The y-intercepts, P and Q, of two perpendicular lines intersecting at

Question

The y-intercepts, P and Q, of two perpendicular lines intersecting at the point A(6, 8) have a sum of zero. What is the area of triangle APQ ?

(A) 45
(B) 48
(C) 54
(D) 60
(E) 72

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PRamesh2008 · Accepted Answer

This is another way of solving the Problem..

Hope it helps..!!

Asmita9502 · Answer

Hi @bunnel Can you please explain the solution for this question?

Thanks!!

gasoline · Answer

Greetings!
I will appreciate if someone could provide an easier and faster method to solve.

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AnanidzeZitze · Answer

the 2nd part need not be done -

the area = 1/2*b*h = 1/2*(distance between p & q = 2c =20)*(draw a perpendicular from A to y-axis= h = 6) = 60

NidhiR · Answer

The y-intercepts, P and Q, of two perpendicular lines intersecting at the point A(6, 8) have a sum of zero. What is the area of triangle APQ ?

(A) 45
(B) 48
(C) 54
(D) 60
(E) 72

Let A(6,8) alongwith P (0,y) and Q (0,-y) form a right angled triangle APQ.

Perpendicular from A on y-axis say, R (with coordinates as (0,8))

Then triangle PRA and ARP are similar.

Hence, PR/RA = AR/RP
 (y-8)/6= 6/(8-(-y))
 (y-8)/6= 6/(8+y)

Solving, we get y =10

Therefore, Area of the Triangle = 1/2 * Base * Height
 = 1/2 * (10 - (-10)) * 6
 = 1/2 * 20 * 6
 = 60

Correct option is Option D

HarshavardhanR · Answer

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Kinshook · Answer

The y-intercepts, P and Q, of two perpendicular lines intersecting at the point A(6, 8) have a sum of zero.

What is the area of triangle APQ ?

Equation of first line: 
x/a1 + y/b = 1; m1 = -b/a1

Equation of second line:
x/a2 - y/b = 1; m2 = b/a2

Since lines are perpendicular: 
m1*m2 = - 1 = (-b/a1)(b/a2) = - b^2/a1*a2
b^2 = a1*a2

Since both lines pass through point (6,8): - 
8/a1 + 6/b = 1; 8/a1 = 1 - 6/b; a1 = 8/(1-6/b)
8/a2 - 6/b = 1 ; 8/a2 = 1 + 6/b; a2 = 8/(1+6/b)

a1*a2 = 8/(1-6/b) * 8/(1+6/b) = b^2
64 = b^2(1-6/b)(1+6/b) = (b - 6)(b+6) = b^2 - 36
b^2 = 100
b = 10 or b = -10

Area of triangle P(0,10), Q(0,-10) & A(6,8) = 1/2 * 20 * 6 = 60

IMO D

Adit_ · Answer

Produce of slopes = -1 for perpendicular lines.

(8+y/6*(8-y/6)) = -1
Solving we get y = 10

So we have (0,10) and (0,-10).
Area of triangle = 1/2*base*height.

Base*Height = Distance from (6,8) to (0,10) * Distance from (6,8) to (0,-10)
= root(360) * root(40)
= root(14400) 
= 120
Thus area = 1/2*120 = 60.

Answer: Option D

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The y-intercepts, P and Q, of two perpendicular lines intersecting at

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