The image below shows the steps I took to solve the problem. Note, b as represented in the work shown below can equal any number since the problem does not specify where the perpendicular line M must cross with Line L.

Step by step:
1) Identify that we need to form the equation in a way to identify the slope (y = mx + b)
2) Identify the slope (m) using "rise over run", which is basically dividing the difference in y coordinates by the difference in x coordinates.
3) Find the slope of the perpendicular line (negative inverse of the first slope): m1
4) simplify the equation to match one of the answer choices

we need m=p-2, so we should get "y=mx+b" form in answer choices

(A) 2x+y=px+7 => y=px-2x+7 => y=x(p-2)+7. It is what we need
(B) 2x + y = –px => y=-px-2x => y=x(-p-2). Wrong
(C) x + 2y = px + 7 => y=x(p-1/2)+7/2. Wrong
(D) y – 7 = x ÷ (p – 2) => y=(1/p-2)x+7. Wrong
(E) 2x + y = 7 – px => y=-px-2x+7 => y=x(-p-2)+7. Wrong

A

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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