Question

Find the perpendicular distance of the point (1,2) from the line 3x+4y-13=0

Answer 1

Answer:

As you mention, points (-3,-4) to the line

3x-4y-1=0,

• let x=-3 and y=-4

• So comparing the line with equation

• Ax+By+C=0

• A=3, B=-4 and C=-1

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Answer 2

The perpendicular distance between the line 3x+4y-13 and the point (1,2) is 2/5 units.

Given,

A line, 3x+4y-13 = 0

A Point, P (1,2)

To find,

The perpendicular distance of point P from the given line.

Solution,

We can easily solve this question by using the formula for distance from a line to a point.

When we put the equation in the given to the form,

Ax + By + C = 0, we get

A = 3, B = 4 and C = -13

The formula for the distance 'd' between a line and a point is,

d = $\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

where ($x_1,y_1$) are the coordinate of the given point.

so, we get

d = |3×1 + 4×2 -13|/$\sqrt{3^2+4^2}$

⇒ d = 2/5 units

∴ The perpendicular distance between the given line and point is 2/5 units.

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