Algebra> Find the coordinates of foot of perpendic...

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

samrat , 11 Years ago

Grade

2 Answers

Latika Leekha

Last Activity: 10 Years ago

Hello student,
Let the required foot of perpendicular be P(x₁, y₁).
Let O be the point O(-1, 3) and AB be the line 3x – 4y – 16 = 0.
Slope of AB = -3/-4 = ¾.
Since OP is perpendicular to AB, so slope of OP = -4/3 (using m₁.m₂ = -1)
$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{-4}{3}$
$\frac{3-y_{1}}{-1-x_{1}}= \frac{-4}{3}$
This gives, 4x₁+3y₁ = 5 …. (1)
Also, the point P lies on AB and hence it will satisfy the equation and so we have
3x₁- 4y₁ -16 = 0 …. (2)
Now, solving equations (1) and (2), we have
x₁ = 68/25
Substituting this value in equation (1) we get, y₁ = -49/25.
Hence, the required foot of perpendicular is (68/25, -49/25).

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

Let the required foot of perpendicular be P(x1, y1).
Let O be the point O(-1, 3) and AB be the line 3x – 4y – 16 = 0.
Slope of AB = -3/-4 = ¾.
Since OP is perpendicular to AB, so slope of OP = -4/3 (using m1.m2 = -1)
(y2-y1)/(x2-x1) = -4/3
(3-y1)/(-1-x1) = -4/3
This gives, 4x1+3y1 = 5 …. (1)
Also, the point P lies on AB and hence it will satisfy the equation and so we have
3x1 - 4y1 -16 = 0 …. (2)
Now, solving equations (1) and (2), we have
x1 = 68/25
Substituting this value in equation (1) we get,
y1 = -49/25.
Hence, the required foot of perpendicular is (68/25, -49/25).

Thanks and Regards