Question

Sum of the distance of a point from two perpendicular lines is 3 the area enclosed by the locus of the point is t then sum of the digits of t i​

Answer 1

Step-by-step explanation:

Let us assume that mutually perpendicular axis x and y are axis of the cartesians co-ordinate.

∴x+y=3

A straight line does not have any area enclosed.

But the area with perpendicular axis is a right angled triangle with base 3 and height 3.

∴Area=

2

3×3

=4.5sq.units

Since, locus exists in all 4 quadrants of the areas

∴4.5×4=18sq.units

Hope it's helpful

Answer 2

Answer:

Area of the square is $18 cm^{2}$  and the sum of digit of 't' is 9

Step-by-step explanation:

Explanation :

Given , sum of the distance of a point from two perpendicular lines is 3 ,

and the area enclosed by point is 't'

Let two perpendicular line be x= 0  and y= 0

If  the equation of line is ax +by + c = 0 at point ($x_{1} , y_1$) then the perpendicular distance is    d = $|\frac{ax_{1}+by_1+c }{\sqrt{a^{2} +b^{2} } } |$

Step1:

Let the distance point be (h ,k)

So  from the distance  formula we get ,

$|\frac{h}{1} | +|\frac{k}{1} | = 3$

|h|+|k | = 3

Therefore , h +k = 3 , -h +k= 3 , -h-k = 3 and h - k = 3 these are the four possible lines .

Step2:

for h + k = 3  when we put h= 0 we get k = 3 and put k=0 , we get h = 3

So the points of this line are  (0,3) and (3,0)

For -h +k = 3 ,

when we put h = 0 than k =3 and when put k = 0 than h = -3

So , the points  are (0,3) and (-3,0)

For line , -h-k=3

Put h= 0 then k =-3 and when k = 0 than h = -3

So , the points are (0,-3) and (-3,0)

Similarly , for h-k = 3

Put h = 0 we get k = -3 and when k = 0 we get h = 3

So the required points are (0,-3) and (3,0)

Now ,plot all these points on the graph and join all the points we get a square .

Step3:

Distance of points   (0,3) and (3,0) = $\sqrt{3^{2}+3^{2} } = 3\sqrt{2}$

So the length of the square is  $3\sqrt{2}$cm

Now, are of the square is (t) = $(3\sqrt{2)} ^{2} = 18 cm^{2}$

Therefore , t = 18 and the sum of digit of t = 1+8 = 9 .

Final answer :

Hence , the area of the square is $18 cm^{2}$  and the sum of digit of 't' is 9 .

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