<span lang="EN-US

Given: (x1,y1) = A(3, 5)

To find:

The distance of a point from the line parallel to another line.

Explanation:

It is given that the required line is parallel to x – 2y = 1

2y = x – 1

y

sin θ and cos θ

So, the equation of the line is

Formula Used:

x – 2y + 7 = 0

Let x – 2y + 7 = 0 intersect the line 2x + 3y = 14 at point P.

Let AP = r

Then, the coordinate of P is given by

x and y

Thus, the coordinate of P is

Clearly, P lies on the line 2x + 3y = 14

r

Hence, the distance of the point (3, 5) from the line 2x + 3y = 14 is

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

<span lang="EN-USRS Aggarwal - Mathematics

The distance betwRD Sharma - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

Find the equationRD Sharma - Mathematics

Find the equationRD Sharma - Mathematics

Area of the trianRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

A point equidistaRD Sharma - Mathematics