Coordinate Geometry
Mathematics Class Tenth
Solution of NCERT exercise 7.1 part3
Question (7) Find the point on the x–axis which is equidistant from (2, –5) and (–2, 9).
Solution
Given, points are (2, –5) and (–2, 9)
To find a point on the x–axis which is equidistant from the given points.
Let the given points are A(2, –5) and B(–2, 9)
Let the equidistant point on x–axis is P(x, 0)
Now, we know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between A(2, –5) and P(x, 0)
Here we have, x1 = 2, y1 = -5
And, x2 = x, and y2 = 0
Thus, by using distance formula, we have
AP `=sqrt((x-2)^2+(0-(-5))^2)`
`=sqrt((x-2)^2+(0+5)^2)`
`=sqrt((x^2+4-4x)+25)`
`=sqrt((x^2-4x+4+25)`
⇒ AP `=sqrt(x^2-4x+29)` - - - - - (i)
(b) Calculation of distance between B(–2, 9) and P(x, 0)
Here we have, x1 = –2, y1 = 9
And, x2 = x, and y2 = 0
Thus, by using distance formula we have
BP `=sqrt((x-(-2))^2+(0-9)^2)`
`=sqrt((x+2)^2+(-9)^2)`
`=sqrt(x^2+4+4x+81)`
⇒ BP `=sqrt(x^2+4x+85)` - - - - (ii)
Since, the point P(x, 0) is equidistant from the given points A, and B
Thus, AP = BP
Thus, by equation (i) and (ii), we have
`sqrt(x^2-4x+29)` `=sqrt(x^2+4x+85)`
After squaring both sides, we get
x2 – 4 x + 29 = x2 + 4 x + 85
⇒ x2 – 4 + 29 – (x2 + 4 x + 85) = 0
⇒ x2 – 4 + 29 – x2 – 4 x – 85 =0
⇒ –8 x – 56 = 0
⇒ – 8 x = 56
`=> x = 56/(-8)`
⇒ x = –7
Thus, after substituting the value of `x` in P(x, 0)
We have, p(–7, 0)
Thus, the point on x–axis which is equidistant from given point is (–7, 0) Answer
Question (8) Find the value of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units.
Solution
Given, points are P(2, –3) and Q(10, y)
And, PQ = 10 unit
Thus, value of `y` = ?
Now, we know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
Here, we have, x1 = 2, y1 = –3
And, x2 = 10 and y2 = y
Thus, using distance formula we have,
PQ `=sqrt((10-2)^2+(y-(-3))^2)`
`=>10=sqrt(8^2+(y+3)^2)`
`=>10=sqrt(64+y^2+9+6y)`
`=>10=sqrt(y^2+6y+73)`
After squaring both sides, we get
100 = y2 + 6 y + 73
⇒ y2 + 6 y + 73 – 100 = 0
⇒ y2 + 6 y – 27 = 0
⇒ y2 + 9 y – 3 y – 27 = 0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9) (y – 3) = 0
Now, if y + 9 = 0
Therefore, y = –9
And, if y – 3 = 0
Therefore, y = 3
Thus, y = 3 or – 9 Answer
Question (9) If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the value of x. Also find the distance QR and PR.
Solution
Given, points are P(5, –3) and R(x, 6)
And Q(0, 1) is equidistant from P(5, –3) and R(x, 6)
This means PQ = QR
Thus, value of x = ?
And, QR and PR = ?
We know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between P(5, –3) and Q(0, 1)
Here we have, x1 = 5, y1 = –3
And, x2 = 0 and y2 = 1
Thus, after using distance formula we have
PQ `=sqrt((0-5)^2+(1-(-3))^2)`
`=sqrt((-5)^2+(1+3)^2)`
`=sqrt(25+4^2)`
`=sqrt(25+16)`
⇒ PQ `=sqrt(41)` - - - - - (i)
(b) Calculation of distance between Q(0,1) and R(x, 6)
Here we have, x1 = 0 and y1 = 1
And, x2 = x and y2 = 6
Thus, by using distance formula we have
QR `=sqrt((x-0)^2+(6-1)^2)`
`=sqrt(x^2+5^2)`
⇒ QR `=sqrt(x^2+25)` - - - - (ii)
Now, since PQ = QR
Thus, from equation (i) and (ii), we get
`sqrt(41) = sqrt(x^2+25)`
After squaring both sides, we get
41 = x2 + 25
⇒ x2 = 41 – 25
⇒ x2 = 16
Thus, x `=sqrt(16)`
⇒ x = ±4
(c) Calculation of value of QR
Now, after substituting the value of x in equation (ii), we get
QR `=sqrt(4^2+25)`
⇒ QR `= sqrt(16+25)`
⇒ QR `= sqrt(41)` - - - - - (iii)
(d) Calculation of distance between P(5, –3) and R(x, 6)
Here R = (±4, 6) [∵ x = ±4 as calculated above]
Thus, here we have, x1 = 5, y1 = –3
And, x2 = 4 and y2 = 6
Calculation of PR when x = 4
Thus, by using distance formula we get
PR `=sqrt((4-5)^2 + (6-(-3))^2)`
`=sqrt((-1)^2+(6+3)^2)`
`=sqrt(1+9^2)`
`=sqrt(1+81)`
⇒ PR `=sqrt(82)`
Calculation of PR, when x = –4
Thus, by using distance formula we get
PR `=sqrt((-4-5)^2 + (6-(-3))^2)`
`=sqrt((-9)^2+(6+3)^2)`
`=sqrt(81+9^2)`
`=sqrt(81+81)`
`=sqrt(81xx2)`
⇒ PR `= 9sqrt2`
Thus, PR `=sqrt(82)` or `9sqrt2`
Thus, value of x = 4, QR `=sqrt(41)` and PR `= sqrt(82)` or `9sqrt2` Answer
Question (10) Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).
Solution
Given, points are (3, 6) and (–3, 4) which is equidistant from (x, y)
Thus, relation between x, and y = ?
Let the given points are P(3, 6) and Q(–3, 4)
And the equidistant point is R(x, y)
This, means that, PR = PQ
We know that, according to distance formula, in a plane the distance between two points
PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
(a) Calculation of distance between P(3, 6) and R(x, y)
Here, we have, x1 = 3 and y1 = 6
And, x2 = x and y2 = y
Thus, by using distance formula we get
PR`=sqrt((x-3)^2+(y-6)^2`
`=sqrt((x^2+9-6x)+(y^2+36-12y))`
`=sqrt(x^2+9-6x+y^2+36-12y)`
⇒ PR `=sqrt(x^2+y^2-6x-12y+45)` - - - - (i)
(b) Calculation of distance between Q(–3, 4) and R(x, y)
Here we have, x1 = –3, y1 = 4
And, x2 = x, and y2 = y
Thus, by using distance formula, we get
QR `=sqrt((x-(-3))^2+(y-4)^2)`
`=sqrt((x+3)^2+ y^2+16-8y)`
`=sqrt(x^2+9+6x+y^2+16-8y)`
⇒ QR `=sqrt(x^2+y^2+6x-8y+25)` - - - - - (ii)
Now, since PR = QR (as per question)
Thus, by equation (i) and (ii), we get
`sqrt(x^2+y^2-6x-12y+45)` `=sqrt(x^2+y^2+6x-8y+25)`
After squaring on both sides we get,
x2 + y2 – 6x – 12y + 45 = x2 + y2 + 6x – 8y + 25
⇒ x2 + y2 – 6x – 12y + 45 – (x2 + y2 + 6x – 8y + 25) = 0
⇒ x2 + y2 – x2 – y2 – 6x – 6x – 12y + 8y + 45 – 25 = 0
⇒ – 12x – 4 y + 20 = 0
⇒ – 4(3x + y – 5) = 0
⇒ 3x + y – 5 = 0
⇒ 3x + y = 5
Thus, relation between x and y is 3x + y – 5 = 0 OR 3x + y = 5 Answer
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