Home
Class 12
MATHS
Find the least value of (x-1)^(2) + (y-2...

Find the least value of `(x-1)^(2) + (y-2)^(2)` under the condition 3x+4y -2 = 0.

Text Solution

Verified by Experts

`(x-1)^(2) + (y-2)^(2)` is square of distances between the points A(1,2) and P(x,y), where P lies on the line 3x +4y-2=0.
Least value of `(x-1)^(2) + (y-2)^(2)`
=Square of shortest distance of point (1,2) from the line 3x + 4y-2=0
`=((|3(1)+4(2)-2|)/(sqrt(3^(2) +4^(2))))^(2) = (9)/(25)`
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    CENGAGE PUBLICATION|Exercise EXAMPLE|12 Videos

  • STRAIGHT LINES

    CENGAGE PUBLICATION|Exercise CONCEPT APPLICATION EXERCISE 2.1|23 Videos

  • STRAIGHT LINE

    CENGAGE PUBLICATION|Exercise Multiple Correct Answers Type|8 Videos

  • THEORY OF EQUATIONS

    CENGAGE PUBLICATION|Exercise JEE ADVANCED (Numerical Value Type )|1 Videos

Similar Questions

Explore conceptually related problems

Find the values of x and y if x^(2)+y^(2)-2x+4y=-5 .

The least value of 2x^(2)+y^(2)+2xy+2x-3y+8 for real numbers x and y is -

Find the least positive values of x and y so that 2(sinx + siny)-2cos(x-y)=3 .

Find the least positive values of x and y: sin(x-y)=1/2, cos(x+y)=1/2 .

The least value of 2x^2+y^2+2xy+2x-3y+8 for realnumbers x and y is

If (3x-5y)/(3x+5y) = 1/2 . Then find the value of (3x^(2)-5y^(2))/(3x^(2) + 5y^(2)) .

The equation of the circle which has normals (x-1).(y-2)=0 and a tangent 3x+4y=6 is (a) x^2+y^2-2x-4y+4=0 (b) x^2+y^2-2x-4y+5=0 (c) x^2+y^2=5 (d) (x-3)^2+(y-4)^2=5

If y=3^(x-1)+3^(-x-1) , then the least value of y is (a) 2 (b) 6 (c) 2//3 (d) 3//2

If (x+1, y-2)=(3,1), find the values of x and y.

(a) Draw the graph of the equation 2y-3x=7 Also find the value of y when x=2(1)/(2) and the value of x when y=3(1)/(2) from the graph.