Q. 53.6( 11 Votes )

Find the points o

Answer :

Let (x1,y1) be any point lying in the equation x+ y = 4

x1 + y1 = 4 …(i)


Distance of the point (x1,y1) from the equation 4x + 3y = 10



[given]




4x1 + 3y1 – 10 = ±5


either 4x1 + 3y1 – 10 = 5 or 4x1 + 3y1 – 10 = -5


4x1 + 3y1 = 5 + 10 or 4x1 + 3y1 = -5 + 10


4x1 + 3y1 = 15 …(ii) or 4x1 + 3y1 = 5 …(iii)


From eq. (i), we have y1 = 4 – x1 …(iv)


Putting the value of y1 in eq. (ii), we get


4x1 + 3(4 – x1) = 15


4x1 + 12 – 3x1 = 15


x1 = 15 – 12


x1 = 3


Putting the value of x1 in eq. (iv), we get


y1 = 4 – 3


y1 = 1


Putting the value of y1 = 4 – x1 in eq. (iii), we get


4x1 + 3(4 – x1) = 5


4x1 + 12 – 3x1 = 5


x1 = 5 – 12


x1 = - 7


Putting the value of x1 in eq. (iv), we get


y1 = 4 – (-7)


y1 = 4 + 7


y1 = 11


Hence, the required points on the given line are (3,1) and (-7,11)


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