 / 3] sq.units.

The equation of line passes through (3, 2) and of slope 3 / 4 is 3x − 4y − 1 = 0

Let the point be (h, k)then

3h − 4k − 1 = 0 …….(i) and

(h − 3)² + (y − 2)² = 5² (ii)

On solving the equations, we get h = −1, 7 and k = −1, 5.

Hence, points are (1, 1) and (7, 5).

Any line through (1, 10) is given by y + 10 = m (x − 1)

Since it makes equal angle say α with the given lines 7x − y + 3 = 0 and x + y − 3 = 0, therefore tan α = [m − 7] / [1 + 7m]

= [m − (−1)] / [1 + m (−1)]

⇒ m = [1 / 3] or 3

Hence, the two possible equations of the third side are 3x + y + 7 = 0 and x − 3y − 31 = 0.

The given lines are ± x ± y = 1 i.e. x + y = 1, x − y = 1, x + y = −1 and x − y = −1.

These lines form a quadrilateral whose vertices are A (−1, 0), B (0, −1), C (1, 0) and D (0, 1) Obviously ABCD is a square.

Length of each side of this square is √1² + 1² = √2

Hence, the area of square is √2 * √2 = 2 sq.units

Trick: Required area = 2c² / |ab| = [2 * 1²] / [|1 * 1|] = 2.

Solution:

The equation of any straight line passing through (3, 2) is y + 2 = m (x − 3) …….(i) The slope of the given line is −√3.

So, tan 60^o = ± [m − (−√3)] / [1 + m (−√3)]

On solving, we get m = 0 or √3

Putting the values of m in (i), the required equation of lines are

y + 2 = 0 and √3x − y = 2 + 3√3.