Best Answer

By using the formula y-y1 = m(x-x1) form an equation from (2, 4) with a slope of -1/2x which is the negative reciprocal of 2x:-

y-4 = -1/2(x-2)

y = -1/2x+5 => 2y = -x+10

Then find the point of intersection of 2y = -x+10 and y = 2x+10 which works out as (-2, 6)

The perpendicular distance from (2, 4) to (-2, 6) is:-

(x1-x2)2 + (y1-y2)2 = distance2

(2-(-2))2 + (4-6)2 = 20 and the square root of this is the distance between the points.

Therefore the distance is the square root of 20 which works out as 2 times the square root of 5.

Q: How do you work out the perpendicular distance from the point 2 4 to the straight line of y equals 2x plus 10?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5

Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point of intersection: (4, 1) Distance: (7-4)2+(5-1)2 = 25 and the square root of this is the perpendicular distance which is 5 units of measurement

Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5

It the point is on the line the distance is 0. If the point is not on the line, then it is possible to draw a unique line from the point to the line which is perpendicular to the line. The distance from the point to the line is the distance along this perpendicular to the line.

Straight line equation: y = 2x+10 Perpendicular equation: y = -1/2x+5 They intersect at: (-2, 6) Length of perpendicular2: (2--2)2+(4-6)2 = 20 and the square root of this is 4.472135955 Therefore distance is 4.472135955 units in length

Related questions

Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5

Its perpendicular distance.

The perpendicular distance from (2, 4) to the equation works out as the square root of 20 or 2 times the square root of 5

Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point of intersection: (4, 1) Distance: (7-4)2+(5-1)2 = 25 and the square root of this is the perpendicular distance which is 5 units of measurement

If you mean the perpendicular distance from the coordinate of (7, 5) to the straight line 3x+4y-16 = 0 then it works out as 5 units.

Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5

It works out as: 2 times the square root of 5

It the point is on the line the distance is 0. If the point is not on the line, then it is possible to draw a unique line from the point to the line which is perpendicular to the line. The distance from the point to the line is the distance along this perpendicular to the line.

Straight line equation: y = 2x+10 Perpendicular equation: y = -1/2x+5 They intersect at: (-2, 6) Length of perpendicular2: (2--2)2+(4-6)2 = 20 and the square root of this is 4.472135955 Therefore distance is 4.472135955 units in length

Given a straight line joining the points A and B, the perpendicular bisector is a straight line that passes through the mid-point of AB and is perpendicular to AB.

The mid-point is needed when the perpendicular bisector equation of a straight line is required. The distance formula is used when the length of a line is required.

1 Equation: y = 2x+10 2 Perpendicular equation works out as: 2y = -x+10 3 Point of intersection: (-2, 6) 4 Distance is the square root of: (-2-2)2+(6-4)2 = 2 times sq rt of 5