Question

ES What are the points on the y-axis whose perpendicular distance from the
line-2-1 is 3 units?​

Answer 1

Correct Question :-

What are the points on the y-axis whose perpendicular distance from the line (x/3) - (y/4) = 1 is 3 units. ?

Solution :-

Given equation of line is (x/3) - (y/4) = 1

→ 4x - 3y = 12

→ 4x - 3y - 12 = 0

Let (0,b) is the point of the y-axis whose distance from given line is 3 unit.

When we compare equation of line with general form of the equation Ax+By+C = 0, we get

→ A = 4

→ B = -3

→ C = -12 ...

Now perpendicular distance of a line Ax+By+C = 0 from a point (x1, y1) is :-

→ D = |Ax1 + By1 + C|/√(A² + B² )

So ,, perpendicular distance of a line 4x - 3y -12 = 0 from a point (0 ,b) is 3 units .

→ 3 = |4*0 - 3*b -12|/√(4² + 3²)

→ 3 = |-3b -12|/√(16 + 9)

→ 3 = |-3(b +4) |/√25

→ 3 = |-3(b +4) |/5

→ 15 = 3| - (b+4) |

→ | - (b+4) | = 5

→ (b + 4) = 5 or, (b + 4) = (-5)

→ b = 5 - 4 = 1 .

Or,

→ b = (-5) + (-4) = (-9).

Hence, The points are (0, 1) and (0, -9).

Answer 2

Correct Question :- What are the points on the y-axis whose perpendicular distance from the line (x/3) - (y/4) = 1 is 3 units. ?

Solution :-

Given equation of line is (x/3) - (y/4) = 1

=>4x - 3y = 12

=> 4x - 3y - 12 = 0

Let (0,b) is the point of the y-axis whose distance from given line is 3 unit.

When we compare equation of line with general form of the equation Ax+By+C = 0, we get

=> A = 4

=> B = -3

=> C = -12 ...

Now perpendicular distance of a line Ax+By+C = 0 from a point (x1, y1) is :- D = |Ax1 + By1 + C|/√(A² + B² )

So , perpendicular distance of a line 4x - 3y -12 = 0 from a point (0 ,b) is 3 units .

=> 3 = |4*0 - 3*b -12|/√(4² + 3²)

=> 3 = |-3b -12|/√(16 + 9)

=> 3 = |-3(b +4) |/√25

=> 3 = |-3(b +4) |/5

=> 15 = 3| - (b+4) |

=> | - (b+4) | = 5

=> (b + 4) = 5 or, (b + 4) = (-5)

=> b = 5 - 4 = 1 .

or,

=> b = (-5) + (-4) = (-9).

Hence, The points are (0, 1) and (0, -9).