Find the point on the line 2x+4y−3=0 which is closest to the point (2,0)

1 answer:

7 0

L1: 2x+4y-3=0 ..........(1)
P: (2,0)
The point on the line L1 closest to the given point P is at the intersection of L1 with L2, which is the perpendicular passing through P.

Slope of L1=-2/4=-1/2
Slope of L2=-1/(-1/2)=2
Since it passes throug P(2,0), we can use the point-slope formula:
(y-0)=2(x-2) =>
L2: 2x-y-4=0.............(2)

Solve for x & y using (1) and (2) to get intersection point required:
(1)-(2)
2x-2x + 4y-(-y) -3 -(-4) =0
5y=-1, y=-1/5
Substitute y=1/5 in equation (1)
2x+4(-1/5)-3=0 =>
2x-19/5=0
x=19/10

=> the point on L1 closest to (2,0) is (19/10, -1/5)

You might be interested in

SOS HELPPPP

Hunter-Best [27]

Answer:

Step-by-step explanation:

5 0

3 years ago

X-y+z=-4 3x+2y-z=5 -2x+3y-z=15 How do I solve this?

Lera25 [3.4K]

$1)\ \ \ x-y+z=-4\ \ \ \Rightarrow\ \ \ z=-4-x+y\\\\2)\ \ \ 3x+2y-z=5 \\.\ \ \ \ \Rightarrow\ \ 3x+2y-(-4-x+y)=5 \\.\ \ \ \ \Rightarrow\ \ 3x+2y+4+x-y=5 \\.\ \ \ \ \Rightarrow\ \ 4x+y=1\\\\3)\ \ \ -2x+3y-z=15\\.\ \ \ \ \Rightarrow\ \ -2x+3y-(-4-x+y)=15\\.\ \ \ \ \Rightarrow\ \ -2x+3y+4+x-y=15\\.\ \ \ \ \Rightarrow\ \ -x+2y=11\\--------------------\\$

$z=-4-x+y\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\\ \left \{ {{4x+y=1\ \ \ \ } \atop {-x+2y=11\ /\cdot4}} \right. \\\\\left \{ {{4x+y=1\ \ \ \ } \atop {-4x+8y=44}} \right. \\-------\\y+8y=1+44\\9y=45\ /:9\\y=5\\\\-x+2y=11\ \ \ \Rightarrow\ \ \ x=2y-11\ \ \ \Rightarrow\ \ \ x=2\cdot5-11=-1\\\\z=-4-x+y=-4-(-1)+5=-4+1+5=2\\\\Ans.\ x=-1\ \ \ and\ \ \ y=5\ \ \ and\ \ \ z=2$

5 0

3 years ago

What is the opposite of 1/4

nirvana33 [79]

Answer:

opposite number of 1/4=-1/4

8 0

3 years ago

The graph below shows the average Valentine’s Day spending between 2003 and 2012

Lapatulllka [165]

The average rate of change of a graph between two intervals is given by the difference in value of the values on the graph of the two interval divided by the difference between the two intervals.

Part A.

From the graph the average Valentine's day spending in 2005 is 98 while the average Valentine's day spending in 2007 is 120.

The average rate of change in spending between 2005 and 2007 is given by

$\frac{120-98}{2007-2005} = \frac{22}{2} =\$11/year$

Part B

From the graph the average Valentine's day spending in 2004 is 100 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2004 and 2010 is given by

$\frac{103-100}{2010-2004} = \frac{3}{6} =\$0.5/year$

Part C:

From the graph the average Valentine's day spending in 2009 is 102 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2009 and 2010 is given by

$\frac{103-102}{2010-2009} = \$1/year$

7 0

4 years ago

Open the picture from the other question and do it here

Alik [6]

$k(x) - g(x) + f(x)$