Question

Consider the points A(1, 2), B(-3,0), C(5, 3), and D(3, k). Find k such that: 1. [AB] is perpendicular to [CD] 2. [AD] is perpendicular to [BC].

Answer 1

We will solve as follows:

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We first determine the slope of AB, that is:

`m_(AB)=(0-2)/(-3-1)\Rightarrow m_(AB)=(1)/(2)`

And using the properties of perpendicular lines, we have the following:

`m_(CD)=-(1)/(m_(AB))`

Where mCD is the slope of the points C & D, therefore:

`m_(CD)=-(1)/((1)/(2))\Rightarrow m_(CD)=-2`

So, now we find the slope of CD using its coordinates, that is:

`m_(CD)=(k-3)/(3-5)\Rightarrow m_(CD)=-(k-3)/(2)`

Now, we equal both values of mCD:

`-(k-3)/(2)=-2`

Now, we solve for k:

`\Rightarrow k-3=4\Rightarrow k=7`

So, the points such that AB is perpendicular to CD are A(1,2), B(-3, 0), C(5, 3) & D(3, 5).

k = 7.

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We determine the slope of BC:

`m_(BC)=(3-0)/(5-(-3))\Rightarrow m_(BC)=(3)/(8)`

From this, we have that the slope of AD is:

`m_(AD)=-(1)/((3)/(8))\Rightarrow m_(AD)=-(8)/(3)`

And now, we calculate the slope of AD using its coordinates, that is:

`m_(AD)=(k-2)/(3-1)\Rightarrow m_(AD)=(k-2)/(2)`

Now, we equal both values of the slope of AD, that is:

`(k-2)/(2)=-(8)/(3)`

Now, we solve for k:

`k-2=-(16)/(3)\Rightarrow k=-(10)/(3)`

So, we have that the points such that AD is perpendicular to BC are A(1, 2), B(-3, 0), C(5, 3) & D(3, -10/3).

k = -10/3.