129k views
12 votes
Find the equation of the perpendicular from the point P(-1,-2) on the line

3x + 4y - 12 = 0. also find the coordinates of the foot of the perpendicular.



User Laksh
by
8.9k points

2 Answers

8 votes

(44/25 , 42/25)

these are the points

User Alice Girard
by
8.4k points
5 votes

eq :- 4x - 3y - 2 =0

and coordinates :- (44/25,42/25)

Step-by-step explanation:-

The given line is 3x + 4y - 12 = 0

=> 4y = -3x + 12 ....(i)

=> y = -3x/4 + 3,

therefore, the slope of the line (i) = -3/4.

From P(-1,-2), draw PN perpendicular to the given line. (see attachment)

As we know,, m2 = -1/m1

So, the slope of the line PN = 4/3

the equation of the line through P(-1,-2) and having a slope 4/3 is :-

= > y - (-2) = 4/3 * (x - (-1))

= > 3y + 6 = 4x + 4

= > 4x - 3y - 2 = 0....(ii)

which is the required equation of the perpendicular from P to the given line. To find the coordinates of N (the foot of perpendicular), solve (i) and (ii) simultaneously. Multiplying (i) by 3 and

(ii) by 4, and on adding, we get

= > 25x - 44 = 0

= > x = 44/25

Multiplying (i) by 4 and (ii) by 3, and on subtracting, we get :-

= > 25y - 42 = 0

= > y = 42/25

Hence, the coordinates of the foot of the perpendicular are (44/25 , 42/25)

Hope it helps you!!

Find the equation of the perpendicular from the point P(-1,-2) on the line 3x + 4y-example-1
User Wattskemov
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories