# Find the points o

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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