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

Regular engagement with Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(j) can boost students’ confidence in the subject.

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

Question 1.
The lines x – 2y + 6 = 0 and 2x – y – 10 = 0 intersect at P. Without finding the co-ordinates of P prove that the equation of the line through P and the origin of co-ordinates is perpendicular to 39x + 33y – 580 = 0.
Solution:
Let given eqns. of lines are
x – 2y + 6 = 0 …(1)
2x – y – 10 = 0 …(2)
Let the coordinates of point of intersection of lines (1) and (2) be P(h, k).
and P(h, k) lies on (1) and (2); we get
h – 2k = – 6 …(3)
and 2h – k = 10 …(4)
On dividing (3) and (4); we have
\(\frac{h-2 k}{2 h-k}\) = \(\frac { -3 }{ 5 }\)
⇒ 5h – 10k = -6h + 3k
⇒ 11h = 13k ⇒ \(\frac { k }{ h }\) = \(\frac { 11 }{ 13 }\)
Now, slope of line OP = \(\frac{k-0}{h-0}\) = \(\frac{k}{h}\) = \(\frac { 11 }{ 13 }\) = m₁
Also, slope of given line 39x + 33y – 580 = 0
=m₂ = \(\frac { -39 }{ 33 }\) = \(\frac { -13 }{ 11 }\)
Here m₁m₂ = \(\frac { 11 }{ 13 }\) × \(\left(\frac{-13}{11}\right)\) = -1
Hence the line joining OP is ⊥ to the line 39x + 33y – 580 = 0

Question 2.
A point P moves so that its distance from the line given x = – 3 is equal to its distance from the point (3, 0). Show that the locus of P is y² = 12x.
Solution:
Let P(h, k) be any point on locus Then by given condition, we have
| h + 3 | = \(\sqrt{(h-3)^2+(k-0)^2}\)
On squaring both sides; we have
(h + 3)² = (h – 3)² + k²
⇒ h² + 6h + 9 = h² – 6h + 9 + k²
⇒ 12h = k²
Thus locus of P(h, k) be given by y² = 12x

Question 3.
A (2, 5), B(4,- 11) are two fixed points and C is a point which moves on the line 3x + 4y + 5 = 0. Find the locus of the centroid of the triangle ABC.
Solution:
Let the coordinates of point C are (h, k) and also point C(h, k) lies on given line
3x + 4y + 5 = 0
∴ 3h + 4k + 5 = 0 …(1)
∴ Coordinates of centroid G are

Also let the centroid G of △ABC be (α, β) whose locus is to be find out.
∴ α = \(\frac{6+h}{3}\) ⇒ 3α – 6 = h
and β = \(\frac{-6+h}{3}\) ⇒ 3β + 6 = k
∴ from (1) ; we have
3(3α – 6) + 4(3β + 6) + 5 = 0
⇒ 9α + 12β + 11 = 0
Thus the locus of point (α, β) be given by 9x + 12y + 11 = 0

Question 4.
Find the cartesian equation of the curve whose parametric equations are :
(i) x = t, y = 3 t + 5;
(ii) x = t, y = t²;
(iii) x = 4 cos θ, y = 4 sin θ;
(iv) x = 4 cos θ; y = 3 sin θ.
Solution:
(i) Given parametric eqns. of curve are
x = t …(1)
and y = 3t + 5 …(2)
For locus, we eliminate t from eqns. (1) and (2);
y = 3x + 5 which is the required locus.

(ii) Given parametric eqns. are
x = t …(1)
y = t² …(2)
For locus, we eliminate t from (1) and (2); we have
y = x² which is the required locus.

(iii) Given parametric eqns. of curve are ;
x = 4 cos θ ⇒ \(\frac { x }{ 4 }\) = cos θ …(1)
and y = 4 sin θ ⇒ \(\frac { y }{ 4 }\) = sin θ …(2)
On squaring and adding eqn. (1) and eqn. (2) ; we have
\(\left(\frac{x}{4}\right)^2\) + \(\left(\frac{y}{4}\right)^2\) = cos² θ + sin² θ = 1
⇒ x² + y² = 16
which is the required locus.

(iv) Given parametric eqns. of curve are
x = 4 cos θ ⇒ \(\frac { x }{ 4 }\) = cos θ …(1)
and y = 3 sin θ ⇒ \(\frac { y }{ 3 }\) = sin θ …(2)
For locus, we eliminate θ from (1) and (2) On squaring and adding (1) and (2); we have
\(\left(\frac{x}{4}\right)^2\) + \( = cos² θ + sin² θ = 1
⇒ [latex]\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1
which is the required eqn. of locus.

Question 5.
Find the locus of the point of intersection of the lines x = \(\frac{a}{m^2}\) and y = \(\frac{2a}{m}\), where m is a parameter.
Solution:
Given eqns. of lines are ;
x = \(\frac{a}{m^2}\) …(1)
and y = \(\frac{2a}{m}\) …(2)
To find the locus of point of intersection of lines (1) and (2), we have to eliminate m from (1) and (2).
From (2); m = \(\frac{2a}{y}\)
∴ from (1); x = \(\frac{a}{\left(\frac{2 a}{y}\right)^2}\)
⇒ x = \(\frac{a y^2}{4 a^2}\)
⇒ ay² = 4a²x
⇒ y² = 4ax, which is the required locus.

Question 6.
Find the intersection S of the lines x – ty + t² = 0, tx + y – t³ – 2t = 0 Show that S lies on the curve whose equation is y² = 4x. Sketch this curve.
Solution:
Given eqn. of lines are
x – ty = t² = 0 …(1)
tx + y – t³ – 2t = 0 …(2)
Multiplying eqn. (2) by t + eqn. (1) ; we have
(1 + t²) x + t² – t⁴ – 2t² = 0
⇒ (1 + t²) x – t⁴ – t² = 0
⇒ (1 + t²) x = t²(t² + 1) ⇒ x = t²
∴ from (1); t² – ty + t² = 0 ⇒ ty = + 2t²
⇒ y = 2t

Thus, required point of intersection of lines (1) and (2) be (t², 2t)
Let S(h, k) be any point on the locus
∴ h = t² …(3)
and k = 2t …(4)
For locus, we eliminate t from (3) and (4) ; we have
h = \(\left(\frac{k}{2}\right)^2\) ⇒ h = \(\frac{k^2}{4}\) ⇒ k² = 4h
Thus the required locus of S(h, k) be given by y² = 4x

Question 7.
Find the locus of the middle point of the portion of the line x cos α + y sin α = p, where p is a constant, intercepted between the axes.
Solution:
The eqn. of given line be
x cos α + y sin α = p …(1)
eqn. (1) meets x-axis at y = 0
∴ x = \(\frac{p}{\cos \alpha}\)
and eqn. (1) meets y-axis at x = 0