Neeraj's house is 1 km in the east of origin(0,0), While going to the school first he takes auto till hospital at B(4,4). From the hospital (4,4) to church (4,8) he travels by city bus. From Church C(4,8) he rides in a metro train and he reaches the school at D(-5,8). All the units are in km.
What is the slope of Neeraj's journey from home to Hospital?
What is the distance of School from Hospital?
What is the equation of the straight linejoining the points A and D?
What is the equation of the straight linejoining church and hospital?
What is the equation of the straight linejoining the points A (House) and C(church)?
Answers
Answer:
slope=8/(-6)=-4/3=y/x
distance=√(4^2+9^2)=√16+81=√977&
eq uation y=(4/9)x
9y=4x
eq of AC: x=4
Given : Neeraj's house is 1 km in the east of origin(0,0)
House A ( 1, 0)
Hospital B(4,4).
Church (4,8)
School at D(-5,8)
All the units are in km.
To Find :
slope of Neeraj's journey from home to Hospital
distance of School from Hospital
equation of the straight line joining the points A and D
equation of the straight line joining church and hospital
equation of the straight line joining the points A (House) and C(church)
Solution:
slope of Neeraj's journey from home to Hospital
House A ( 1, 0) Hospital B(4,4).
Slope = ( 4 - 0)/ (4 -1 ) = 4/3
distance of School from Hospital
Hospital B(4,4) , School at D(-5,8)
= √(-5 - 4)² + (8 - 4)²
= √81 + 16
= √97 km
equation of the straight line joining the points A and D
A ( 1, 0) , D(-5,8)
Slope = ( 8 - 0)/(-5 - 1) = 8/(-6) = -4/3
y - 0 = (-4/3)(x - 1)
=> 3y = -4x + 4
=> 4x + 3y = 4
equation of the straight line joining church and hospital
Hospital B(4,4) Church (4,8)
Slope = ( 8 - 4)/(4 - 4) = 4/0
y - 4 = (4/0) (x - 4)
=> 0 = x - 4
=> x= 4
equation of the straight line joining the points A (House) and C(church)
House A ( 1, 0) Church (4,8)
Slope = ( 8 - 0)/(4 - 1) = 8/3
y - 0 = (8/3)(x - 1)
=> 3y = 8x - 8
=> 8x - 3y = 8
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