Answer :
Given: (x1,y1) = A(3, 5), tanθ
⇒ sin θ and cos θ
To find:
The distance of a point from the line parallel to another line.
Explanation:
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 Expertsview all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
RELATED QUESTIONS :
<span lang="EN-US
RS Aggarwal - MathematicsThe distance betw
RD Sharma - Mathematics<span lang="EN-US
RS Aggarwal - MathematicsFind the equation
RD Sharma - MathematicsFind the equation
RD Sharma - MathematicsArea of the trian
RD Sharma - Mathematics<span lang="EN-US
RD Sharma - Mathematics<span lang="EN-US
RD Sharma - Mathematics<span lang="EN-US
RD Sharma - MathematicsA point equidista
RD Sharma - Mathematics