The Straight Line Class 11 OP Malhotra Exe-16J ISC Maths Ch-16 Solutions. In this article you would learn about Locus of Points of Two Lines whose Equation Involve One Parameter. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
The Straight Line Class 11 OP Malhotra Exe-16J ISC Maths Solutions Ch-16
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-16 | The Straight Line |
| Writer | O.P. Malhotra |
| Exe-16(J) | Locus of Points of Two Lines whose Equation Involve One Parameter.. |
Locus of Points of Two Lines whose Equation Involve One Parameter.
The Straight Line Class 11 OP Malhotra Exe-16J ISC Maths Ch-16 Solutions.
Que-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.
Sol: 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
(h−2k)/(2h−k) = −3/5
⇒ 5h – 10k = -6h + 3k
⇒ 11h = 13k ⇒ k/h = 11/13
Now, slope of line OP = (k−0)/(h−0) = k/h = 11/13 = m1
Also, slope of given line 39x + 33y – 580 = 0
=m2 = −39/33 = −13/11
Here m1m2 = (11/13) × (−13/11) = -1
Hence the line joining OP is ⊥ to the line 39x + 33y – 580 = 0
Que-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.
Sol: Let P(h, k) be any point on locus Then by given condition, we have
| h + 3 | = √{(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
Que-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.
Sol: 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
[{(2+4+h)/3}, {(5-11+k)/3}]
[{(6+h)/3}, {(-6+k)/3}]
Also let the centroid G of △ABC be (α, β) whose locus is to be find out.
∴ α = {(6+h)/3}
⇒ 3α – 6 = h
and β = {(−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
Que-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.
Sol: (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 ⇒ (x/4) = cos θ …(1)
and y = 4 sin θ ⇒ (y/4) = sin θ …(2)
On squaring and adding eqn. (1) and eqn. (2) ; we have
(x/4)² + (y/4)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 ⇒ x/4 = cos θ …(1)
and y = 3 sin θ ⇒ y/3 = sin θ …(2)
For locus, we eliminate θ from (1) and (2) On squaring and adding (1) and (2); we have
(x/4)² + cos² θ + sin² θ = 1
⇒ (x²/16) + (y²/9) = 1
which is the required eqn. of locus.
Que-5: Find the locus of the point of intersection of the lines x = a/m² and y = 2a/m, where m is a parameter
Sol: Given eqns. of lines are ;
x = a/m² …(1)
and y = 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 = 2a/y
∴ from (1); x = a/(2a/y)²
⇒ x = (ay²)/(4a²)
⇒ ay² = 4a²x
⇒ y² = 4ax, which is the required locus.
Que-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.
Sol: 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 = (k/2)² ⇒ h = k²/4 ⇒ k2 = 4h
Thus the required locus of S(h, k) be given by y2 = 4x
Que-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.
Sol: The eqn. of given line be
x cos α + y sin α = p …(1)
eqn. (1) meets x-axis at y = 0
∴ x = p/cosα
and eqn. (1) meets y-axis at x = 0
∴ y = p/cosα
Thus line (1) meets coordinate axes at
A [{p/cosα}, 0] and B [0, {p/cosα}]
Let P (h,k) be the mid-point of AB whose locus is to be find out.
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