The availability of Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(i) encourages students to tackle difficult exercises.

## S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(i)

Question 1.

Its distance from the x-axis is c^{2}.

Solution:

Let (h, k) be any point on the locus. then by given condition, | k | = c^{2}

∴ Locus of point (h, k) be given by | y | = c^{2}

Question 2.

It distance from the origin is 5 .

Solution:

Let (h, k) be any point on the locus Then according to given condition, we have

\(\sqrt{(h-0)^2+(k-0)^2}\) = 5

On squaring both sides; we have

h^{2} + k^{2} = 25

∴ Locus of point (h, k) be given by

x^{2} + y^{2} = 25

Question 3.

The sum of the squares of its distances from the points (2, 4) and (- 3, – 1) is 30 .

Solution:

Let P(h, k) be any point on the locus and given points are A(2, 4) and B(- 3, – 1)

Then by given condition, we have

PA^{2} + PB^{2} = 30

⇒ (h – 2)^{2} + (k – 4)^{2} + (h + 3)^{2} + (k + 1)^{2} = 30

⇒ 2h^{2} + 2k^{2} + 2h – 6k = 0

⇒ h^{2} + k^{2} + h – 3k = 0

Thus the locus of point (h, k) be given by

x^{2} + y^{2} + x – 3y = 0

Question 4.

Its distance from the x-axis is half its distance from the y-axis.

Solution:

Let P(h, k) be any point on the locus and equations of x-axis and y-axis are given by

y = 0 and x = 0 .

Then by given condition, we have

| k | = \(\frac { 1 }{ 2 }\) | h |

Thus required locus of point P(h, k) be given by | y | = \(\frac { 1 }{ 2 }\) | x | ⇒ 2y = ± x

Question 5.

Its distance from the y-axis is equal to its distance from the point (1, 1).

Solution:

Let P(h, k) be any point on the locus and given point be A(1, 1).

eqn. of y-axis be x = 0

∴ distance of P(h, k) from y-axis = | h |

distance of P(h, k) from A(1, 1)

\(=\sqrt{(h-1)^2+(k-1)^2}\)

Then according to given condition, we have

| h | = \(\sqrt{(h-1)^2+(k-1)^2}\);

On squaring both sides ; we have

h^{2} = (h – 1)^{2} + (k – 1)^{2}

⇒ h^{2} = h^{2} – 2h = 1 + k^{2} – 2k + 1

⇒ k^{2} – 2k – 2h + 2 = 0

Thus the required locus of P(h, k) be given by y^{2} – 2y – 2x + 2 = 0

Question 6.

Find the locus of a point which is equidistant from the points (1, 0) and (- 1, 0).

Solution:

Let P(h, k) be any point on the locus and given points are A(1, 0) and B(- 1, 0).

Then by given condition, we have

| PA| = | PB |

⇒ PA^{2} = PB^{2}

(h – 1)^{2} + (k – 0)^{2} = (h + 1)^{2} + (k – 0)^{2}

⇒ h^{2} – 2h + 1 = h^{2} + 2h = 1

⇒ 4h = 0 ⇒ h = 0

Hence the required locus of P(h, k) be given by x = 0

Question 7.

A(2, 0) and B(4, 0) are two given points. A point P moves so that P^{2} = PB^{2} = 10. Find the locus of P.

Solution:

Let P(h, k) be the required locus.

given PA^{2} + PB^{2} = 10

∴ (h – 2)^{2} + (k – 0)^{2} + (h – 4)^{2} + (k – 0)^{2} = 10

⇒ h^{2} – 4h + 4 + k^{2} + h^{2} – 8k + 16 + k^{2} = 10

⇒ 2h^{2} + 2k^{2} – 4h – 8k + 10 = 0

⇒ h^{2} + k^{2} – 2h – 4k + 5 = 0

Thus required locus of P(h, k) be given by

x^{2} +y^{2} – 2x – 4y + 5 = 0

Question 8.

Find the locus of a point such that the sum of its distances from the points (0, 2) and (0, – 2) is 6 .

Solution:

Let P(h, k) be any point on the locus and the given points are A (0, 2) and B(0, – 2).

Then by given condition, we have

On squaring both sides; we have

9[h^{2} + (k – 2)^{2}] = (9 – 2k)^{2}

⇒ 9[h^{2} + k^{2} – 4k + 4] = 81 + 4k^{2} – 36k

⇒ 9h^{2} + 5k^{2} – 45 = 0

Thus the required locus of P(h, k) be given by 9x^{2} + 5y^{2} – 45 = 0

Question 9.

Find the locus of a point, so that the join of points (- 5, 1) and (3, 2) subtends a right angle at the moving point.

Solution:

Let P(h, k) be any point on the locus.

Then by given condition, we have

PA^{2} + PB^{2} = AB^{2} [using pythagoras theorem]

(h + 5)^{2} + (k – 1)^{2} + (h – 3)^{2} + (k – 2)^{2} = (3 + 5)^{2} + (2 – 1)^{2}

⇒ 2h^{2} + 2k^{2} + 4h – 6k – 26 = 0

⇒ h^{2} + k^{2} + 2h – 3k – 13 = 0

Thus the required locus of P(h, k) be given by x^{2} + y^{2} + 2x – 3y – 13 = 0

Question 10.

Two points A and B with co-ordinates (5, 3) and (3, – 2) re given. A point P moves so that the area of △PAB is constant and equal to 9 square units. Find the equation to the locus of the point P.

Solution:

Let P(h, k) be any point on the locus and given points are A(5, 3) and B(3, – 2)

∴ area of △PAB = \(\frac { 1 }{ 2 }\) | (3h – 5k) + (- 10 – 9) + 3k + 2h |

= \(\frac { 1 }{ 2 }\) | 5h – 2k – 9 |

Then by given condition, we have

area of △PAB = 9

\(\frac { 1 }{ 2 }\) | 5h – 2k – 19 | = 9

⇒ 5h – 2k – 19 = ± 18

⇒ 5h – 2k – 1 = 0 and 5h – 2k – 37 = 0

Hence, the locus of P(h, k) be given by

3x – 2y – 1 = 0 and 5x – 2y – 37 = 0

Question 11.

Show that (1, 2) lies on the locus

x^{2} + y^{2} – 4x – 6y + 11 = 0 .

Solution:

eqn. of given locus be

x^{2} + y^{2} – 4x – 6y + 11 = 0 …(1)

putting x =1 and y = 2 in L.H.S of eqn. (1)

L.H.S. =1^{2} + 2^{2} – 4 – 12 + 11 = 0 = R.H.S

Thus, the point (1, 2) lies on the locus.

Question 12.

Does the point (3, 0) lie on the curve

3x^{2} + y^{2} – 4x + 7 = 0?

Solution:

eqn. of given locus be,

3x^{2} + y^{2} – 4x + 7 = 0 …(1)

putting x = 3 and y = 0 in L.H.S of eqn. (1).

L.H.S =3 × 3^{2} + 0 – 4 × 3 + 7

= 22 ≠ 0 = R.H.S

Thus point (3, 0) does not lies on given locus.

Question 13.

Find the condition that the point (h, k) may lie on the curve x^{2} + y^{2} + 5x + 11y – 2 = 0.

Solution:

eqn. of given curve be

x^{2} + y^{2} + 5x + 11y – 2 = 0

The point (h, k) lies on given curve (1)

∴ h^{2} + k^{2} + 5h + 11k – 2 = 0

which is the required condition.

Question 14.

If the line (2 + k) x -(2 – k) y +(4k + 14) = 0 passes through the point (- 1, 21), find k.

Solution:

eqn. of given line be

(2 + k) x – (2 – k) y + 4k + 14 = 0 …(1)

eqn. (1) passes through the point (- 1, 21).

(2 + k)(- 1) – (2 – k) 21 + 4k + 14 = 0

⇒ – 2 – k – 42 + 21k + 4k + 14 = 0

⇒ 24k – 30 = 0 ⇒ k = \(\frac { 30 }{ 24 }\) = \(\frac { 5 }{ 4 }\)

Question 15.

A is the point (- 1, 0) and B is the point (1, 1). Find a point on the line 4x + 5y = 4, which is equidistant from A and B.

Solution:

Let P(h, k) be any point on line 4x + 5y = 4

∴ 4h + 5k = 4

Also by given condition, we have

| PA | = | PB | ⇒ PA^{2} = PB^{2}

⇒ (h + 1)^{2} + (k – 0)^{2} = (h – 1)^{2} + (k – 1)^{2}

⇒ h^{2} + 2h + 1 + k^{2} = h^{2} – 2h + k^{2} – 2k + 2

⇒ 4h + 2k – 1 = 0 …(2)

eqn. (1) – eqn. (2) gives;

3k = 3 ⇒ k = 1

∴ from (1); 4h + 5 = 4 ⇒ 4h = – 1

⇒ h = –\(\frac { 1 }{ 4 }\)

Hence the coordinates of required point be \(\left(-\frac{1}{4}, 1\right)\).

Question 16.

The co-ordinates of the point S are (4, 0) and a point P has coordinates (x, y). Express PS^{2} in terms of x and y. Given that M is the foot of the perpendicular from P to the y-axis and that the point P moves so that the lengths PS and PM are equal, prove that the locus of P is 8 x = y^{2} + 16. Find the co-ordinates of one of the two points on the curve whose distance from S is 20 units.

Solution:

Given coordinates of point S are (4, 0) and point P are (x, y).

∴ PS^{2} = (x – 4)^{2} + (y – 0)^{2} = (x – 4)^{2} + y^{2}

eqn. of y-axis be x = 0

Since M be the foot of ⊥ from P(x, y) on y-axis. ∴ Coordinates of M are (0, y).

Then by given condition, we have

PS = PM ⇒ PS^{2} = PM^{2}

(x – 4)^{2} + y^{2} = (x – 0)^{2} + (y – y)^{2}

⇒ x^{2} – 8x + 16 + y^{2} = x^{2}

⇒ y^{2} – 8x + 16 = 0

which is the required locus.

Let Q(h, k) be any point on curve (1)

∴ k^{2} – 8h + 16 = 0 …(1)

Also by given condition QS = 20

⇒ QS^{2} = 400

⇒ (h – 4)^{2} + (k – 0)^{2} = 400

⇒ h^{2} – 8h + 16 + 8h – 16 = 400 [using eqn. (2)]

⇒ h^{2} = 400

⇒ h = ± 20

When h = 20 ∴ from (2); we have

k^{2} – 144 = 0 ⇒ k = ± 12

When h = – 20 ∴ from (2); k^{2} + 176 = 0

which does not gives real values of k.

Thus required points on curve be (20, ± 12).

Question 17.

Find the ratio in which the line joining the points (6, 12) and (4, 9) is divided by the curve x^{2} + y^{2} = 4.

Solution:

Let the point P divides the line segment joining the points A(6, 12) and B(4, 9) in the ratio k : 1

Then by section formula, coordinates of P are \(\left(\frac{4 k+6}{k+1}, \frac{9 k+12}{k+1}\right) \text {. }\) Further point P lies on given curve x^{2} + y^{2} = 4.

Thus the required ratio be 4 : 3 and 44 : 31 externally

i.e. the curve divides the line segment externally in the ratio 4 : 3 and 44 : 31.

Question 18.

AB is a line of fixed length, 6 units, joining the points A(t, 0) and B which lies on the positive y-axis. P is a point on AB distant 2 units from A. Express the co-ordinates of B and P in terms of t. Find the locus of P as t varies.

Solution:

Let the coordinates of B are (0, k) and given coordinates of A are (t, 0).

since | AB | = 6

⇒ AB^{2} = 36

⇒ (t – 0)^{2} + (0 – k)^{2} = 36

⇒ t^{2} + k^{2} = 36

⇒ k = \(\pm \sqrt{36-t^2}\)

But B lies on +ve y-axis.

∴ Coordinates of B are \(\left(0,+\sqrt{36-t^2}\right)\)

Thus P(α, β) divides the line segment AB in the ratio 4 : 2 i.e. 2 : 1.

Then by section formula, the coordinates of P are \(\left(\frac{22}{3}, \frac{\sqrt{36-t^2}}{3}\right)\)

Question 19.

A rod of length l slides with its ends on two perpendicular lines. Find the locus of its mid-point.

Solution:

Let AB be a rod of length l and suppose it intersects on x-axis at A(a, 0) and y-axis at (0, b).

Let its mid-point whose locus is to be find out be P(h, k).

∴ h = \(\frac { a }{ 2 }\), k =\(\frac { b }{ 2 }\)

Also In right angled △AOB, we have

OA^{2} + OB^{2} = AB^{2}

⇒ a^{2} + b^{2} = l^{2}

⇒ (2h)^{2} + (2k)^{2} = l^{2}

⇒ 4h^{2} + 4k^{2} = l^{2}

Thus the locus of P(h, k) be given by

4x^{2} + 4y^{2} = l^{2}.

Question 20.

If O is the origin and Q is a variable, point on x^{2} = 4y, find the locus of the mid-point of OQ.

Solution:

eqn. of given curve be

x^{2} = 4y …(1)

Let Q(h, k) be any point on curve.

∴ h^{2} = 4k …(2)

Now Mid-point of OQ be \(\left(\frac{h}{2}, \frac{k}{2}\right)\)

Also mid-point of OQ be (α, β), whose locus is to be find out.

∴ α = \(\frac { h }{ 2 }\) ⇒ h = 2α

and β = \(\frac { k }{ 2 }\) ⇒ k = 2β

∴ from (2) ; (2α)^{2} = 4(2β) ⇒ α^{2} = 2β

Thus locus of (α, β) be given by x^{2} = 2y