# Class 11 RD Sharma Solutions – Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates- Exercise 22.2

**Question 1. Find the locus of a point equidistant from the point (2, 4) and the y-axis.**

**Solution:**

Let C (a, b) be any point on the locus and let A (2, 4) and B (0, b). We are given,

=> CA = CB

=> CA

^{2}= CB^{2}Using distance formula, we get,

=> (a − 2)

^{2}+ (b − 4)^{2}= (a − 0)^{2}+ (b − b)^{2}=> a

^{2}+ 4 − 4a + b^{2}+ 16 − 8b = a^{2}=> b

^{2}− 4a − 8b + 20 = 0Replacing (a, b) with (x, y), we get the locus of our point,

=> y

^{2}− 4x − 8y + 20 = 0

Therefore, locus of the point is y^{2}− 4x − 8y + 20 = 0.

**Question 2. Find the equation of the locus of a point **that** moves such that the ratio of its distance from (2, 0) and (1, 3) is 5:4.**

**Solution:**

Let C (a, b) be any point on the locus and let A (2, 0) and B (1, 3). We are given,

=> CA/CB = 5/4

=> CA

^{2}/CB^{2}= 25/16Using distance formula, we get,

=>

=>

=> 16 (a

^{2 }+ 4 − 4a + b^{2}) = 25 (a^{2}+ 1 − 2a + b^{2}+ 9 − 6b)=> 9a

^{2}+ 9b^{2}+ 14a − 150b + 186 = 0Replacing (a, b) with (x, y), we get the equation of the locus of our point,

=> 9x

^{2}+ 9y^{2}+ 14x − 150y + 186 = 0

Therefore the locus of the point is 9x^{2}+ 9y^{2}+ 14x − 150y + 186 = 0.

**Question 3. A point moves as so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a, prove that the equation to its locus is x**^{2}/a^{2} − y^{2}/b^{2} = 1, where b^{2} = a^{2} (e^{2} − 1).

^{2}/a

^{2}− y

^{2}/b

^{2}= 1, where b

^{2}= a

^{2}(e

^{2}− 1).

**Solution:**

Let C (h, k) be any point on the locus and let A (ae, 0) and B (−ae, 0). We are given,

=> CA − CB = 2a

Using distance formula, we get,

=>

=>

Squaring both sides, we get,

=>

=> h

^{2}+ a^{2}e^{2}− 2aeh + k^{2}= 4a^{2}+ (h + ae)^{2}+ k^{2}+=> h

^{2}+ a^{2}e^{2}− 2aeh + k^{2}= 4a^{2}+ h^{2}+ a^{2}e^{2}+ 2aeh + k^{2}+=> −4aeh − 4a

^{2}=Squaring both sides again, we get,

=> −(eh + a) = (h + ae)

^{2}+ k^{2}=> e

^{2}h^{2}+ a^{2}+ 2aeh = h^{2 }+ a^{2}e^{2}+ 2aeh + k^{2}=> h

^{2}(e^{2}– 1) – k^{2}= a^{2}(e^{2}– 1)=>

As we are given, b

^{2}= a^{2}(e^{2}− 1), we get,=> h

^{2}/a^{2}− k^{2}/b^{2}= 1Replacing (h, k) with (x, y), we get the equation of the locus of our point,

=> x

^{2}/a^{2}− y^{2}/b^{2}= 1

Hence proved.

**Question 4. Find the locus of a point such that the sum of its distances from (0, 2) and (0, −2) is 6.**

**Solution:**

Let C (a, b) be any point on the locus and let A (0, 2) and B (0, −2). We are given,

=> CA + CB = 6

Using distance formula, we get,

=>

=>

Squaring both sides, we get,

=> a

^{2 }+ b^{2}+ 4 − 4b= 36 + a^{2}+ b^{2}+ 4 + 4b −=> −8b − 36 =

=> −4 (2b + 9) =

Squaring both sides again, we get,

=> (2b + 9)

^{2}==> 4b

^{2}+ 81 + 36b = 9a^{2}+ 9b^{2}+ 36b + 36=> 9a

^{2 }+ 5b^{2}= 45Replacing (a, b) with (x, y), we get the locus of our point,

=> 9x

^{2}+ 5y^{2 }= 45

Therefore the locus of the point is 9x^{2}+ 5y^{2}= 45.

**Question 5. Find the locus of a point which is equidistant from (1, 3) and x-axis.**

**Solution:**

Let C (a, b) be any point on the locus and let A (1, 3) and B (a, 0). We are given,

=> CA = CB

=> CA

^{2}= CB^{2}Using distance formula, we get,

=> (a − 1)

^{2 }+ (b − 3)^{2}= (a − a)^{2}+ (b − 0)^{2}=> a

^{2}+ 1 − 2a + b^{2}+ 9 − 6b = b^{2}=> a

^{2}− 2a − 6b + 10 = 0Replacing (a, b) with (x, y), we get the locus of our point,

=> x

^{2}− 2x − 6y + 10 = 0

Therefore the locus of the point is x^{2}− 2x − 6y + 10 = 0.

**Question 6. Find the locus of a point **that** moves such that its distance from the origin is three times is distance from x-axis.**

**Solution:**

Let C (a, b) be any point on the locus and let A (0, 0) and B (a, 0). We are given,

=> CA = 3 CB

=> CA

^{2}= 9 CB^{2}Using distance formula, we get,

=> (a − 0)

^{2}+ (b − 0)^{2 }= 9 [(a − a)^{2}+ (b − 0)^{2}]=> a

^{2}+ b^{2}= 9b^{2}=> a

^{2}= 8b^{2}Replacing (a, b) with (x, y), we get the locus of our point,

=> x

^{2}= 8y^{2}

Therefore the locus of the point is x^{2 }= 8y^{2}.

**Question 7. A (5, 3), B (3, −2) are two fixed points, find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 sq. units.**

**Solution:**

Let P (a, b) be any point on the locus and we have A (5, 3) and B (3, −2). We are given,

=> Area of the triangle PAB = 9

=>

=> |5(−2−b) + 3(b−3) + h(3+2)| = 18

=> |5a − 2b − 19| = 18

=> 5a − 2b − 19 = ±18

=> 5a − 2b − 37 = 0 or 5a − 2b − 1 = 0

Replacing (a, b) with (x, y), we get the locus of our point,

=> 5x − 2y − 37 = 0 or 5x − 2y − 1 = 0

Therefore the equation to the locus of the point is 5x − 2y − 37 = 0 or 5x − 2y − 1 = 0.

**Question 8. Find the locus of a point such that the line segment having **endpoints** (2, 0) and (−2, 0) subtend a right angle at that point.**

**Solution:**

Let C (a, b) be any point on the locus and let A (2, 0) and B (−2, 0).

We are given ∠ ACB = 90

^{o}=> AB

^{2}= CA^{2}+ CB^{2}Using distance formula, we get,

=> (2+2)

^{2}+ (0−0)^{2}= (a−2)^{2}+ (b−0)^{2}+ (a+2)^{2}+ (b−0)^{2}=> 16 = a

^{2}+ 4 − 4a + b^{2}+ a^{2}+ 4 + 4a + b^{2}=> 2a

^{2}+ 2b^{2}+ 8 = 16=> a

^{2}+ b^{2}= 4Replacing (a, b) with (x, y), we get the locus of our point,

=> x

^{2}+ y^{2}= 4

Therefore the locus of the point is x^{2}+ y^{2}= 4.

**Question 9. A (−1, 1), B (2, 3) are two fixed points, find the locus of a point P which moves so that the area of the triangle PAB is 8 sq. units.**

**Solution:**

Let P (a, b) be any point on the locus and we have A (−1, 1) and B (2, 3). We are given,

=> Area of the triangle PAB = 8

=>

=> |−1(3−b) + 2(b−1) + a(1−3)| = 16

=> |−2a + 3b − 5| = 16

=> −2a + 3b − 5 = ±16

=> 2a − 3b + 21 = 0 or 2a − 3b − 11 = 0

Replacing (a, b) with (x, y), we get the locus of our point,

=> 2x − 3y + 21 = 0 or 2x − 3y − 11 = 0

Therefore the locus of the point is 2x − 3y + 21 = 0 or 2x − 3y − 11 = 0.

**Question 10. A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1:2.**

**Solution:**

Let C (h, k) be any point on the locus and let AB = l (given) be the length of the rod. Suppose, coordinates of A and B are (a, 0) and (0, b) respectively.

According to the question,

=> h = 2a/3

=> a = 3h/2 . . . . (1)

And k = b/3

=> b = 3k . . . . (2)

Let the origin be O (0, 0). Now we know △ AOB is right-angled.

=> AB

^{2}= OA^{2}+ OB^{2}=> l

^{2}= [(a−0)^{2}+ (0−0)^{2}] + [(0−0)^{2}+ (b−0)^{2}]=> a

^{2}+ b^{2}= l^{2}Using (1) and (2), we get,

=> (3h/2)

^{2}+ (3k)^{2}= l^{2}=> 9h

^{2}/4 + 9k^{2}= l^{2}=> 9h

^{2}+ 36k^{2}= 4l^{2}Replacing (h, k) with (x, y), we get the locus of our point,

=> 9x

^{2}+ 36y^{2}= 4l^{2}

Therefore the locus of the point is 9x^{2}+ 36y^{2}= 4l^{2}.

**Question 11. Find the locus of the mid-point of the portion of the line x cos α + y sin α = p which is intercepted between the axes. **

**Solution:**

We are given,

=> x cos α + y sin α = p

=>

Intercepts on x-axis and y -axis are p/cos α and p/sin α respectively.

Suppose (x, y) is the mid-point of the portion of the given line which is intercepted between the axes.

=> (x, y) =

=> x = p/2 cos α and y = p/2 sin α

=> 2 cos α = p/x and 2 sin α = p/y

Squaring both sides of these, we get,

=> 4 cos

^{2}α = p^{2}/x^{2}. . . . (1)=> 4 sin

^{2}α = p^{2}/y^{2}. . . . (2)Adding (1) and (2), we get,

=> 4 cos

^{2}α + 4 sin^{2}α = p^{2}/x^{2}+ p^{2}/y^{2 }=> p

^{2}/x^{2}+ p^{2}/y^{2}= 4=> p

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

Therefore the locus of the mid-point is p^{2}(x^{2}+ y^{2}) = 4x^{2}y^{2}.

**Question 12. If O is the origin and Q is the variable point on y**^{2} = x. Find the locus of the mid-point of OQ.

^{2}= x. Find the locus of the mid-point of OQ.

**Solution:**

Let P (h, k) be the point on the locus and let Q (a, b).

According to the question,

=> h = (a+0)/2 and k = (b+0)/2

=> h = a/2 and k = b/2

=> a = 2h and b = 2k

As point Q lies on y

^{2}= x, we get,=> (2k)

^{2}= 2h=> 4k

^{2}= 2h=> 2k

^{2}= hReplacing (h, k) with (x, y), we get the locus of our point,

=> 2y

^{2}= x

Therefore the locus of the mid-point is 2y^{2}= x.

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