OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Ex 16(i)

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 c².
Solution:
Let (h, k) be any point on the locus. then by given condition, | k | = c²
∴ Locus of point (h, k) be given by | y | = c²

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
h² + k² = 25
∴ Locus of point (h, k) be given by
x² + y² = 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
PA² + PB² = 30
⇒ (h – 2)² + (k – 4)² + (h + 3)² + (k + 1)² = 30
⇒ 2h² + 2k² + 2h – 6k = 0
⇒ h² + k² + h – 3k = 0
Thus the locus of point (h, k) be given by
x² + y² + 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
h² = (h – 1)² + (k – 1)²
⇒ h² = h² – 2h = 1 + k² – 2k + 1
⇒ k² – 2k – 2h + 2 = 0
Thus the required locus of P(h, k) be given by y² – 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 |
⇒ PA² = PB²
(h – 1)² + (k – 0)² = (h + 1)² + (k – 0)²
⇒ h² – 2h + 1 = h² + 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 P² = PB² = 10. Find the locus of P.
Solution:
Let P(h, k) be the required locus.
given PA² + PB² = 10
∴ (h – 2)² + (k – 0)² + (h – 4)² + (k – 0)² = 10
⇒ h² – 4h + 4 + k² + h² – 8k + 16 + k² = 10
⇒ 2h² + 2k² – 4h – 8k + 10 = 0
⇒ h² + k² – 2h – 4k + 5 = 0
Thus required locus of P(h, k) be given by
x² +y² – 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[h² + (k – 2)²] = (9 – 2k)²
⇒ 9[h² + k² – 4k + 4] = 81 + 4k² – 36k
⇒ 9h² + 5k² – 45 = 0
Thus the required locus of P(h, k) be given by 9x² + 5y² – 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
PA² + PB² = AB² [using pythagoras theorem]
(h + 5)² + (k – 1)² + (h – 3)² + (k – 2)² = (3 + 5)² + (2 – 1)²
⇒ 2h² + 2k² + 4h – 6k – 26 = 0
⇒ h² + k² + 2h – 3k – 13 = 0
Thus the required locus of P(h, k) be given by x² + y² + 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
x² + y² – 4x – 6y + 11 = 0 .
Solution:
eqn. of given locus be
x² + y² – 4x – 6y + 11 = 0 …(1)
putting x =1 and y = 2 in L.H.S of eqn. (1)
L.H.S. =1² + 2² – 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
3x² + y² – 4x + 7 = 0?
Solution:
eqn. of given locus be,
3x² + y² – 4x + 7 = 0 …(1)
putting x = 3 and y = 0 in L.H.S of eqn. (1).
L.H.S =3 × 3² + 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 x² + y² + 5x + 11y – 2 = 0.
Solution:
eqn. of given curve be
x² + y² + 5x + 11y – 2 = 0
The point (h, k) lies on given curve (1)
∴ h² + k² + 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 | ⇒ PA² = PB²
⇒ (h + 1)² + (k – 0)² = (h – 1)² + (k – 1)²
⇒ h² + 2h + 1 + k² = h² – 2h + k² – 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 PS² 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 = y² + 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).
∴ PS² = (x – 4)² + (y – 0)² = (x – 4)² + y²
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 ⇒ PS² = PM²
(x – 4)² + y² = (x – 0)² + (y – y)²
⇒ x² – 8x + 16 + y² = x²
⇒ y² – 8x + 16 = 0
which is the required locus.
Let Q(h, k) be any point on curve (1)
∴ k² – 8h + 16 = 0 …(1)
Also by given condition QS = 20
⇒ QS² = 400
⇒ (h – 4)² + (k – 0)² = 400
⇒ h² – 8h + 16 + 8h – 16 = 400 [using eqn. (2)]
⇒ h² = 400
⇒ h = ± 20
When h = 20 ∴ from (2); we have
k² – 144 = 0 ⇒ k = ± 12
When h = – 20 ∴ from (2); k² + 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 x² + y² = 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 x² + y² = 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
⇒ AB² = 36
⇒ (t – 0)² + (0 – k)² = 36
⇒ t² + k² = 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
OA² + OB² = AB²
⇒ a² + b² = l²
⇒ (2h)² + (2k)² = l²
⇒ 4h² + 4k² = l²
Thus the locus of P(h, k) be given by
4x² + 4y² = l².

Question 20.
If O is the origin and Q is a variable, point on x² = 4y, find the locus of the mid-point of OQ.
Solution:
eqn. of given curve be
x² = 4y …(1)
Let Q(h, k) be any point on curve.
∴ h² = 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) ⇒ α² = 2β
Thus locus of (α, β) be given by x² = 2y