Coordinate Geometry
NCERT Exercise
7.1 Part 5
Question 7: Find the point on the x-axis which is equidistant from (2, –5) and (– 2, 9).
Solution: Let given coordinates are A(2, –5), BC (–2, 9) and P(x, 0) is on the x-axis.
Since, point P is equidistanct from the given coordinates,
thus AP = BP
`AP=sqrt((x_2-x_2)^2+(y_2-y_1)^2)`
Here, x1 = 2, x2 = x, y1 = -5 and y2 = 0
So, `AP=sqrt((x-2)^2+(0+5)^2)`
`=sqrt(x^2+4-4x+25)`
`=sqrt(x^2-4x+29)`
`BP= sqrt((x_2-x_2)^2+(y_2-y_1)^2)`
Here, x1 = -2, x2 = x, y1 = 9 and y2 = 0
So, `BP=sqrt((x+2)^2+(0-9)^2)`
`=sqrt(x^2+4+4x+81)`
`=sqrt(x^2+4x+85)`
Now, since `AP=BP`
So, `sqrt(x^2-4x+29)=sqrt(x^2+4x+85)`
After squaring both sides we get:
`x^2-4x+29=x^2+4x+85`
Or, `-4x+29=4x+85`
Or, `-8x+29=85`
Or, `-8x=85-29=56`
Or, `-x=56/8=7`
Or, `x=-7`
Thus, required point is (-7, 0)
Question 8: Find the value of y for the distance between the points P(2, – 3) and Q (10, y) is 10 units.
Solution: Given, P(2, –3) and Q(10, y) and QP = 10 unit
`PQ=sqrt((x_2-x_2)^2+(y_2-y_1)^2)`
Here, x1 = 2, x2 = 10, y1 = -3 and y2 = y
So, `PQ=sqrt((10-2)^2+(y+3)^2)`
Or `10=sqrt(8^2+(y+3)^2)`
Or `10=sqrt(64+y^2+9+6y)`
After squaring both sides we get;
`10^2=64+9+y^2+6y`
Or, `100=73+y^2+6y`
Or, `y^2+6y=100-73=27`
Or, `y^2+6y-27=0`
Or, `y^2+9y-3y-27=0`
Or, `y(y+9)-3(y+9)=0`
Or, `(y-3)(y+9)=0`
Thus, if `y-3=0`
Then `y=3`
And if `y+9=0`
Then `y=-9`
Thus, the required value = 3 or -9
Thus, required value = 3 or – 9