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)`
And`BP=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`