Find the equation of the perpendicular from the point
P(1, -2) on the line 4x - 3y - 5 = 0.
Also, find the co-ordinates of the foot of the perpendicular.
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Answers
Given line equation: 4x – 3y – 5 = 0
3y = 4x – 5
y = (4/3) x – 5
Slope of the line (m1) = 4/3
Let the slope of the line perpendicular to the given line be m2
Then, m1 x m2 = -1
(4/3) x m2 = -1
m2 = -3/4
Now, the equation of the line having slope m2 and passing through the point (1, -2) will be
y + 2 = (-3/4) (x – 1)
4y + 8 = -3x + 3
3x + 4y + 5 = 0
Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines
4x – 3y – 5 = 0 …. (1) and
3x + 4y + 5 = 0 …. (2)
On multiplying (1) by 4 and (2) by 3, we get
16x – 12y – 20 = 0
9x + 12y + 15 = 0
Adding we get,
25x – 5 = 0
x = 5/25
x = 1/5
Putting the value of x in (1), we have
4(1/5) – 3y – 5 = 0
4/5 – 3y – 5 = 0
3y = 4/5 – 5 = (4 – 25)/5
3y = -21/5
y = -7/5
Thus, the co-ordinates are (1/5, -7/5)
Answer:
Given the equation : 4x−3y−5=0
3y=4x−5
y=( 4 ) x - 5
3
Slope of the line (m^1 )= 4
3
Let the slope of the line perpendicular to the given line be m^2
Then, m^1 ×m ^2 =−1
(4)
3 ×m^2 =−1
m^2 = -3
4
Now, the equation of the line having slope m^2
and passing through the point (1,−2) will be
y+2=( -3) (x-1)
4
4y+8=−3x+3
3x+4y+5=0
Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines
4x−3y−5=0 ..(1) and
3x+4y+5=0 .. (2)
On multiplying (1) by 4 and (2) by 3, we get
16x−12y−20=0
9x+12y+15=0
Adding we get,
25x−5=0
x= 5
25
x= 1
5
Putting the value of x in (1), we have
4( 1 ). −3y−5=0
5
4
5 − 3y−5=0
3y= 4
5
−5=( 4 -25 )
5
3y= -21
5
y= -7
5
Thus, the co-ordinates are ( 1 , -75 )
5
Step-by-step explanation: