Class 10 Maths

# Coordinate Geometry

## NCERT Exercise

### 7.1 Part 6

Question 9: If Q (0, 1) is equidistant from P(5, – 3) and R (x, 6), find the value of x. Also find the distances QR and PR.

Answer: Given, Q(0, 1), P(5, –3) and R(x, 6) and QP = QR

QP=sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = 0, x2 = 5, y1 = 1 and y2 = -3

So, QP=sqrt((5-0)^2+(-3-1)^2)

Or =sqrt(5^2+(-4)^2)

Or =sqrt(25+16)=sqrt(41)

QR= sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = 0, x2 = x, y1 =  and y2 = 6

So, QR=sqrt((x-0)^2+(6-1)^2)

=sqrt(x^2+5^2)

=sqrt(x^2+25) -----------(1)

Now, since QP=QR

So, sqrt(41)=sqrt(x^2+25)

After squaring both sides we get:

41=x^2+25

Or, x^2=41-25=16

Or, x=sqrt(16)=4

Now, from equation (1)

QR=sqrt(x^2+25)

If x=+4
Then, QR=sqrt(4^2+25)

Or, QR=sqrt(16+25)=sqrt(41)

If x=-4
Then, QR=sqrt((-4)^2+25)

Or, QR=sqrt(16+25)=sqrt(41)

Now, PR=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

Here, x1 = 5, x2 = x, y1 = -3 and y2 = 6

So, PR=sqrt((x-5)^2+(6+3)^2)

When, x=+4
Then, PR=sqrt((4-5)^2+(6+3)^2)

=sqrt((-1)^2+9^2)=sqrt(1+81)

Or, PR=sqrt(82)

When x=-4
Then, PR=sqrt((-4-5)^2+(6+3)^2)

=sqrt((-9)^2+9^2)=sqrt(81+81)

Or, PR=sqrt(162)=9sqrt2

Hence, the answer is x=±4, QR=sqrt(41), PR=sqrt(82) or 9sqrt2

Question 10: Find the relation between x and y such that the point (x, y) is equidistance from the point (3, 6) and (–3, 4).

Answer: Let P(x, y), A (3, 6) and B (– 3, 4)
Here AP = BP

AP=sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = 3, x2 = x, y1 = 6 and y2 = y

So, AP=sqrt((x-3)^2+(y-6)^2

=sqrt(x^2+9-6x+y^2+36-12y)

Or, AP=sqrt(x^2+y^2-6x-12y+45)

AndBP=sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = -3, x2 = x, y1 = 4 and y2 = y

So, BP=sqrt((x+3)^2+(y-4)^2))

=sqrt(x^2+9+6x+y^2+16-8y)

Or, BP=sqrt(x^2+y^2+6x-8y+25)

Now, since AP=BP

So, sqrt(x^2+y^2-6x-12y+45)=sqrt(x^2+y^2+6x-8y+25)

After squaring on both sides we have:

x^2+y^2-6x-12y+45=x^2+y^2+6x-8y+25
Or, -6x-12y+45=6x-8y+25
Or, -6x-6x-12y+8y+45-25=0
Or, -12x-4y+20=0
Or, -4(3x+y-5)=0
Or, 3x+y-5=0
Or, 3x+y=5`