Video: CBSE Class X • Pack 4 • 2015 • Question 29

CBSE Class X • Pack 4 • 2015 • Question 29

03:49

Video Transcript

Find the values of 𝑘 so that the area of the triangle with vertices one, negative one; negative four, two 𝑘; and negative 𝑘, negative five is 24 square units.

The formula we can use to help us calculate the area of a triangle with coordinates at 𝑥 one, 𝑦 one; 𝑥 two, 𝑦 two; and 𝑥 three, 𝑦 three is 𝑥 one multiplied by 𝑦 two minus 𝑦 three plus 𝑥 two multiplied by 𝑦 three minus 𝑦 one plus 𝑥 three multiplied by 𝑦 one minus 𝑦 two all over two.

Now, it doesn’t really matter which values we choose to be 𝑥 one, 𝑦 one and so on. Let’s call one, negative one 𝑥 one, 𝑦 one; negative four, two 𝑘 𝑥 two, 𝑦 two; and negative 𝑘, negative five 𝑥 three, 𝑦 three.

The first part of this formula 𝑥 one multiplied by 𝑦 two minus 𝑦 three becomes one multiplied by two 𝑘 minus negative five, which is simply two 𝑘 plus five. The next part 𝑥 two multiplied by 𝑦 three minus 𝑦 one becomes negative four multiplied by negative five minus negative one. That’s negative four multiplied by negative four which is equal to 16. The final part becomes negative 𝑘 multiplied by negative one minus two 𝑘, which is 𝑘 plus two 𝑘 squared.

Substituting each of these individual parts into the formula for the area and we get two 𝑘 plus five plus 16 plus 𝑘 plus two 𝑘 squared all over two. That simplifies to two 𝑘 squared plus three 𝑘 plus 21 all over two. Remember that we were told that the area of the triangle needed to be 24 square units. So we can equate this expression to 24.

Let’s begin to solve by multiplying both sides by two. That gives us 48 is equal to two 𝑘 squared plus three 𝑘 plus 21. Then subtracting 48 from both sides and we get that zero is equal to two 𝑘 squared plus three 𝑘 minus 27. And we’ll need to factorize the expression two 𝑘 squared plus three 𝑘 minus 27. Since the coefficient of 𝑘 squared is two, we know that the front of one of the brackets must contain two 𝑘 and the other must contain 𝑘.

Next, we’ll list the factor pairs of negative 27. They are negative one and 27, one and negative 27, negative three and nine, and three and negative nine. We could use a bit of trial and error to figure out the factor pairs we’re interested in. However, we should spot that one of these factors will be multiplied by two 𝑘.

The factors we’re interested in that will give us a sum of three when one is doubled are nine and negative three. Two multiplied by negative three is negative six and nine minus six is three. So our equation factorizes to two 𝑘 plus nine multiplied by 𝑘 minus three equals zero.

The product of these two brackets is zero. That means that at least one of the brackets must be equal to zero: either two 𝑘 plus nine is equal to zero or 𝑘 minus three is equal to zero.

To solve this first equation, we’ll subtract nine from both sides and then divide by two to give us a value of 𝑘 as negative nine over two. To solve the other equation, we’ll add three to both sides, giving us that 𝑘 is equal to three.

𝑘 is equal to negative nine over two or three.

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