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Bunuel
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Bunuel

Tough and Tricky questions: Coordinate Geometry.



Line L contains the points (2,3) and (p,q). If q = 2, which of the following could be the equation of line m, which is perpendicular to line L?

(A) 2x + y = px + 7
(B) 2x + y = –px
(C) x + 2y = px + 7
(D) y – 7 = x ÷ (p – 2)
(E) 2x + y = 7 – px


Hey Bunuel,

In the above solution instead of y=mx+c how can we get the solution using equation (y-y1)=m(x-x1),if not can you please elaborate why.
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aby0007
Bunuel

Tough and Tricky questions: Coordinate Geometry.



Line L contains the points (2,3) and (p,q). If q = 2, which of the following could be the equation of line m, which is perpendicular to line L?

(A) 2x + y = px + 7
(B) 2x + y = –px
(C) x + 2y = px + 7
(D) y – 7 = x ÷ (p – 2)
(E) 2x + y = 7 – px


Hey Bunuel,

In the above solution instead of y=mx+c how can we get the solution using equation (y-y1)=m(x-x1),if not can you please elaborate why.

Hi aby0007,

We surely can.
Let (x1, y1) = (2,3) and (x2, y2) = (p,q)
m = \(\frac{y2 - y1}{x2 - x1}\) = \(\frac{q - 3}{p - 2}\)

Given q = 2
Therefore, m = \(\frac{-1}{p - 2}\)
This is the slope of the line L
We need to find the perpendicular line, hence the slope = p -2
Only Option A satisfies this.
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Bunuel

Tough and Tricky questions: Coordinate Geometry.



Line L contains the points (2,3) and (p,q). If q = 2, which of the following could be the equation of line m, which is perpendicular to line L?

(A) 2x + y = px + 7
(B) 2x + y = –px
(C) x + 2y = px + 7
(D) y – 7 = x ÷ (p – 2)
(E) 2x + y = 7 – px


Hey Bunuel,

In the above solution instead of y=mx+c how can we get the solution using equation (y-y1)=m(x-x1),if not can you please elaborate why.

Hi aby0007,

We surely can.
Let (x1, y1) = (2,3) and (x2, y2) = (p,q)
m = \(\frac{y2 - y1}{x2 - x1}\) = \(\frac{q - 3}{p - 2}\)

Given q = 2
Therefore, m = \(\frac{-1}{p - 2}\)
This is the slope of the line L
We need to find the perpendicular line, hence the slope = p -2
Only Option A satisfies this.


Hey,

Sorry for not being clear I mean to say after finding the slope as (p-2).

When we are again putting values in y = mx +c can we use (y-y1)=m(x-x1); Thus giving us the required answer.

Thanks
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The key to solving the problem is substituing and getting the value by verification with the given option
y=mx+c
m=(3-2/2-p) of line l
=>m1= p-2 since it's perpendicular
therefore the required equation equals
=>y=(p-2)*x +c
=>2x+y=px+c
and a satisfies the condition hence IMO A
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