Another Geo App of Calc Question (1 Viewer)

[redorange]

Hi again.. stuck on this Q..... totally lost

 

[Riviet]

For part a), find the equation of MN, then use the perpendicular distance formula: d=|(ax₁+by₁+c)/sqrt(a²+b²)|, where a, b, and c are the values from the line in the form ax+by+c=0. This will be your perpendicular height for triangle MNP in part b).
For part b), you can find the area of the triangle using 1/2 x base x height. To get the base, find the distance from point M to point N. The height is the expression that you get in terms of p from part a. Hope that helps.

 

[abcd9146]

lineMN=
y=x+6

part A
perpendicular distance

x-y+6=0
(p,p²)

d=(|ax₁+by₁+c|)/√(a²+b²)
d=(|p-p²+6|)/√(1²+(-1)²)
d=(p-p²+6)/√2

part B
distanceMN= M(3,9) N(-2,4)
=√[(x₂-x₁)² + (y₂-y₁)²]
=√[(-5)²+(-5)²]
=√(50)

area of triangle = 1/2.B.H

=1/2.√(50).(p-p²+6)/√2
=√(50).(p-p²+6)/2√2 <- rationlise
=10(p-p²+6)/4
=5(p-p²+6)/2

 

[SoulSearcher]

(a) you have 3 points here, N(-2, 4), M(3, 9), and P(p, p²)
find the equaion of the line MN
which will require point-gradient formula
i.e. (9-4) / (3+2) = 1, which is gradient of the line
therefore y - 4 = x + 2
x - y + 6 = 0
using perpendicular distance formula from the line to point P,
| p - p² + 6 | / sqrt(1 + 1) = | p - p² + 6 | / sqrt (2)

(b) use the formula for area of a triangle, 1/2 * base * perpendicular height
base length is
sqrt{ (3+2)² + (9-4)² } = sqrt (50) = 5 sqrt (2)
perpendicular height is | p - p² + 6 | / sqrt (2)
therefore area of triangle is
1/2 * | p - p² + 6 | / sqrt (2) * 5 sqrt (2)
= 1/2 * 5(p - p² + 6)
= 5(p - p² + 6) / 2

 

[Riviet]

Wow, fast responses guys, nice work! =D

 

[redorange]

ah.. thanks alot guys

yeah.. nice and fast replies!