logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Straight Lines
0 votes

Find the coordinates of the foot of perpendicular from the point (-1,3) to the line $3x-4y-16=0$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Slope of a line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\large\frac{(y_2-y_1)}{(x_2-x_1)}$
  • If two lines are perpendicular then the product of their slopes is -1.
Let (a,b) be the coordinates of the foot of the perpendicular from the point (-1,3) to the line $3x-4y-16=0$
Slope of the line joining (-1,3) and (a,b) is $ m_1 = \large\frac{b-3}{a+1}$
Slope of the line $3x-4y-16$ is
$m_2 = \large\frac{3}{4}$
Since the lines are perpendicular $m_1m_2=-1$
(i.e.,) $ \large\frac{b-3}{a+1} $$ \times \large\frac{3}{4}$$=1$
$3(b-3)=-4(a+1)$
$ \Rightarrow 4a+3b=5$------------(1)
Since the point (a,b) lies on the line $3x-4y=16$
$\qquad 3a-4b=16$------------(2)
On solving eq(1) and (2) we get,
$ \qquad (\times 3) 4a+3b=5$
$ \qquad ( \times 4) 3a-4b=16$
$\qquad \qquad 12a+9b=15$
$ \qquad \quad -12a-16b=-64$
$ \qquad \qquad \qquad 25b=-49$
$ \Rightarrow b = -\large\frac{49}{25}$
Substituting for b in eqn (1)we get,
$4a+3 \bigg( \large\frac{-49}{25} \bigg)$$=5$
$ \Rightarrow 4a=5+\large\frac{147}{25}$
$ \Rightarrow 4a=\large\frac{125+147}{25}$$= \large\frac{272}{25}$
$ \therefore a = \large\frac{272}{100}$
$ \Rightarrow a = \large\frac{68}{25}$
Hence $a = \large\frac{68}{25}$ and $b=-\large\frac{49}{25}$
Hence the coordinate of the foot of the perpendicular are $ \bigg( \large\frac{68}{25}$$, \large\frac{-49}{25} \bigg)$
answered May 13, 2014 by thanvigandhi_1
edited May 24, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...