find the distance of point (1,2) from the straight line with slope 5 and passing through the point of intersection of x+2y = 7 and x - 3y = 7
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Point of intersection of x + 2y = 7 and x - 3y = 7 is
x + 2y = x - 3y⇒ 5y = 0 ⇒ y = 0
Now, x + 2y = 7 ⇒ x + 2(0) = 7 ⇒ x = 7
The point of intersection is (7, 0)
slope of the line = 5
Equation of the line is (y - 0) = 5( x - 7) ⇒ y = 5x - 35 ⇒ 5x - y - 35 = 0
The distance between the point (1, 2) and the line 5x - y - 35 = 0 is
|5(1) - 2 - 35|/ sqrt ((5)^2 + (-1)^2)
|-32|/sqrt 26
32/sqrt 26
Note: As the math type is not working properly today, hence the fractions are given in this way.
x + 2y = x - 3y⇒ 5y = 0 ⇒ y = 0
Now, x + 2y = 7 ⇒ x + 2(0) = 7 ⇒ x = 7
The point of intersection is (7, 0)
slope of the line = 5
Equation of the line is (y - 0) = 5( x - 7) ⇒ y = 5x - 35 ⇒ 5x - y - 35 = 0
The distance between the point (1, 2) and the line 5x - y - 35 = 0 is
|5(1) - 2 - 35|/ sqrt ((5)^2 + (-1)^2)
|-32|/sqrt 26
32/sqrt 26
Note: As the math type is not working properly today, hence the fractions are given in this way.
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